Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 2 - Functions

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Page No 2.31:

Question 1:

Give an example of a function
(i) which is one-one but not onto
(ii) which is not one-one but onto
(iii) which is neither one-one nor onto

Answer:

(i) which is one-one but not onto.

f: Z → Z given by f(x)=3x+2

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x)=f(y)
$\Rightarrow$ 3x + 2 =3y + 2
$\Rightarrow$ 3x = 3y
$\Rightarrow$ x = y
$\Rightarrow$ f(x) = f(y) $\Rightarrow$ x = y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
$\Rightarrow$ 3x + 2 = y
$\Rightarrow$ 3x = y - 2

$\Rightarrow x = \frac{y - 2}{3} . It may not be in the domain (Z) because if we take y = 3, x = \frac{y - 2}{3} = \frac{3 - 2}{3} = \frac{1}{3} ∉ domain Z .$
So, for every element in the co domain there need not be any element in the domain such that f(x) = y.
Thus, f is not onto.

(ii) which is not one-one but onto.
f: Z → N $\cup$ {0} given by f(x) = |x|

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
$\Rightarrow$ |x| = |y|
$\Rightarrow$ x= $\pm$ y
So, different elements of domain f may give the same image.
So, f is not one-one.

Surjectivity:
Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
$\Rightarrow$ |x| = y
$\Rightarrow$ x = $\pm$ y, which is an element in Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.

(iii) which is neither one-one nor onto.

f: Z → Z given by f(x) = 2x² + 1

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
$f (x) = f (y) \Rightarrow 2 x^{2} + 1 = 2 y^{2} + 1 \Rightarrow 2 x^{2} = 2 y^{2} \Rightarrow x^{2} = y^{2} \Rightarrow x = \pm y$
So, different elements of domain f may give the same image.
Thus, f is not one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
$\Rightarrow 2 x^{2} + 1 = y \Rightarrow 2 x^{2} = y - 1 \Rightarrow x^{2} = \frac{y - 1}{2} \Rightarrow x = \pm \sqrt{\frac{y - 1}{2}}, ∉ Z always. For example, if we take, y = 4, x = \pm \sqrt{\frac{y - 1}{2}} = \pm \sqrt{\frac{4 - 1}{2}} = \pm \sqrt{\frac{3}{2}}, ∉ Z So, x may not be in Z (domain).$

Thus, f is not onto.

Page No 2.31:

Question 2:

Which of the following functions from A to B are one-one and onto?
(i) f₁ = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}
(ii) f₂ = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}
(iii) f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Answer:

(i) f₁ = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:
f₁ (1) = 3
f₁(2) = 5
f₁ (3) = 7
$\Rightarrow$ Every element of A has different images in B.
So, f₁ is one-one.

Surjectivity:
Co-domain of f₁ = {3, 5, 7}
Range of f₁ =set of images = {3, 5, 7}
$\Rightarrow$ Co-domain = range
So, f₁ is onto.

(ii) f₂ = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}

Injectivity:
f₂ (2) = a
f₂ (3) = b
f₂ (4) = c
$\Rightarrow$ Every element of A has different images in B.
So, f₂ is one-one.

Surjectivity:
Co-domain of f₂ = {a, b, c}
Range of f₂ = set of images = {a, b, c}
$\Rightarrow$ Co-domain = range
So, f₂ is onto.

(iii) f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:
f₃ (a) = x
f₃ (b) = x
f₃ (c) = z
f₃ (d) = z
$\Rightarrow$ a and b have the same image x. (Also c and d have the same image z)
So, f₃ is not one-one.

Surjectivity:
Co-domain of f₁ ={x, y, z}
Range of f₁ =set of images = {x, z}
So, the co-domain is not same as the range.
So, f₃ is not onto.

Page No 2.31:

Question 3:

Prove that the function f : N → N, defined by f(x) = x² + x + 1, is one-one but not onto.

Answer:

f : N → N, defined by f(x) = x² + x + 1

Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
$\Rightarrow x^{2} + x + 1 = y^{2} + y + 1 \Rightarrow (x^{2} - y^{2}) + (x - y) = 0 \Rightarrow (x + y) (x - y) + (x - y) = 0 \Rightarrow (x - y) (x + y + 1) = 0 \Rightarrow x - y = 0 [(x+y+1) cannot be zero because x and y are natural numbers] \Rightarrow x = y$
So, f is one-one.

Surjectivity:
$The minimum number in N is 1. When x =1, x^{2} + x + 1 = 1 + 1 + 1 = 3 \Rightarrow x^{2} + x + 1 \geq 3, for every x in N . ⇒ f (x) will not assume the values 1 and 2. So, f is not onto.$

Page No 2.31:

Question 4:

Let A = {−1, 0, 1} and f = {(x, x²) : x ∈ A}. Show that f : A → A is neither one-one nor onto.

Answer:

A = {−1, 0, 1} and f = {(x, x²) : x ∈ A}
Given, f(x) = x²

Injectivity:
f(1) = 1²=1 and
f(-1)=(-1)²=1

$\Rightarrow$ 1 and -1 have the same images.
So, f is not one-one.

Surjectivity:
Co-domain of f = {-1, 0, 1}

f(1) = 1² = 1,
f(-1) = (-1)² = 1 and
f(0) = 0
$\Rightarrow$ Range of f = {0, 1}
So, both are not same.
Hence, f is not onto.

Page No 2.31:

Question 5:

Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : N → N given by f(x) = x³
(iv) f : Z → Z given by f(x) = x³
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x² + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x³ + 1
(x) f : R → R, defined by f(x) = x³ − x
(xi) f : R → R, defined by f(x) = sin²x + cos²x
(xii) f : Q − {3} → Q, defined by $f (x) = \frac{2 x + 3}{x - 3}$
(xiii) f : Q → Q, defined by f(x) = x³ + 1
(xiv) f : R → R, defined by f(x) = 5x³ + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x²
(xvii) f : R → R, defined by f(x) = $\frac{x}{x^{2} + 1}$ [NCERT EXEMPLAR]

Answer:

(i) f : N → N, given by f(x) = x²

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x)=f(y)

$x^{2} = y^{2} x = y (We do not get ± because x and y are in N)$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

$x^{2} = y x = \sqrt{y}, which may not be in N . For example, if y =3, x = \sqrt{3} is not in N .$

So, f is not a surjection.

So, f is not a bijection.

(ii) f : Z → Z, given by f(x) = x²

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

$x^{2} = y^{2} x = \pm y$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

$x^{2} = y x = \pm \sqrt{y} which may not be in Z . For example, if y =3, x = \pm \sqrt{3} is not in Z .$

So, f is not a surjection.

So, f is not a bijection.

(iii) f : N → N, given by f(x) = x³

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

$x^{3} = y^{3} x = y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

$x^{3} = y x = \sqrt[3]{y} which may not be in N . For example, if y =3, x = \sqrt[3]{3} is not in N .$

So, f is not a surjection and f is not a bijection.

(iv) f : Z → Z, given by f(x) = x³

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x) = f(y)

$x^{3} = y^{3} x = y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

$x^{3} = y x = \sqrt[3]{y} which may not be in Z . For example, if y =3, x = \sqrt[3]{3} is not in Z .$

So, f is not a surjection and f is not a bijection.

(v) f : R → R, defined by f(x) = |x|

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

f(x) = f(y)

$|x| = |y| x = \pm y$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$|x| = y x = \pm y \in Z$

So, f is a surjection and f is not a bijection.

(vi) f : Z → Z, defined by f(x) = x² + x

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

$x^{2} + x = y^{2} + y Here, we cannot say that x = y . For example, x = 2 and y = - 3 Then, x^{2} + x = 2^{2} + 2 = 6 y^{2} + y = {(- 3)}^{2} - 3 = 6 So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

$x^{2} + x = y Here, we cannot say x \in Z . For example, y =-4. x^{2} + x = - 4 x^{2} + x + 4 = 0 x = \frac{- 1 \pm \sqrt{- 15}}{2} = \frac{- 1 \pm i \sqrt{15}}{2} which is not in Z .$

So, f is not a surjection and f is not a bijection.

(vii) f : Z → Z, defined by f(x) = x − 5

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x $-$ 5 = y $-$ 5

x = y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x $-$ 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

(viii) f : R → R, defined by f(x) = sinx

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$\sin x = \sin y Here, x may not be equal to y because \sin 0 = sinπ . So, 0 and π have the same image 0.$

So, f is not an injection .

Surjection test:

Range of f = [ $-$ 1, 1]

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) f : R → R, defined by f(x) = x³ + 1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$x^{3} + 1 = y^{3} + 1 x^{3} = y^{3} x = y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$x^{3} + 1 = y x = \sqrt[3]{y - 1} \in R$

So, f is a surjection.

So, f is a bijection.

(x) f : R → R, defined by f(x) = x³ − x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$x^{3} - x = y^{3} - y Here, we cannot say x = y . For example, x =1 and y =-1 x^{3} - x = 1 - 1 = 0 y^{3} - y = {(- 1)}^{3} - (- 1) - 1 + 1 = 0 So, 1 and -1 have the same image 0.$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$x^{3} - x = y By observation we can say that there exist some x in R, such that x^{3} -x=y.$

So, f is a surjection and f is not a bijection.

(xi) f : R → R, defined by f(x) = sin²x + cos²x

f(x) = sin²x + cos²x = 1

So, f(x) = 1 for every x in R.

So, for all elements in the domain, the image is 1.

So, f is not an injection.

Range of f = {1}

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(xii) f : Q − {3} → Q, defined by $f (x) = \frac{2 x + 3}{x - 3}$

Injection test:

Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).

f(x) = f(y)

$\frac{2 x + 3}{x - 3} = \frac{2 y + 3}{y - 3} (2 x + 3) (y - 3) = (2 y + 3) (x - 3) 2 x y - 6 x + 3 y - 9 = 2 x y - 6 y + 3 x - 9 9 x = 9 y x = y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

$\frac{2 x + 3}{x - 3} = y 2 x + 3 = x y - 3 y 2 x - x y = - 3 y - 3 x (2 - y) = - 3 (y + 1) x = \frac{3 (y + 1)}{y - 2}, which is not defined at y =2.$

So, f is not a surjection and f is not a bijection.

(xiii) f : Q → Q, defined by f(x) = x³ + 1

Injection test:

Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x) = f(y)

$x^{3} + 1 = y^{3} + 1 x^{3} = y^{3} x = y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).

f(x) = y

$x^{3} + 1 = y x = \sqrt[3]{y - 1,} which may not be in Q . For example, if y = 8, x^{3} + 1 = 8 x^{3} = 7 x = \sqrt[3]{7}, which is not in Q .$

So, f is not a surjection and f is not a bijection.

So, f is a surjection and f is a bijection.

(xiv) f : R → R, defined by f(x) = 5x³ + 4

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$5 x^{3} + 4 = 5 y^{3} + 4 5 x^{3} = 5 y^{3} x^{3} = y^{3} x = y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$5 x^{3} + 4 = y 5 x^{3} = y - 4 x^{3} = \frac{y - 4}{5} x = \sqrt[3]{\frac{y - 4}{5}} \in R$

So, f is a surjection and f is a bijection.

(xv) f : R → R, defined by f(x) = 3 − 4x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$3 - 4 x = 3 - 4 y - 4 x = - 4 y x = y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$3 - 4 x = y 4 x = 3 - y x = \frac{3 - y}{4} \in R$

So, f is a surjection and f is a bijection.

(xvi) f : R → R, defined by f(x) = 1 + x²

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$1 + x^{2} = 1 + y^{2} x^{2} = y^{2} x = \pm y$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$1 + x^{2} = y x^{2} = y - 1 x = \pm \sqrt{y - 1} which may not be in R For example, if y =0, x = \pm \sqrt{- 1} = \pm i is not in R .$

So, f is not a surjection and f is not a bijection.

(xvii) f : R → R, defined by f(x) = $\frac{x}{x^{2} + 1}$

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$\frac{x}{x^{2} + 1} = \frac{y}{y^{2} + 1} x y^{2} + x = x^{2} y + y x y^{2} - x^{2} y + x - y = 0 - x y (- y + x) + 1 (x - y) = 0 (x - y) (1 - x y) = 0 x = y or x = \frac{1}{y}$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$\frac{x}{x^{2} + 1} = y y x^{2} - x + y = 0 x = \frac{- (- 1) \pm \sqrt{1 - 4 y^{2}}}{2 y}, if y \neq 0 = \frac{1 \pm \sqrt{1 - 4 y^{2}}}{2 y}, which may not be in R For example, if y = 1, then x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i \sqrt{3}}{2}, which is not in R So, f is not surjection and f is not bijection .$

So, f is not a surjection and f is not a bijection.

Page No 2.31:

Question 6:

If f : A → B is an injection, such that range of f = {a}, determine the number of elements in A.

Answer:

Range of f = {a}
So, the number of images of f = 1
Since, f is an injection, there will be exactly one image for each element of f .
So, number of elements in A = 1.

Page No 2.31:

Question 7:

Show that the function f : R − {3} → R − {2} given by $f (x) = \frac{x - 2}{x - 3}$ is a bijection.

Answer:

f : R − {3} → R − {2} given by
$f (x) = \frac{x - 2}{x - 3}$
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
f(x) = f(y)
$\Rightarrow \frac{x - 2}{x - 3} = \frac{y - 2}{y - 3} \Rightarrow (x - 2) (y - 3) = (y - 2) (x - 3) \Rightarrow x y - 3 x - 2 y + 6 = x y - 3 y - 2 x + 6 \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).
f(x) = y
$\Rightarrow \frac{x - 2}{x - 3} = y \Rightarrow x - 2 = x y - 3 y \Rightarrow x y - x = 3 y - 2 \Rightarrow x (y - 1) = 3 y - 2 \Rightarrow x = \frac{3 y - 2}{y - 1}, which is in R - {3}$
So, for every element in the co-domain, there exists some pre-image in the domain.
$\Rightarrow$ f is onto.
Since, f is both one-one and onto, it is a bijection.

Page No 2.32:

Question 8:

Let A = [ $-$ 1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:

(i) f(x) = $\frac{x}{2}$ (ii) g(x) = |x| (iii) h(x) = x² [NCERT EXEMPLAR]

Answer:

(i) f : A $\to$ A, given by f(x) = $\frac{x}{2}$

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

$\frac{x}{2}$ = $\frac{y}{2}$

x = y

So, f is one-one.

Surjection test:

Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain)

f(x) = y

$\frac{x}{2}$ = y

x = 2y, which may not be in A.

For example, if y = 1, then

x = 2, which is not in A.

So, f is not onto.

So, f is not bijective.

(ii) g(x) = |x|

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

|x| = |y|

x = $\pm$ y

So, f is not one-one.

Surjection test:

For y = $-$ 1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

(iii) h(x) = x²

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = $\pm$ y

So, f is not one-one.

Surjection test:

For y = $-$ 1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

Page No 2.32:

Question 9:

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:

(i) {(x, y) : x is a person, y is the mother of x}
(ii) {(a, b) : a is a person, b is an ancestor of a} [NCERT EXEMPLAR]

Answer:

(i) f = {(x, y) : x is a person, y is the mother of x}

As, for each element x in domain set, there is a unique related element y in co-domain set.

So, f is the function.

Injection test:

As, y can be mother of two or more persons

So, f is not injective.

Surjection test:

For every mother y defined by (x, y), there exists a person x for whom y is mother.

So, f is surjective.

Therefore, f is surjective function.

(ii) g = {(a, b) : a is a person, b is an ancestor of a}

Since, the ordered map (a, b) does not map 'a' - a person to a living person.
So, g is not a function.

Page No 2.32:

Question 10:

Let A = {1, 2, 3}. Write all one-one from A to itself.

Answer:

A ={1, 2, 3}
Number of elements in A = 3
Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}

Page No 2.32:

Question 11:

If f : R → R be the function defined by f(x) = 4x³ + 7, show that f is a bijection.

Answer:

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
$\Rightarrow 4 x^{3} + 7 = 4 y^{3} + 7 \Rightarrow 4 x^{3} = 4 y^{3} \Rightarrow x^{3} = y^{3} \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
$\Rightarrow 4 x^{3} + 7 = y \Rightarrow 4 x^{3} = y - 7 \Rightarrow x^{3} = \frac{y - 7}{4} \Rightarrow x = \sqrt[3]{\frac{y - 7}{4}} \in R$
So, for every element in the co-domain, there exists some pre-image in the domain.
$\Rightarrow$ f is onto.
Since, f is both one-to-one and onto, it is a bijection.

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Question 12:

Show that the exponential function f : R → R, given by f(x) = e^x, is one-one but not onto. What happens if the co-domain is replaced by $R \overset{+}{0}$ (set of all positive real numbers)?

Answer:

f : R → R, given by f(x) = e^x

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
f(x)=f(y)
$\Rightarrow e^{x} = e^{y} \Rightarrow x = y$
So, f is one-one.

Surjectivity:
We know that range of e^x is (0, ∞) = R⁺
Co-domain = R
Both are not same.
So, f is not onto.

If the co-domain is replaced by R⁺, then the co-domain and range become the same and in that case, f is onto and hence, it is a bijection.

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Question 13:

Show that the logarithmic function $f : R \overset{+}{0} \to R given by f (x) = \log_{a} x, a > 0$ is a bijection.

Answer:

f:R⁺→R given by $f (x) = \log_{a} x$ , a>0
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
$\log_{a} x = \log_{a} y \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R⁺ (domain).
f(x) = y
$\log_{a} x = y \Rightarrow x = a^{y} \in R^{+}$
So, for every element in the co-domain, there exists some pre-image in the domain.
$\Rightarrow$ f is onto.
Since f is one-one and onto, it is a bijection.

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Question 14:

If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.

Answer:

A ={1, 2, 3}
Number of elements in A = 3
Number of one - one functions = number of ways of arranging 3 elements = 3! = 6
So, the possible one -one functions can be the following:

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}
Here, in each function, range = {1, 2, 3}, which is same as the co-domain.
So, all the functions are onto.

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Question 15:

If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.

Answer:

A ={1, 2, 3}
Possible onto functions from A to A can be the following:

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}

Here, in each function, different elements of the domain have different images.
So, all the functions are one-one.

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Question 16:

Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.

Answer:

We know that every onto function from A to itself is one-one.
So, the number of one-one functions = number of bijections = n!

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Question 17:

Give examples of two one-one functions f₁ and f₂ from R to R, such that f₁ + f₂ : R → R. defined by (f₁ + f₂) (x) = f₁ (x) + f₂ (x) is not one-one.

Answer:

We know that f₁: R → R, given by f₁(x)=x, and f₂(x)=-x are one-one.
Proving f₁is one-one:
$Let f_{1} (x) = f_{1} (y) \Rightarrow x = y$
So, f₁is one-one.

Proving f₂is one-one:
$Let f_{2} (x) = f_{2} (y) \Rightarrow - x = - y \Rightarrow x = y$
So, f₂ is one-one.

Proving (f₁ + f₂) is not one-one:
Given:
(f₁ + f₂) (x) = f₁ (x) + f₂ (x)= x + (-x) =0
So, for every real number x, (f₁ + f₂) (x)=0
So, the image of ever number in the domain is same as 0.
Thus, (f₁ + f₂) is not one-one.

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Question 18:

Give examples of two surjective functions f₁ and f₂ from Z to Z such that f₁ + f₂ is not surjective.

Answer:

We know that f₁: R → R, given by f₁(x) = x, and f₂(x) = -x are surjective functions.
Proving f₁is surjective :
Let y be an element in the co-domain (R), such that f₁(x) = y.
f₁(x) = y
$\Rightarrow$ x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f₁is surjective .

Proving f₂ is surjective :Let f2(x)=f2(y)−x=−yx=y
Let y be an element in the co domain (R) such that f₂(x) = y.
f₂(x) = y
$\Rightarrow$ x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f₂is surjective .

Proving (f₁ + f₂) is not surjective :
Given:
(f₁ + f₂) (x) = f₁ (x) + f₂ (x)= x + (-x) =0
So, for every real number x, (f₁ + f₂) (x)=0
So, the image of every number in the domain is same as 0.
$\Rightarrow$ Range = {0}
Co-domain = R
So, both are not same.
So, f₁ + f₂is not surjective.

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Question 19:

Show that if f₁ and f₂ are one-one maps from R to R, then the product $f_{1} \times f_{2} : R \to R$ defined by $(f_{1} \times f_{2}) (x) = f_{1} (x) f_{2} (x)$ need not be one-one.

Answer:

We know that f₁: R → R, given by f₁(x) = x, and f₂(x) = x are one-one.
Proving f₁is one-one:
Let x and y be two elements in the domain R, such that
f₁(x) = f₁(y)
$\Rightarrow$ x = yet f1(x)=f1(y)x=y
So, f₁is one-one.

Proving f₂is one-one:
Let x and y be two elements in the domain R, such that
f₂(x) = f₂(y)
$\Rightarrow$ x = yet f1(x)=f1(y)x=y
So, f₂is one-one.

Proving $f_{1} \times f_{2}$ is not one-one:
Given:
$(f_{1} \times f_{2}) (x) = f_{1} (x) \times f_{2} (x) = x \times x = x^{2} Let x and y be two elements in the domain R, such that (f_{1} \times f_{2}) (x) = (f_{1} \times f_{2}) (y) \Rightarrow x^{2} = y^{2} \Rightarrow x = \pm y So, (f_{1} \times f_{2}) is not one-one.$ f1×f2

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Question 20:

Suppose f₁ and f₂ are non-zero one-one functions from R to R. Is $\frac{f_{1}}{f_{2}}$ necessarily one-one? Justify your answer. Here, $\frac{f_{1}}{f_{2}} : R \to R$ is given by $(\frac{f_{1}}{f_{2}}) (x) = \frac{f_{1} (x)}{f_{2} (x)}$ for all $x \in R$ .

Answer:

We know that f₁: R → R, given by f₁(x)=x³ and f₂(x)=x are one-one.
Injectivity of f₁:
Let x and y be two elements in the domain R, such that
$f_{1} (x) = f_{2} (y) \Rightarrow x^{3} = y \Rightarrow x = \sqrt[3]{y} \in R$ Let f1(x)=f1(y)x=y
So, f₁is one-one.

Injectivity of f₂:
Let x and y be two elements in the domain R, such that
$f_{2} (x) = f_{2} (y) \Rightarrow x = y ⇒ x \in R .$ Let f2(x)=f2(y)−x=−yx=y
So, f₂ is one-one.

Proving $\frac{f_{1}}{f_{2}}$ is not one-one:
Given that $\frac{f_{1}}{f_{2}} (x) = \frac{f_{1} (x)}{f_{2} (x)} = \frac{x^{3}}{x} = x^{2}$
Let x and y be two elements in the domain R, such that

$\frac{f_{1}}{f_{2}} (x) = \frac{f_{1}}{f_{2}} (y) \Rightarrow x^{2} = y^{2} \Rightarrow x = \pm y$
So, $\frac{f_{1}}{f_{2}}$ is not one-one.

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Question 21:

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(i) an injective map from A to B
(ii) a mapping from A to B which is not injective
(iii) a mapping from A to B.

Answer:

(i) {(2, 7), (3, 6), (4, 5)}

(ii) {(2, 2), (3, 2), (4, 5)}

(iii) {(2, 5), (3, 6), (4, 7)}

Disclaimer: There are many more possibilities of each case.

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Question 22:

Show that f : R $\to$ R, given by f(x) = x $-$ [x], is neither one-one nor onto.

Answer:

We have, f(x) = x $-$ [x]

Injection test:

f(x) = 0 for all x $\in$ Z

So, f is a many-one function.

Surjection test:

Range (f) = [0, 1) $\neq$ R.

So, f is an into function.

Therefore, f is neither one-one nor onto.

Page No 2.32:

Question 23:

Let f : N $\to$ N be defined by

$f (n) = \{\begin{cases} n + 1, if n is odd \\ n - 1, if n is even \end{cases}$

Show that f is a bijection. [CBSE 2012, NCERT]

Answer:

$We have, f (n) = \{\begin{cases} n + 1, if n is odd \\ n - 1, if n is even \end{cases} Injection test : Case I : If n is odd, Let x, y \in N such that f (x) = f (y) As, f (x) = f (y) \Rightarrow x + 1 = y + 1 \Rightarrow x = y Case II : If n is even, Let x, y \in N such that f (x) = f (y) As, f (x) = f (y) \Rightarrow x - 1 = y - 1 \Rightarrow x = y So, f is injective . Surjection test : Case I : If n is odd, As, for every n \in N, there exists y = n - 1 in N such that f (y) = f (n - 1) = n - 1 + 1 = n Case II : If n is even, As, for every n \in N, there exists y = n + 1 in N such that f (y) = f (n + 1) = n + 1 - 1 = n So, f is surjective . So, f is a bijection .$

Page No 2.46:

Question 1:

Find gof and fog when f : R → R and g : R → R are defined by
(i) f(x) = 2x + 3        and g(x) = x² + 5
(ii) f(x) = 2x + x²           and       g(x) = x³
(iii) f(x) = x² + 8              and       g(x) = 3x³ + 1
(iv) f(x) = x                       and       g(x) = |x|
(v) f(x) = x² + 2x − 3       and       g(x) = 3x − 4
(vi) f(x) = 8x³                   and       g(x) = x¹^/3

Answer:

Given, f : R → R and g : R → R
So, gof : R → R and fog : R → R

(i) f(x) = 2x + 3 and g(x) = x² + 5
Now, (gof) (x)
= g (f (x))
= g (2x +3)
= (2x + 3)² + 5
= 4x²+ 9 + 12x +5
=4x²+ 12x + 14

(fog) (x)
=f (g (x))
= f (x² + 5)
= 2 (x² + 5) +3
= 2 x²+ 10 + 3
= 2x² + 13

(ii) f(x) = 2x + x² and g(x) = x³
^{$(g o f) (x) = g (f (x)) = g (2 x + x^{2}) = {(2 x + x^{2})}^{3} (f o g) (x) = f (g (x)) = f (x^{3}) = 2 (x^{3}) + {(x^{3})}^{2} = 2 x^{3} + x^{6}$}

(iii) f(x) = x² + 8 and g(x) = 3x³ + 1
$(g o f) (x) = g (f (x)) = g (x^{2} + 8) = 3 {(x^{2} + 8)}^{3} + 1 (f o g) (x) = f (g (x)) = f (3 x^{3} + 1) = {(3 x^{3} + 1)}^{2} + 8 = 9 x^{6} + 6 x^{3} + 1 + 8 = 9 x^{6} + 6 x^{3} + 9$

(iv) f(x) = x and g(x) = |x|
$(g o f) (x) = g (f (x)) = g (x) = |x| (f o g) (x) = f (g (x)) = f (|x|) = |x|$

(v) f(x) = x² + 2x − 3 and g(x) = 3x − 4
$(g o f) (x) = g (f (x)) = g (x^{2} + 2 x - 3) = 3 (x^{2} + 2 x - 3) - 4 = 3 x^{2} + 6 x - 9 - 4 = 3 x^{2} + 6 x - 13 (f o g) (x) = f (g (x)) = f (3 x - 4) = {(3 x - 4)}^{2} + 2 (3 x - 4) - 3 = 9 x^{2} + 16 - 24 x + 6 x - 8 - 3 = 9 x^{2} - 18 x + 5$

(vi) f(x) = 8x³ and g(x) = x¹^/3
$(g o f) (x) = g (f (x)) = g (8 x^{3}) = {(8 x^{3})}^{\frac{1}{3}} = {[{(2 x)}^{3}]}^{\frac{1}{3}} = 2 x (f o g) (x) = f (g (x)) = f (x^{\frac{1}{3}}) = 8 {(x^{\frac{1}{3}})}^{3} = 8 x$

Page No 2.46:

Question 2:

Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.

Answer:

f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}

f : {3, 9, 12} → {1, 3,4} and g : {1, 3, 4, 5} → {3, 9}

Co-domain of f is a subset of the domain of g.
So, gof exists and gof : {3, 9, 12} → {3, 9}
$(g o f) (3) = g (f (3)) = g (1) = 3 (g o f) (9) = g (f (9)) = g (3) = 3 (g o f) (12) = g (f (12)) = g (4) = 9 \Rightarrow g o f = \{(3, 3), (9, 3), (12, 9)\}$
Co-domain of g is a subset of the domain of f.
So, fog exists and fog : {1, 3, 4, 5} → {3, 9, 12}
$(f o g) (1) = f (g (1)) = f (3) = 1 (f o g) (3) = f (g (3)) = f (3) = 1 (f o g) (4) = f (g (4)) = f (9) = 3 (f o g) (5) = f (g (5)) = f (9) = 3 \Rightarrow f o g = \{(1, 1), (3, 1), (4, 3), (5, 3)\}$

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Question 3:

Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.

Answer:

f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f : {1, 4, 9, 16} → {-1, -2, -3, 4} and g : {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof : {1, 4, 9, 16} → {-2, -4, -6, 8}
$(g o f) (1) = g (f (1)) = g (- 1) = - 2 (g o f) (4) = g (f (4)) = g (- 2) = - 4 (g o f) (9) = g (f (9)) = g (- 3) = - 6 (g o f) (16) = g (f (16)) = g (4) = 8 So, g o f = \{(1, - 2), (4, - 4), (9, - 6), (16, 8)\}$

But the co-domain of g is not same as the domain of f.
So, fog does not exist.

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Question 4:

Let A = {a, b, c}, B = {u v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as :

f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.

Show that f and g both are bijections and find fog and gof.

Answer:

Proving f is a bijection:
f = {(a, v), (b, u), (c, w)} and f : A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u v, w}
Range of f = {u v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.

Proving g is a bijection:
g = {(u, b), (v, a), (w, c)} and g : B → A
Injectivity of g: No two elements of B have the same image in A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.

Finding fog:
Co-domain of g is same as the domain of f.
So, fog exists and fog : {u v, w} → {u v, w}
$(f o g) (u) = f (g (u)) = f (b) = u (f o g) (v) = f (g (v)) = f (a) = v (f o g) (w) = f (g (w)) = f (c) = w So, f o g = \{(u, u), (v, v), (w, w)\}$

Finding gof:
Co-domain of f is same as the domain of g.
So, fog exists and gof : {a, b, c} → {a, b, c}
$(g o f) (a) = g (f (a)) = g (v) = a (g o f) (b) = g (f (b)) = g (u) = b (g o f) (c) = g (f (c)) = g (w) = c So, g o f = \{(a, a), (b, b), (c, c)\}$

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Question 5:

Find fog (2) and gof (1) when : f : R → R ; f(x) = x² + 8 and g : R → R; g(x) = 3x³ + 1.

Answer:

$(f o g) (2) = f (g (2)) = f (3 \times 2^{3} + 1) = f (25) = 25^{2} + 8 = 633 (g o f) (1) = g (f (1)) = g (1^{2} + 8) = g (9) = 3 \times 9^{3} + 1 = 2188$

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Question 6:

Let R⁺ be the set of all non-negative real numbers. If f : R⁺ → R⁺ and g : R⁺ → R⁺ are defined as $f (x) = x^{2} and g (x) = + \sqrt{x}$ , find fog and gof. Are they equal functions?

Answer:

Given, f : R⁺ → R⁺ and g : R⁺ → R⁺
So, fog : R⁺ → R⁺ and gof : R⁺ → R⁺
Domains of fog and gof are the same.
$(f o g) (x) = f (g (x)) = f (\sqrt{x}) = {(\sqrt{x})}^{2} = x (g o f) (x) = g (f (x)) = g (x^{2}) = \sqrt{x^{2}} = x So, (f o g) (x) = (g o f) (x), \forall x \in R^{+}$
Hence, fog = gof

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Question 7:

Let f : R → R and g : R → R be defined by f(x) = x² and g(x) = x + 1. Show that fog ≠ gof.

Answer:

Given, f : R → R and g : R → R.
So, the domains of f and g are the same.

$(f o g) (x) = f (g (x)) = f (x + 1) = {(x + 1)}^{2} = x^{2} + 1 + 2 x (g o f) (x) = g (f (x)) = g (x^{2}) = x^{2} + 1$
So, fog ≠ gof

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Question 8:

Let f : R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x − 1. Show that fog = gof = I_R.

Answer:

Given, f : R → R and g : R → R
$\Rightarrow$ fog : R → R and gof : R → R (Also, we know that I_R : R → R)
So, the domains of all fog, gof and I_Rare the same.
$(f o g) (x) = f (g (x)) = f (x - 1) = x - 1 + 1 = x = I_{R} (x) . . . (1) (g o f) (x) = g (f (x)) = g (x + 1) = x + 1 - 1 = x = I_{R} (x) . . . (2) From (1) and (2), (f o g) (x) = (g o f) (x) = I_{R} (x), \forall x \in R Hence, f o g = g o f = I_{R}$

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Question 9:

Verify associativity for the following three mappings : f : N → Z₀ (the set of non-zero integers), g : Z₀ → Q and h : Q → R given by f(x) = 2x, g(x) = 1/x and h(x) = e^x.

Answer:

Given that f : N → Z₀ , g : Z₀ → Q and h : Q → R .
gof : N → Qand hog : Z₀ → R
$\Rightarrow$ h o (gof ) : N → R and (hog) o f: N → R
So, both have the same domains.
$(g o f) (x) = g (f (x)) = g (2 x) = \frac{1}{2 x} . . . (1) (h o g) (x) = h (g (x)) = h (\frac{1}{x}) = e^{\frac{1}{x}} . . . (2) Now, (h o (g o f)) (x) = h ((g o f) (x)) = h (\frac{1}{2 x}) = e^{\frac{1}{2 x}} [from (1)] ((h o g) o f) (x) = (h o g) (f (x)) = (h o g) (2 x) = e^{\frac{1}{2 x}} [from (2)] ⇒ (h o (g o f)) (x) = ((h o g) o f) (x), \forall x \in N So, h o (g o f) = (h o g) o f$
Hence, the associative property has been verified.

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Question 10:

Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N. Show that ho (gof) = (hog) of.

Answer:

Given, f : N → N, g : N → N and h : N → R
$\Rightarrow$ gof : N → N and hog : N → R
$\Rightarrow$ ho (gof) : N → R and (hog) of : N → R
So, both have the same domains.
$(g o f) (x) = g (f (x)) = g (2 x) = 3 (2 x) + 4 = 6 x + 4 . . . (1) (h o g) (x) = h (g (x)) = h (3 x + 4) = \sin (3 x + 4) . . . (2) Now, (h o (g o f)) (x) = h ((g o f) (x)) = h (6 x + 4) = \sin (6 x + 4) [from (1)] ((h o g) o f) (x) = (h o g) (f (x)) = (h o g) (2 x) = \sin (6 x + 4) [from (2)] So, (h o (g o f)) (x) = ((h o g) o f) (x), \forall x \in N Hence, h o (g o f) = (h o g) o f$

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Question 11:

Give examples of two functions f : N → N and g : N → N, such that gof is onto but f is not onto.

Answer:

Let us consider a function f : N → N given by f(x) = x +1 , which is not onto.
[This not onto because if we take 0 in N (co-domain), then,
0=x+1
$\Rightarrow$ x=-1 $\notin N$ ]

Let us consider g : N → N given by
$g (x) = \{\begin{cases} x - 1, if x > 1 \\ 1, if x = 1 \end{cases} Now, let us find (g o f) (x) Case 1: x > 1 (g o f) (x) = g (f (x)) = g (x + 1) = x + 1 - 1 = x Case 2: x = 1 (g o f) (x) = g (f (x)) = g (x + 1) = 1 From case-1 and case-2, (g o f) (x) = x, \forall x \in N, which is an identity function and, hence, it is onto.$

Page No 2.46:

Question 12:

Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but g is not injective.

Answer:

Let f : N → Z be given by f (x) = x, which is injective.
(If we take f(x) = f(y), then it gives x = y)

Let g : Z → Z be given by g (x) = |x|, which is not injective.
If we take f(x) = f(y), we get:
|x| = |y|
$\Rightarrow$ x = $\pm$ y

Now, gof : N → Z.
$(g o f) (x) = g (f (x)) = g (x) = |x|$
Let us take two elements x and y in the domain of gof , such that
$(g o f) (x) = (g o f) (y) \Rightarrow |x| = |y| \Rightarrow x = y (We don't get ± here because x, y ∈N)$
So, gof is injective.

Page No 2.46:

Question 13:

If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.

Answer:

Given, f : A → B and g : B → C are one - one.
Then, gof : A → B
Let us take two elements x and y from A, such that
$(g o f) (x) = (g o f) (y) \Rightarrow g (f (x)) = g (f (y)) \Rightarrow f (x) = f (y) (As, g is one-one) \Rightarrow x = y (As, f is one-one)$
Hence, gof is one-one.

Page No 2.46:

Question 14:

If f : A → B and g : B → C are onto functions, show that gof is a onto function.

Answer:

Given, f : A → B and g : B → C are onto.
Then, gof : A → C
Let us take an element z in the co-domain (C).
Now, z is in C and g : B → C is onto.
So, there exists some element y in B, such that g (y) = z ... (1)
Now, y is in B and f : A → B is onto.
So, there exists some x in A, such that f (x) = y ... (2)
From (1) and (2),
z = g (y) = g (f (x)) = (gof) (x)
So, z = (gof) (x), where x is in A.
Hence, gof is onto.

Page No 2.54:

Question 1:

Find fog and gof if
(i) $f (x) = e^{x}, g (x) = \log_{e} x$
(ii) $f (x) = x^{2}, g (x) = \cos x$
(iii) $f (x) = | x |, g (x) = \sin x$
(iv) $f (x) = x + 1, g (x) = e^{x}$
(v) $f (x) = \sin^{- 1} x, g (x) = x^{2}$
(vi) $f (x) = x + 1, g (x) = \sin x$
(vii) $f (x) = x + 1, g (x) = 2 x + 3$
(viii) $f (x) = c, c \in R, g (x) = \sin x^{2}$
(ix) $f (x) = x^{2} + 2, g (x) = 1 - \frac{1}{1 - x}$

Answer:

$(i) f (x) = e^{x}, g (x) = \log_{e} x f : R \to (0, \infty); g : (0, \infty) \to R Computing f o g : Clearly, the range of g is a subset of the domain of f . f o g : (0, \infty) \to R (f o g) (x) = f (g (x)) = f (\log_{e} x) = \log_{e} e^{x} = x Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (e^{x}) = \log_{e} e^{x} = x$

$(i i) f (x) = x^{2}, g (x) = \cos x f : R \to [0, \infty); g : R \to [- 1, 1] Computing fog : Clearly, the range of g is not a subset of the domain of f . \Rightarrow Domain (f o g) = \{x : x \in domain of g and g (x) \in domain of f\} \Rightarrow Domain (f o g) = x : x \in R and \cos x \in R} \Rightarrow Domain of (f o g) = R f o g : R \to R (f o g) (x) = f (g (x)) = f (\cos x) = \cos^{2} x Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (x^{2}) = \cos (x^{2})$

$(i i i) f (x) = |x|, g (x) = \sin x f : R \to (0, \infty); g : R \to [- 1, 1] Computing f o g : Clearly, the range of g is a subset of the domain of f . \Rightarrow f o g : R \to R (f o g) (x) = f (g (x)) = f (\sin x) = |\sin x| Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (|x|) = \sin |x|$

$(i v) f (x) = x + 1, g (x) = e^{x} f : R \to R; g : R \to [1, \infty) Computing f o g : Clearly, range of g is a subset of domain of f . \Rightarrow f o g : R \to R (f o g) (x) = f (g (x)) = f (e^{x}) = e^{x} + 1 Computing g o f : Clearly, range of f is a subset of domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (x + 1) = e^{x + 1}$

$(v) f (x) = \sin^{- 1} x, g (x) = x^{2} f : [- 1, 1] \to [\frac{- π}{2}, \frac{π}{2}]; g : R \to [0, \infty) Computing f o g : Clearly, the range of g is not a subset of the domain of f . Domain (f o g) = \{x : x \in domain of g and g (x) \in domain of f\} Domain (f o g) = \{x : x \in R and x^{2} \in [- 1, 1]\} Domain (f o g) = \{x : x \in R and x \in [- 1, 1]\} Domain of (f o g) = [- 1, 1] f o g : [- 1, 1] \to R (f o g) (x) = f (g (x)) = f (x^{2}) = \sin^{- 1} (x^{2}) Computing g o f : Clearly, the range of f is a subset of the domain of g . f o g : [- 1, 1] \to R (g o f) (x) = g (f (x)) = g (\sin^{- 1} x) = {(\sin^{- 1} x)}^{2}$

$(vi) f (x) = x + 1, g (x) = \sin x f : R \to R; g : R \to [- 1, 1] Computing f o g : Clearly, the range of g is a subset of the domain of f . \Rightarrow f o g : R \to R (f o g) (x) = f (g (x)) = f (\sin x) = \sin x + 1 Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (x + 1) = \sin (x + 1)$

$(v i i) f (x) = x + 1, g (x) = 2 x + 3 f : R \to R; g : R \to R Computing f o g : Clearly, the range of g is a subset of the domain of f . \Rightarrow f o g : R \to R (f o g) (x) = f (g (x)) = f (2 x + 3) = 2 x + 3 + 1 = 2 x + 4 Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (x + 1) = 2 (x + 1) + 3 = 2 x + 5$

$(v i i i) f (x) = c, g (x) = \sin x^{2} f : R \to \{c\}; g : R \to [0, 1] Computing f o g : Clearly, the range of g is a subset of the domain of f . f o g : R \to R (f o g) (x) = f (g (x)) = f (\sin x^{2}) = c Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow f o g : R \to R (g o f) (x) = g (f (x)) = g (c) = \sin c^{2}$

$(i x) f (x) = x^{2} + 2 f : R \to [2, \infty) g (x) = 1 - \frac{1}{1 - x} For domain of g : 1 - x \neq 0 \Rightarrow x \neq 1 \Rightarrow Domain of g= R - \{1\} g (x) = 1 - \frac{1}{1 - x} = \frac{1 - x - 1}{1 - x} = \frac{- x}{1 - x} For range of g : y = \frac{- x}{1 - x} \Rightarrow y - x y = - x \Rightarrow y = x y - x \Rightarrow y = x (y - 1) \Rightarrow x = \frac{y}{y - 1} Range of g = R - \{1\} So, g: R - \{1\} \to R - \{1\} Computing f o g : Clearly, the range of g is a subset of the domain of f . \Rightarrow f o g : R - \{1\} \to R (f o g) (x) = f (g (x)) = f (\frac{- x}{x - 1}) = {(\frac{- x}{x - 1})}^{2} + 2 = \frac{x^{2} + 2 x^{2} + 2 - 4 x}{{(1 - x)}^{2}} = \frac{3 x^{2} - 4 x + 2}{{(1 - x)}^{2}} Computing g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow g o f : R \to R (g o f) (x) = g (f (x)) = g (x^{2} + 2) = 1 - \frac{1}{1 - (x^{2} + 2)} = 1 - \frac{1}{- (x^{2} + 1)} = \frac{x^{2} + 2}{x^{2} + 1}$

Page No 2.54:

Question 2:

Let f(x) = x² + x + 1 and g(x) = sin x. Show that fog ≠ gof.

Answer:

$(f o g) (x) = f (g (x)) = f (\sin x) = \sin^{2} x + \sin x + 1 and (g o f) (x) = g (f (x)) = g (x^{2} + x + 1) = \sin (x^{2} + x + 1) So, f o g \neq g o f .$

Page No 2.54:

Question 3:

If f(x) = |x|, prove that fof = f.

Answer:

Domains of f and fof are same as R.
$(f o f) (x) = f (f (x)) = f (|x|) = ||x|| = |x| = f (x) So, (f o f) (x) = f (x), \forall x \in R Hence, f o f = f$

Page No 2.54:

Question 4:

If f(x) = 2x + 5 and g(x) = x² + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f²

Also, show that fof ≠ f²

Answer:

f(x) and g(x) are polynomials.
$\Rightarrow$ f : R → R and g : R → R.
So, fog : R → R and gof : R → R.
$(i) (f o g) (x) = f (g (x)) = f (x^{2} + 1) = 2 (x^{2} + 1) + 5 = 2 x^{2} + 2 + 5 = 2 x^{2} + 7$

$(ii) (g o f) (x) = g (f (x)) = g (2 x + 5) = {(2 x + 5)}^{2} + 1 = 4 x^{2} + 20 x + 26$

$(iii) (f o f) (x) = f (f (x)) = f (2 x + 5) = 2 (2 x + 5) + 5 = 4 x + 10 + 5 = 4 x + 15$

$(iv) f^{2} (x) = f (x) \times f (x) = (2 x + 5) (2 x + 5) = {(2 x + 5)}^{2} = 4 x^{2} + 20 x + 25$

→→ →

Page No 2.54:

Question 5:

If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?

Answer:

$We know that f : R \to [- 1, 1] and g : R \to R Clearly, the range of f is a subset of the domain of g . g o f : R \to R (g o f) (x) = g (f (x)) = g (\sin x) = 2 \sin x$

$Clearly, the range of g is a subset of the domain of f . f o g : R \to R So, (f o g) (x) = f (g (x)) = f (2 x) = \sin (2 x)$

Clearly, fog $\neq$ gof

Page No 2.54:

Question 6:

Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (fh).

Answer:

$We know that f : R \to [- 1, 1] and g : R \to R Clearly, the range of g is a subset of the domain of f . f o g : R \to R Now, (f h) (x) = f (x) h (x) = (\sin x) (\cos x) = \frac{1}{2} \sin (2 x) Domain of f h is R . Since range of sin x is [-1,1], - 1 \leq \sin 2 x \leq 1 \Rightarrow \frac{- 1}{2} \leq \frac{\sin x}{2} \leq \frac{1}{2} Range of f h = [\frac{- 1}{2}, \frac{1}{2}] So, (f h) : R \to [\frac{- 1}{2}, \frac{1}{2}] Clearly, range of f h is a subset of g . ⇒ g o (f h) : R \to R ⇒domains of f o g and g o (f h) are the same. So, (f o g) (x) = f (g (x)) = f (2 x) = \sin (2 x) and (g o (f h)) (x) = g ((f h) (x)) = g (\sin x \cos x) = 2 \sin x \cos x = \sin (2 x) ⇒ (f o g) (x) = (g o (f h)) (x), \forall x \in R Hence, f o g = g o (f h)$

Page No 2.54:

Question 7:

Let f be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.

Answer:

$Given, f : R \to R Since g (x) = 2 x is a polynomial, g : R \to R Clearly, g o f : R \to R a n d f + f : R \to R So, domains of gof and f+f are the same. (g o f) (x) = g (f (x)) = 2 f (x) (f + f) (x) = f (x) + f (x) = 2 f (x) ⇒ (g o f) (x) = (f + f) (x), \forall x \in R Hence, g o f = f + f$

Page No 2.54:

Question 8:

If $f (x) = \sqrt{1 - x} and g (x) = \log_{e} x$ are two real functions, then describe functions fog and gof.

Answer:

$f (x) = \sqrt{1 - x} For domain, 1-x≥0 \Rightarrow x \leq 1 \Rightarrow domain of f =(- \infty, 1] \Rightarrow f : (- \infty, 1] \to (0, \infty) g (x) = \log_{e} x Clearly, g : (0, \infty) \to R Computation of f o g : Clearly, the range of g is not a subset of the domain of f . So, we need to compute the domain of f o g . \Rightarrow Domain (f o g) = \{x : x \in Domain (g) and g (x) \in Domain of f\} \Rightarrow Domain (f o g) = \{x : x \in (0, \infty) {and log}_{e} x ∈ (- \infty, 1]\} \Rightarrow Domain (f o g) = \{x : x \in (0, \infty) and x \in (0, e]\} \Rightarrow Domain (f o g) = \{x : x \in (0, e]\} \Rightarrow Domain (f o g) =(0, e] \Rightarrow f o g : (0, e) \to R So, (f o g) (x) = f (g (x)) = f (\log_{e} x) = \sqrt{1 - \log_{e} x} Computation of g o f : Clearly, the range of f is a subset of the domain of g . \Rightarrow g o f : (- \infty, 1] \to R \Rightarrow (g o f) (x) = g (f (x)) = g (\sqrt{1 - x}) = \log_{e} \sqrt{1 - x} = \log_{e} {(1 - x)}^{\frac{1}{2}} = \frac{1}{2} \log_{e} (1 - x)$

Page No 2.54:

Question 9:

If $f : (- \frac{π}{2}, \frac{π}{2}) \to R and g : [- 1, 1] \to R$ be defined as
$f (x) = \tan x and g (x) = \sqrt{1 - x^{2}}$ respectively, describe fog and gof.

Answer:

$g (x) = \sqrt{1 - x^{2}} \Rightarrow x^{2} \geq 0, \forall x \in [- 1, 1] \Rightarrow - x^{2} \leq 0, \forall x \in [- 1, 1] \Rightarrow 1 - x^{2} \leq 1, \forall x \in [- 1, 1] We know that 1- x^{2} ≥0 ⇒0≤1- x^{2} ≤1 ⇒Range of g (x) = [0, 1] So, f : (\frac{- π}{2}, \frac{π}{2}) \to R and g : [- 1, 1] \to [0, 1] Computation of fog: Clearly, the range of g is a subset of the domain of f. So, fog: [- 1, 1] \to R (f o g) (x) = f (g (x)) = f (\sqrt{1 - x^{2}}) = \tan \sqrt{1 - x^{2}} Computation of gof: Clearly, the range of f is not a subset of the domain of g . \Rightarrow Domain (g o f) = \{x \in domain of f and f (x) \in domain of g\} \Rightarrow Domain (g o f) = \{x \in (\frac{- π}{2}, \frac{π}{2}) and \tan x \in [- 1, 1]\} \Rightarrow Domain (g o f) = \{x \in (\frac{- π}{2}, \frac{π}{2}) and x \in (\frac{- π}{4}, \frac{π}{4})\} \Rightarrow Domain (g o f) = \{x \in (\frac{- π}{4}, \frac{π}{4})\} Now, g o f : (\frac{- π}{4}, \frac{π}{4}) \to R So, (g o f) (x) = g (f (x)) = g (\tan x) = \sqrt{1 - \tan^{2} x}$

Page No 2.54:

Question 10:

If $f (x) = \sqrt{x + 3} and g (x) = x^{2} + 1$ be two real functions, then find fog and gof.

Answer:

$f (x) = \sqrt{x + 3} For domain, x + 3 \geq 0 \Rightarrow x \geq - 3 Domain of f =[-3, ∞) Since f is a square root function, range of f =[0, ∞) f : [-3, \infty) \to [0, \infty) g (x) = x^{2} + 1 is a polynomial. \Rightarrow g : R \to R Computation of f o g : Range of g is not a subset of the domain of f . and domain (f o g) = \{x : x \in domain of g and g (x) \in domain of f (x)\} \Rightarrow Domain (f o g) = \{x : x \in R and x^{2} + 1 \in [-3, \infty)\} \Rightarrow Domain (f o g) = \{x : x \in R and x^{2} + 1 \geq - 3\} \Rightarrow Domain (f o g) = \{x : x \in R and x^{2} + 4 \geq 0\} \Rightarrow Domain (f o g) = \{x : x \in R and x \in R\} \Rightarrow Domain (f o g) =R f o g : R \to R (f o g) (x) = f (g (x)) = f (x^{2} + 1) = \sqrt{x^{2} + 1 + 3} = \sqrt{x^{2} + 4} Computation of g o f : Range of f is a subset of the domain of g . g o f : [-3, \infty) \to R \Rightarrow (g o f) (x) = g (f (x)) = g (\sqrt{x + 3}) = {(\sqrt{x + 3})}^{2} + 1 = x + 3 + 1 = x + 4$

Page No 2.54:

Question 11:

Let f be a real function given by $f (x) = \sqrt{x - 2}$ .
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f²

Also, show that fof ≠ f² .

Answer:

$f (x) = \sqrt{x - 2} For domain, x - 2 \geq 0 \Rightarrow x \geq 2 Domain of f = [2, \infty) Since f is a square-root function, range of f = (0, \infty) So, f : [2, \infty) \to (0, \infty) (i) f o f Range of f is not a subset of the domain of f . \Rightarrow Domain (f o f) = \{x : x \in domain of f and f (x) \in domain of f\} \Rightarrow Domain (f o f) = \{x : x \in [2, \infty) and \sqrt{x - 2} \in [2, \infty)\} \Rightarrow Domain (f o f) = \{x : x \in [2, \infty) and \sqrt{x - 2} \geq 2\} \Rightarrow Domain (f o f) = \{x : x \in [2, \infty) and x - 2 \geq 4\} \Rightarrow Domain (f o f) = \{x : x \in [2, \infty) and x \geq 6\} \Rightarrow Domain (f o f) = \{x : x \geq 6\} \Rightarrow Domain (f o f) =[6, ∞) f o f : [6, \infty) \to R (f o f) (x) = f (f (x)) = f (\sqrt{x - 2}) = \sqrt{\sqrt{x - 2} - 2}$

$(ii) f o f o f = (f o f) o f We have, f : [2, \infty) \to (0, \infty) and f o f : [6, \infty) \to R \Rightarrow Range of f is not a subset of the domain of f o f . Then, domain ((f o f) o f) = \{x : x \in domain of f and f (x) \in domain of f o f\} \Rightarrow Domain ((f o f) o f) = \{x : x \in [2, \infty) and \sqrt{x - 2} \in [6, \infty)\} \Rightarrow Domain ((f o f) o f) = \{x : x \in [2, \infty) and \sqrt{x - 2} \geq 6\} \Rightarrow Domain ((f o f) o f) = \{x : x \in [2, \infty) and x - 2 \geq 36\} \Rightarrow Domain ((f o f) o f) = \{x : x \in [2, \infty) and x \geq 38\} \Rightarrow Domain ((f o f) o f) = \{x : x \geq 38\} \Rightarrow Domain ((f o f) o f) =[38, ∞) f o f : [38, \infty) \to R So, ((f o f) o f) (x) = (f o f) (f (x)) = (f o f) (\sqrt{x - 2}) = \sqrt{\sqrt{\sqrt{x - 2} - 2} - 2}$

$(i i i) We have, (f o f o f) (x) = \sqrt{\sqrt{\sqrt{x - 2} - 2} - 2} So, (f o f o f) (38) = \sqrt{\sqrt{\sqrt{38 - 2} - 2} - 2} = \sqrt{\sqrt{\sqrt{36} - 2} - 2} = \sqrt{\sqrt{6 - 2} - 2} = \sqrt{2 - 2} =0$

$(iv) We have, f o f = \sqrt{\sqrt{x - 2} - 2} \Rightarrow f^{2} (x) = f (x) \times f (x) = \sqrt{x - 2} \times \sqrt{x - 2} = x - 2 So, f o f \neq f^{2}$

Page No 2.55:

Question 12:

Let $f (x) = \{\begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$ . Find fof.

Answer:

$f (x) = \{\begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases} It can be written as, f (x) = \{\begin{cases} \begin{array}{l} 1 + x, & 0 \leq x \leq 1 \end{array} \begin{array}{l} 1 + x, & 1 < x \leq 2 \end{array} \begin{array}{l} 3 - x, & 2 < x \leq 3 \end{array} \end{cases} When \begin{array}{l} , & 0 \leq x \leq 1 \end{array} Then, f (x) = 1 + x Now when \begin{array}{l} , & 0 \leq x \leq 1 \end{array} then \begin{array}{l} , & 1 \leq x + 1 \leq 2 \end{array} Then, f (f (x)) = 1 + (1 + x) = 2 + x [∵ 1 \leq f (x) < 2] When \begin{array}{l} , & 1 < x \leq 2 \end{array} Then, f (x) = 1 + x Now when \begin{array}{l} , & 1 < x \leq 2 \end{array} then \begin{array}{l} , & 2 < x + 1 \leq 3 \end{array} Then, f (f (x)) = 3 - (1 + x) = 2 - x [∵ 2 \leq f (x) < 3] When \begin{array}{l} , & 2 < x \leq 3 \end{array} Then, f (x) = 3 - x Now when \begin{array}{l} , & 2 < x \leq 3 \end{array} then \begin{array}{l} , & 0 \leq 3 - x < 1 \end{array} Then, f (f (x)) = 1 + (3 - x) = 4 - x [∵ 0 \leq f (x) < 1] f (f (x)) = \{\begin{cases} \begin{array}{l} 2 + x, & 0 \leq x \leq 1 \end{array} \begin{array}{l} 2 - x, & 1 < x \leq 2 \end{array} \begin{array}{l} 4 - x, & 2 < x \leq 3 \end{array} \end{cases}$

Page No 2.55:

Question 13:

If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| – $x, \forall x \in R$ . Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).

Answer:

Given: f(x) = |x| + x
and g(x) = |x| – $x, \forall x \in R$

$f o g = f (g (x)) = |g (x)| + g (x) = ||x| - x| + (|x| - x)$

Therefore,
$f (g (x)) = \{\begin{cases} 0 x \geq 0 \\ 4 x x < 0 \end{cases} f o g = \{\begin{cases} 4 x x < 0 \\ 0 x \geq 0 \end{cases}$

$g o f = g (f (x)) = |f (x)| - f (x) = ||x| + x| - (|x| + x) g (f (x)) = \{\begin{cases} 0 x \geq 0 \\ 0 x < 0 \end{cases}$
Therefore, g(f(x)) = gof = 0

Now, fog(−3) =(4)(−3) = −12 (since, fog = 4x for x < 0)

fog(5) = 0 (since, fog = 0 for x $\geq$ 0)

gof(−2) = 0 (since, gof = 0 for x < 0)

Page No 2.68:

Question 1:

State with reasons whether the following functions have inverse:
(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Answer:

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have:
f (1) = f (2) = f (3) = f (4) = 10
$\Rightarrow$ f is not one-one.
$\Rightarrow$ f is not a bijection.
So, f does not have an inverse.

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
g (5) = g (7) = 4
$\Rightarrow$ f is not one-one.
$\Rightarrow$ f is not a bijection.
So, f does not have an inverse.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, different elements of the domain have different images in the co-domain.
$\Rightarrow$ h is one-one.
Also, each element in the co-domain has a pre-image in the domain.
$\Rightarrow$ h is onto.
$\Rightarrow$ h is a bijection.
$\Rightarrow$ h has an inverse and it is given by
h^-1={(7, 2), (9, 3), (11, 4), (13, 5)}

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Question 2:

Find f ⁻¹ if it exists : f : A → B, where
(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²

Answer:

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
Given: f(x) = 3 x
So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Clearly, this is one-one.
Range of f = Range of f =B
So, f is a bijection and, thus, f^-1 exists.
Hence, f^-1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²
Given: f(x) = x²
So, f = {(1, 1), (3, 9), (5, 25), (7,49), (9, 81)}
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A) .
$\Rightarrow$ f is not a bijection.
So, f^-1does not exist.

Page No 2.68:

Question 3:

Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f⁻¹, g⁻¹ and gof⁻¹ and show that (gof)⁻¹ = f ⁻¹o g⁻¹.

Answer:

$f = \{(1, a), (2, b), (3, c)\} and g = \{(a, apple), (b, ball), (c, cat)\} Clearly, f and g are bijections. So, f and g are invertible. Now, f^{- 1} = \{(a, 1), (b, 2), (c, 3)\} and g^{- 1} = \{(apple, a), (ball, b), (cat, c)\} So, f^{- 1} o g^{- 1} = \{(apple, 1), (ball, 2), (cat, 3)\} . . . (1) f : \{1, 2, 3\} \to \{a, b, c\} and g : \{a, b, c\} \to \{apple, ball, cat\} So, g o f : \{1, 2, 3\} \to \{apple, ball, cat\} \Rightarrow (g o f) (1) = g (f (1)) = g (a) = apple (g o f) (2) = g (f (2)) = g (b) = ball, and (g o f) (3) = g (f (3)) = g (c) = cat ∴ g o f = \{(1, apple), (2, ball), (3, cat)\} Clearly, g o f is a biject ion. So, g o f is invertible. {(g o f)}^{- 1} = \{(apple, 1), (ball, 2), (cat, 3)\} . . . (2) From (1) and (2), we get: {(g o f)}^{- 1} = f^{- 1} o g^{- 1}$

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Question 4:

Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f : A → B, g : B → C be defined as f(x) = 2x + 1 and g(x) = x² − 2. Express (gof)⁻¹ and f⁻¹ og⁻¹ as the sets of ordered pairs and verify that (gof)⁻¹ = f⁻¹ og⁻¹.

Answer:

$f (x) = 2 x + 1 \Rightarrow f = \{(1, 2 (1) + 1), (2, 2 (2) + 1), (3, 2 (3) + 1), (4, 2 (4) + 1)\} = \{(1, 3), (2, 5), (3, 7), (4, 9)\} g (x) = x^{2} - 2 \Rightarrow g = \{(3, 3^{2} - 2), (5, 5^{2} - 2), (7, 7^{2} - 2), (9, 9^{2} - 2)\} = \{(3, 7), (5, 23), (7, 47), (9, 79)\} Clearly f and g are bijections and, hence, f^{- 1} : B \to A and g^{- 1} : C \to B exist. So, f^{- 1} = \{(3, 1), (5, 2), (7, 3), (9, 4)\} and g^{- 1} = \{(7, 3), (23, 5), (47, 7), (79, 9)\} Now, (f^{- 1} o g^{- 1}) : C \to A f^{- 1} o g^{- 1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\} . . . (1) Also, f : A \to B and g : B \to C, ⇒ g o f : A \to C, {(g o f)}^{- 1} : C \to A So, f^{- 1} o g^{- 1} and {(g o f)}^{- 1} have same domains. (g o f) (x) = g (f (x)) = g (2 x + 1) = {(2 x + 1)}^{2} - 2 \Rightarrow (g o f) (x) = 4 x^{2} + 4 x + 1 - 2 \Rightarrow (g o f) (x) = 4 x^{2} + 4 x - 1 Then, (g o f) (1) = g (f (1)) = 4 + 4 - 1 = 7, (g o f) (2) = g (f (2)) = 4 + 4 - 1 = 23, (g o f) (3) = g (f (3)) = 4 + 4 - 1 = 47 and (g o f) (4) = g (f (4)) = 4 + 4 - 1 = 79 So, g o f = \{(1, 7), (2, 23), (3, 47), (4, 79)\} \Rightarrow {(g o f)}^{- 1} = \{(7, 1), (23, 2), (47, 3), (79, 4)\} . . . (2) From (1) and (2), we get: {(g o f)}^{- 1} = f^{- 1} o g^{- 1}$

Page No 2.68:

Question 5:

Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f⁻¹

Answer:

Injectivity of f:
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)
$\Rightarrow$ 3x + 5 =3y + 5
$\Rightarrow$ 3x = 3y
$\Rightarrow$ x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y

$\Rightarrow 3 x + 5 = y \Rightarrow 3 x = y - 5 \Rightarrow x = \frac{y - 5}{3} \in Q (domain)$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = 3 y + 5 \Rightarrow x - 5 = 3 y \Rightarrow y = \frac{x - 5}{3} So, f^{- 1} (x) = \frac{x - 5}{3} [from (1)]$

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Question 6:

Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer:

Injectivity of f :
Let x and y be two elements of domain (R), such that
f(x) = f(y)
$\Rightarrow$ 4x + 3 = 4y + 3
$\Rightarrow$ 4x = 4y
$\Rightarrow$ x = y
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (R), such that f(x) = y.

$\Rightarrow 4 x + 3 = y \Rightarrow 4 x = y - 3 \Rightarrow x = \frac{y - 3}{4} \in R (Domain)$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = 4 y + 3 \Rightarrow x - 3 = 4 y \Rightarrow y = \frac{x - 3}{4} So, f^{- 1} (x) = \frac{x - 3}{4} [from (1)]$

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Question 7:

Consider f : R → R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with inverse f⁻¹ of f given by f⁻¹ $(x) = \sqrt{x - 4}$ , where R⁺ is the set of all non-negative real numbers.

Answer:

Injectivity of f :
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)
$\Rightarrow x^{2} + 4 = y^{2} + 4 \Rightarrow x^{2} = y^{2} \Rightarrow x = y (as co-domain as R^{+})$
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (Q), such that f(x) = y

$\Rightarrow x^{2} + 4 = y \Rightarrow x^{2} = y - 4 \Rightarrow x = \sqrt{y - 4} \in R$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = y^{2} + 4 \Rightarrow x - 4 = y^{2} \Rightarrow y = \sqrt{x - 4} So, f^{- 1} (x) = \sqrt{x - 4} [from (1)]$

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Question 8:

If $f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3}$ , show that fof(x) = x for all $x \neq \frac{2}{3}$ . What is the inverse of f?

Answer:

$(f o f) (x) = f (f (x)) = f (\frac{4 x + 3}{6 x - 4}) = \frac{4 (\frac{4 x + 3}{6 x - 4}) + 3}{6 (\frac{4 x + 3}{6 x - 4}) - 4} = \frac{16 x + 12 + 18 x - 12}{24 x + 18 - 24 x + 16} = \frac{34 x}{34} = x ⇒ (f o f) (x) = x = I_{X}, where I is an identity function. So, f = f^{- 1} Hence, f^{- 1} = \frac{4 x + 3}{6 x - 4}$

Page No 2.68:

Question 9:

Consider f : R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with $f^{- 1} (x) = \frac{\sqrt{x + 6} - 1}{3}$ .

Answer:

Injectivity of f :
Let x and y be two elements of domain (R⁺), such that
f(x)=f(y)
$\Rightarrow 9 x^{2} + 6 x - 5 = 9 y^{2} + 6 y - 5 \Rightarrow 9 x^{2} + 6 x = 9 y^{2} + 6 y \Rightarrow x = y (As, x, y \in R^{+})$
So, f is one-one.

Surjectivity of f:
Let y is in the co domain (Q) such that f(x) = y

$\Rightarrow 9 x^{2} + 6 x - 5 = y \Rightarrow 9 x^{2} + 6 x = y + 5 \Rightarrow 9 x^{2} + 6 x + 1 = y + 6 (Adding 1 on both sides) \Rightarrow {(3 x + 1)}^{2} = y + 6 \Rightarrow 3 x + 1 = \sqrt{y + 6} \Rightarrow 3 x = \sqrt{y + 6} - 1 \Rightarrow x = \frac{\sqrt{y + 6} - 1}{3} \in R^{+} (domain)$

$\Rightarrow$ f is onto.
So, f is a bijection and hence, it is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = 9 y^{2} + 6 y - 5 \Rightarrow x + 5 = 9 y^{2} + 6 y \Rightarrow x + 6 = 9 y^{2} + 6 y + 1 (adding 1 on both sides) \Rightarrow x + 6 = {(3 y + 1)}^{2} \Rightarrow 3 y + 1 = \sqrt{x + 6} \Rightarrow 3 y = \sqrt{x + 6} - 1 \Rightarrow y = \frac{\sqrt{x + 6} - 1}{3} So, f^{- 1} (x) = \frac{\sqrt{x + 6} - 1}{3} [from (1)]$

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Question 10:

If f : R → R be defined by f(x) = x³ −3, then prove that f⁻¹ exists and find a formula for f⁻¹. Hence, find f⁻¹ (24) and f⁻¹ (5).

Answer:

Injectivity of f :
Let x and y be two elements in domain (R),

$such that, x^{3} - 3 = y^{3} - 3 \Rightarrow x^{3} = y^{3} \Rightarrow x = y$
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (R) such that f(x) = y

$\Rightarrow x^{3} - 3 = y \Rightarrow x^{3} = y + 3 \Rightarrow x = \sqrt[3]{y + 3} \in R$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = y^{3} - 3 \Rightarrow x + 3 = y^{3} \Rightarrow y = \sqrt[3]{x + 3} = f^{- 1} (x) [from (1)] So, f^{- 1} (x) = \sqrt[3]{x + 3} Now, f^{- 1} (24) = \sqrt[3]{24 + 3} = \sqrt[3]{27} = \sqrt[3]{3^{3}} = 3 and f^{- 1} (5) = \sqrt[3]{5 + 3} = \sqrt[3]{8} = \sqrt[3]{2^{3}} = 2$

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Question 11:

A function f : R → R is defined as f(x) = x³ + 4. Is it a bijection or not? In case it is a bijection, find f⁻¹ (3).

Answer:

Injectivity of f:
Let x and y be two elements of domain (R), such that
$f (x) = f (y) \Rightarrow x^{3} + 4 = y^{3} + 4 \Rightarrow x^{3} = y^{3} \Rightarrow x = y$
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (R), such that f(x) = y.

$\Rightarrow x^{3} + 4 = y \Rightarrow x^{3} = y - 4 \Rightarrow x = \sqrt[3]{y - 4} \in R (domain)$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = y^{3} + 4 \Rightarrow x - 4 = y^{3} \Rightarrow y = \sqrt[3]{x - 4} So, f^{- 1} (x) = \sqrt[3]{x - 4} [from (1)] f^{- 1} (3) = \sqrt[3]{3 - 4} = \sqrt[3]{- 1} = - 1$

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Question 12:

If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)⁻¹ = f⁻¹ og ⁻¹.

Answer:

Injectivity of f:
Let x and y be two elements of domain (Q), such that
f(x) = f(y)
$\Rightarrow$ 2x = 2y
$\Rightarrow$ x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y.

$\Rightarrow 2 x = y \Rightarrow x = \frac{y}{2} \in Q (domain)$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f^-¹:
$Let f^{- 1} (x) = y . . . (1) \Rightarrow x = f (y) \Rightarrow x = 2 y \Rightarrow y = \frac{x}{2} So, f^{- 1} (x) = \frac{x}{2} (from (1))$

Injectivity of g:
Let x and y be two elements of domain (Q), such that
g(x) = g(y)
$\Rightarrow$ x + 2 = y + 2
$\Rightarrow$ x = y
So, g is one-one.

Surjectivity of g:
Let y be in the co domain (Q), such that g(x) = y.

$\Rightarrow x + 2 = y \Rightarrow x = 2 - y \in Q (domain)$

$\Rightarrow$ g is onto.
So, g is a bijection and, hence, it is invertible.

Finding g^-¹:
$Let g^{- 1} (x) = y . . . (2) \Rightarrow x = g (y) \Rightarrow x = y + 2 \Rightarrow y = x - 2 So, g^{- 1} (x) = x - 2 (From (2))$

Verification of (gof)⁻¹ = f⁻¹ og ⁻¹:

$f (x) = 2 x; g (x) = x + 2 and f^{- 1} (x) = \frac{x}{2}; g^{- 1} (x) = x - 2 Now, (f^{- 1} o g^{- 1}) (x) = f^{- 1} (g^{- 1} (x)) \Rightarrow (f^{- 1} o g^{- 1}) (x) = f^{- 1} (x - 2) \Rightarrow (f^{- 1} o g^{- 1}) (x) = \frac{x - 2}{2} . . . (3) (g o f) (x) = g (f (x)) = g (2 x) = 2 x + 2 Let {(g o f)}^{- 1} (x) =y .... (4) x = (g o f) (y) \Rightarrow x = 2 y + 2 \Rightarrow 2 y = x - 2 \Rightarrow y = \frac{x - 2}{2} \Rightarrow {(g o f)}^{- 1} (x) = \frac{x - 2}{2} [from (4) . . . (5)] From (3) and (5), {(g o f)}^{- 1} = f^{- 1} o g^{- 1}$

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Question 13:

Let A = R $-$ {3} and B = R $-$ {1}. Consider the function f : A $\to$ B defined by f(x) = $\frac{x - 2}{x - 3}$ . Show that f is one-one and onto and
hence find f^{$-$ 1}. [CBSE 2012, 2014]

Answer:

We have,

A = R $-$ {3} and B = R $-$ {1}
The function f : A $\to$ B defined by f(x) = $\frac{x - 2}{x - 3}$

$Let x, y \in A such that f (x) = f (y) . Then, \frac{x - 2}{x - 3} = \frac{y - 2}{y - 3} \Rightarrow x y - 3 x - 2 y + 6 = x y - 2 x - 3 y + 6 \Rightarrow - x = - y \Rightarrow x = y ∴ f is one - one . Let y \in B . Then, y \neq 1 . The function f is onto if there exists x \in A such that f (x) = y . Now, f (x) = y \Rightarrow \frac{x - 2}{x - 3} = y \Rightarrow x - 2 = x y - 3 y \Rightarrow x - x y = 2 - 3 y \Rightarrow x (1 - y) = 2 - 3 y \Rightarrow x = \frac{2 - 3 y}{1 - y} \in A [y \neq 1] Thus, for any y \in B, there exists \frac{2 - 3 y}{1 - y} \in A such that f (\frac{2 - 3 y}{1 - y}) = \frac{(\frac{2 - 3 y}{1 - y}) - 2}{(\frac{2 - 3 y}{1 - y}) - 3} = \frac{2 - 3 y - 2 + 2 y}{2 - 3 y - 3 + 3 y} = \frac{- y}{- 1} = y ∴ f is onto . So, f is one - one and onto fucntion . Now, As, x = (\frac{2 - 3 y}{1 - y}) So, f^{- 1} (x) = (\frac{2 - 3 x}{1 - x}) = \frac{3 x - 2}{x - 1}$

Page No 2.69:

Question 14:

Consider the function f : R⁺ $\to [- 9, \infty]$ given by f(x) = 5x² + 6x $-$ 9. Prove that f is invertible with f^{$-$ 1}(y) = $\frac{\sqrt{54 + 5 y} - 3}{5}$ . [CBSE 2015]

Answer:

$We have, f (x) = 5 x^{2} + 6 x - 9 Let y = 5 x^{2} + 6 x - 9 = 5 (x^{2} + \frac{6}{5} x - \frac{9}{5}) = 5 (x^{2} + 2 \times x \times \frac{3}{5} + \frac{9}{25} - \frac{9}{25} - \frac{9}{5}) = 5 ({(x + \frac{3}{5})}^{2} - \frac{9}{25} - \frac{9}{5}) = 5 {(x + \frac{3}{5})}^{2} - \frac{9}{5} - 9 = 5 {(x + \frac{3}{5})}^{2} - \frac{54}{5} \Rightarrow y + \frac{54}{5} = 5 {(x + \frac{3}{5})}^{2} \Rightarrow \frac{5 y + 54}{25} = {(x + \frac{3}{5})}^{2} \Rightarrow \sqrt{\frac{5 y + 54}{25}} = x + \frac{3}{5} \Rightarrow x = \frac{\sqrt{5 y + 54}}{5} - \frac{3}{5} \Rightarrow x = \frac{\sqrt{5 y + 54} - 3}{5} Let g (y) = \frac{\sqrt{5 y + 54} - 3}{5} Now, f o g (y) = f (g (y)) = f (\frac{\sqrt{5 y + 54} - 3}{5}) = 5 {(\frac{\sqrt{5 y + 54} - 3}{5})}^{2} + 6 (\frac{\sqrt{5 y + 54} - 3}{5}) - 9 = 5 (\frac{5 y + 54 + 9 - 6 \sqrt{5 y + 54}}{25}) + (\frac{6 \sqrt{5 y + 54} - 18}{5}) - 9 = \frac{5 y + 63 - 6 \sqrt{5 y + 54}}{5} + \frac{6 \sqrt{5 y + 54} - 18}{5} - 9 = \frac{5 y + 63 - 18 - 45}{5} = y = I_{Y}, Identity function Also, g o f (x) = g (f (x)) = g (5 x^{2} + 6 x - 9) = \frac{\sqrt{5 (5 x^{2} + 6 x - 9) + 54} - 3}{5} = \frac{\sqrt{25 x^{2} + 30 x - 45 + 54} - 3}{5} = \frac{\sqrt{25 x^{2} + 30 x + 9} - 3}{5} = \frac{\sqrt{{(5 x + 3)}^{2}} - 3}{5} = \frac{5 x + 3 - 3}{5} = x = I_{X}, Identity function So, f is invertible . Also, f^{- 1} (y) = g (y) = \frac{\sqrt{5 y + 54} - 3}{5}$

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Question 15:

Let f : N $\to$ N be a function defined as f $(x) =$ 9 $x^{2} +$ 6 $x -$ 5. Show that f : N $\to$ S, where S is the range of f, is invertible. find the inverse of f and hence find f ^{$-$ 1}(43) and f ^{$-$ 1}(163).

Answer:

We have,

f : N $\to$ N is a function defined as f (x) = 9x² + 6x $-$ 5.

Let y = f (x) = 9x² + 6x $-$ 5

$\Rightarrow y = 9 x^{2} + 6 x - 5 \Rightarrow y = 9 x^{2} + 6 x + 1 - 1 - 5 \Rightarrow y = (9 x^{2} + 6 x + 1) - 6 \Rightarrow y = {(3 x + 1)}^{2} - 6 \Rightarrow y + 6 = {(3 x + 1)}^{2}$

$\Rightarrow \sqrt{y + 6} = 3 x + 1 (∵ y \in N) \Rightarrow \sqrt{y + 6} - 1 = 3 x \Rightarrow x = \frac{\sqrt{y + 6} - 1}{3} \Rightarrow g (y) = \frac{\sqrt{y + 6} - 1}{3} [Let x = g (y)]$

Now,

$f o g (y) = f [g (y)] = f (\frac{\sqrt{y + 6} - 1}{3}) = 9 {(\frac{\sqrt{y + 6} - 1}{3})}^{2} + 6 (\frac{\sqrt{y + 6} - 1}{3}) - 5 = 9 (\frac{y + 6 - 2 \sqrt{y + 6} + 1}{9}) + 2 (\sqrt{y + 6} - 1) - 5 = y + 6 - 2 \sqrt{y + 6} + 1 + 2 \sqrt{y + 6} - 2 - 5 = y = I_{Y}, Identity function$

$g o f (x) = g [f (x)] = g (9 x^{2} + 6 x - 5) = \frac{\sqrt{(9 x^{2} + 6 x - 5) + 6} - 1}{3} = \frac{\sqrt{(9 x^{2} + 6 x + 1)} - 1}{3} = \frac{\sqrt{{(3 x + 1)}^{2}} - 1}{3} = \frac{(3 x + 1) - 1}{3} = \frac{3 x}{3} = x = I_{X}, Identity function$

Since, fog(y) and gof(x) are identity function.

Thus, f is invertible.

So, $f^{- 1} (x) = g (x) = \frac{\sqrt{x + 6} - 1}{3}$ .

Now,

f ^{$-$ 1}(43) = $\frac{\sqrt{43 + 6} - 1}{3} = \frac{\sqrt{49} - 1}{3} = \frac{7 - 1}{3} = \frac{6}{3} = 2$

And f ^{$-$ 1}(163) = $\frac{\sqrt{163 + 6} - 1}{3} = \frac{\sqrt{169} - 1}{3} = \frac{13 - 1}{3} = \frac{12}{3} = 4$

Page No 2.69:

Question 16:

Let f : R $- \{- \frac{4}{3}\} \to$ R be a function defined as f $(x)$ $= \frac{4 x}{3 x + 4}$ . Show that
f : R $- \{- \frac{4}{3}\} \to$ Rang (f) is one-one and onto. Hence, find f ^{$-$ 1}.

Answer:

The function $f : R - \{- \frac{4}{3}\} \to R - \{\frac{4}{3}\} is given by f (x) = \frac{4 x}{3 x + 4}$ .
Injectivity: Let $x, y \in R - \{- \frac{4}{3}\}$ be such that
$f (x) = f (y) \Rightarrow \frac{4 x}{3 x + 4} = \frac{4 y}{3 y + 4} \Rightarrow 4 x (3 y + 4) = 4 y (3 x + 4) \Rightarrow 12 x y + 16 x = 12 x y + 16 y \Rightarrow 16 x = 16 y \Rightarrow x = y$
Hence, f is one-one function.
Surjectivity: Let y be an arbitrary element of $R - \{\frac{4}{3}\}$ . Then,
f(x) = y
$\Rightarrow \frac{4 x}{3 x + 4} = y \Rightarrow 4 x = 3 x y + 4 y \Rightarrow 4 x - 3 x y = 4 y \Rightarrow x = \frac{4 y}{4 - 3 y}$
As $y \in R - \{\frac{4}{3}\}, \frac{4 y}{4 - 3 y} \in R$ .
Also, $\frac{4 y}{4 - 3 y} \neq - \frac{4}{3}$ because $\frac{4 y}{4 - 3 y} = - \frac{4}{3} \Rightarrow 12 y = - 16 + 12 y \Rightarrow 0 = - 16$ , which is not possible.
Thus,
$x = \frac{4 y}{4 - 3 y} \in R - \{- \frac{4}{3}\}$ such that
$f (x) = f (\frac{4 x}{3 x + 4}) = \frac{4 (\frac{4 y}{4 - 3 y})}{3 (\frac{4 y}{4 - 3 y}) + 4} = \frac{16 y}{12 y + 16 - 12 y} = \frac{16 y}{16} = y$ , so every element in $R - \{\frac{4}{3}\}$ has pre-image in $R - \{- \frac{4}{3}\}$ .
Hence, f is onto.
Now,
$x = \frac{4 y}{4 - 3 y}$
Replacing x by $f^{- 1} (x)$ and y by x, we have
$f^{- 1} (x) = \frac{4 x}{4 - 3 x}$

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Question 17:

Let A = R – {2} and B = R – {1}. If f : A → B is a function defined by $f (x) = \frac{x - 1}{x - 2},$ show that f is one-one and onto. Find f^–¹.

Answer:

Given: $f (x) = \frac{x - 1}{x - 2}$

To show f is one-one:

$Let f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - 1}{x_{1} - 2} = \frac{x_{2} - 1}{x_{2} - 2} \Rightarrow (x_{1} - 1) (x_{2} - 2) = (x_{2} - 1) (x_{1} - 2) \Rightarrow x_{1} x_{2} - 2 x_{1} - x_{2} + 2 = x_{1} x_{2} - 2 x_{2} - x_{1} + 2 \Rightarrow - 2 x_{1} - x_{2} = - 2 x_{2} - x_{1} \Rightarrow - 2 x_{1} + x_{1} = - 2 x_{2} + x_{2} \Rightarrow - x_{1} = - x_{2} \Rightarrow x_{1} = x_{2} Hence, f is one - one . To show f is onto : Let y \in B ∴ y = f (x) \Rightarrow y = \frac{x - 1}{x - 2} \Rightarrow y (x - 2) = x - 1 \Rightarrow x y - 2 y = x - 1 \Rightarrow x y - x = 2 y - 1 \Rightarrow x (y - 1) = 2 y - 1 \Rightarrow x = \frac{2 y - 1}{y - 1} Thus, for every value of y in R - \{1\}, there exists a pre - image x = \frac{2 y - 1}{y - 1} in R - \{2\} . Hence, f is onto .$

Since, f is one-one and onto
Therefore, f is invertible with $f^{- 1} (y) = \frac{2 y - 1}{y - 1}$ .

$Hence, f^{- 1} (x) = \frac{2 x - 1}{x - 1} .$

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Question 18:

Show that the function f : N → N defined by f(x) = x² + x + 1 is one-one but not onto. Find the inverse of f : N → S, where S is range of f.

Answer:

Given: The function f : N → N defined by f(x) = x² + x + 1

To show f is one-one:

$Let f (x_{1}) = f (x_{2}) \Rightarrow x_{1}^{2} + x_{1} + 1 = x_{2}^{2} + x_{2} + 1 \Rightarrow x_{1}^{2} + x_{1} = x_{2}^{2} + x_{2} \Rightarrow x_{1}^{2} + x_{1} - x_{2}^{2} - x_{2} = 0 \Rightarrow x_{1}^{2} - x_{2}^{2} + x_{1} - x_{2} = 0 \Rightarrow (x_{1} - x_{2}) (x_{1} + x_{2}) + (x_{1} - x_{2}) = 0 \Rightarrow (x_{1} - x_{2}) (x_{1} + x_{2} + 1) = 0 \Rightarrow x_{1} - x_{2} = 0 or x_{1} + x_{2} + 1 = 0 \Rightarrow x_{1} = x_{2} or x_{1} = - (x_{2} + 1) \Rightarrow x_{1} = x_{2} (∵ x_{1}, x_{2} \in N) Hence, f is one - one . To show f is not onto : Since f (x) = x^{2} + x + 1 ∴ f (1) = 3 f (2) = 7 f (3) = 13 and so on Thus, Range of f = \{3, 7, 13, . . .\} \neq N Hence, f is not onto . Now, Let f : N \to Range of f y = x^{2} + x + 1 \Rightarrow x^{2} + x + 1 - y = 0 \Rightarrow x^{2} + x + (1 - y) = 0 \Rightarrow x = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (1 - y)}}{2 (1)} \Rightarrow x = \frac{- 1 \pm \sqrt{1 - 4 + 4 y}}{2} \Rightarrow x = \frac{- 1 \pm \sqrt{4 y - 3}}{2} \Rightarrow x = \frac{- 1 + \sqrt{4 y - 3}}{2} or x = \frac{- 1 - \sqrt{4 y - 3}}{2} \Rightarrow x = \frac{- 1 + \sqrt{4 y - 3}}{2} (∵ x \in N) Hence, f^{- 1} (x) = \frac{- 1 + \sqrt{4 x - 3}}{2} .$

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Question 19:

Let f : [−1, ∞) → [−1, ∞) be given by f(x) = (x + 1)² − 1, x ≥ −1. Show that f is invertible. Also, find the set S = {x : f(x) = f⁻¹ (x)}.

Answer:

$Injectivity : Let x and y \in [- 1, \infty), such that f (x) = f (y) \Rightarrow {(x + 1)}^{2} - 1 = {(y + 1)}^{2} - 1 \Rightarrow {(x + 1)}^{2} = {(y + 1)}^{2} \Rightarrow (x + 1) = (y + 1) \Rightarrow x = y So, f is a injection . Surjectivity : Let y \in [- 1, \infty) . Then, f (x) = y \Rightarrow {(x + 1)}^{2} - 1 = y \Rightarrow x + 1 = \sqrt{y + 1} \Rightarrow x = \sqrt{y + 1} - 1 Clearly, x = \sqrt{y + 1} - 1 is real for all y \geq - 1 . Thus, every element y \in [- 1, \infty) has its pre - image x \in [- 1, \infty) given by x = \sqrt{y + 1} - 1 . \Rightarrow f is a surjection . So, f is a bijection . Hence, f is invertible . Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow {(y + 1)}^{2} - 1 = x \Rightarrow {(y + 1)}^{2} = x + 1 \Rightarrow y + 1 = \sqrt{x + 1} \Rightarrow y = \pm \sqrt{x + 1} - 1 \Rightarrow f^{- 1} (x) = \pm \sqrt{x + 1} - 1 [from (1)] f (x) = f^{- 1} (x) \Rightarrow {(x + 1)}^{2} - 1 = \pm \sqrt{x + 1} - 1 \Rightarrow {(x + 1)}^{2} = \pm \sqrt{x + 1} \Rightarrow {(x + 1)}^{4} = x + 1 \Rightarrow (x + 1) [{(x + 1)}^{3} - 1] = 0 \Rightarrow x + 1 = 0 or {(x + 1)}^{3} - = 0 \Rightarrow x = - 1 or {(x + 1)}^{3} =1 \Rightarrow x = - 1 or x +1=1 \Rightarrow x = - 1 or x=0 \Rightarrow S = \{0, - 1\}$

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Question 20:

Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by f(x) = x² and g(x) = sin (π x/2). Show that g⁻¹ exists but f⁻¹ does not exist. Also, find g−1.

Answer:

f is not one-one because
$f (- 1) = {(- 1)}^{2} = 1 and f (1) = 1^{2} = 1$
$\Rightarrow$ -1 and 1 have the same image under f.
$\Rightarrow$ f is not a bijection.
So, f ^-1does not exist.

Injectivity of g:
Let x and y be any two elements in the domain (A), such that
$g (x) = g (y) \Rightarrow \sin (\frac{πx}{2}) = \sin (\frac{πy}{2}) \Rightarrow (\frac{πx}{2}) = (\frac{πy}{2}) \Rightarrow x = y$
So, g is one-one.

Surjectivity of g:
$Range of g = [\sin (\frac{π (- 1)}{2}), \sin (\frac{π (1)}{2})] = [\sin (\frac{- π}{2}), \sin (\frac{π}{2})] = [- 1, 1] = A (co-domain of g)$
$\Rightarrow$ g is onto.
$\Rightarrow$ g is a bijection.
So, g^-1 exists.

Also,
$let g^{- 1} (x) = y . . . (1) \Rightarrow g (y) = x \Rightarrow \sin (\frac{πy}{2}) = x \Rightarrow (\frac{πy}{2}) = \sin^{- 1} x \Rightarrow y = \frac{2}{π} \sin^{- 1} x \Rightarrow g^{- 1} (x) = \frac{2}{π} \sin^{- 1} x [from (1)]$

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Question 21:

Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.

Answer:

Injectivity:
Let x and y be two elements in the domain (R), such that
$f (x) = f (y) \Rightarrow \cos (x + 2) = \cos (y + 2) \Rightarrow x + 2 = y + 2 or x + 2 = 2 π - (y + 2) \Rightarrow x = y or x +2=2π- y -2 \Rightarrow x = y or x =2π- y -4 So, we cannot say that x = y For example, \cos \frac{π}{2} = \cos \frac{3 π}{2} = 0 So, \frac{π}{2} and \frac{3 π}{2} have the same image 0.$
$\Rightarrow$ f is not one-one.
$\Rightarrow$ f is not a bijection.
Thus, f is not invertible.

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Question 22:

If A = {1, 2, 3, 4} and B = {a, b, c, d}, define any four bijections from A to B. Also give their inverse functions.

Answer:

$f_{1} = \{(1, a), (2, b), (3, c), (4, d)\} \Rightarrow {f_{1}}^{- 1} = \{(a, 1), (b, 2), (c, 3), (d, 4)\} f_{2} = \{(1, b), (2, a), (3, c), (4, d)\} \Rightarrow {f_{2}}^{- 1} = \{(b, 1), (a, 2), (c, 3), (d, 4)\} f_{3} = \{(1, a), (2, b), (4, c), (3, d)\} \Rightarrow {f_{3}}^{- 1} = \{(a, 1), (b, 2), (c, 4), (d, 3)\} f_{4} = \{(1, b), (2, a), (4, c), (3, d)\} \Rightarrow {f_{4}}^{- 1} = \{(b, 1), (a, 2), (c, 4), (d, 3)\}$
Clearly, all these are bijections because they are one-one and onto.

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Question 23:

Let A and B be two sets, each with a finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.

Answer:

$A and B are two non empty sets . Let f be a function from A to B . It is given that there is injective map from A to B . That means f is one - one function . It is also given that there is injective map from B to A . That means every element of set B has its image in set A . \Rightarrow f is onto function or surjective . ∴ f is bijective . (If a function is both injective and surjective, then the function is bijective .)$

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Question 24:

If f : A → A, g : A → A are two bijections, then prove that
(i) fog is an injection
(ii) fog is a surjection

Answer:

Given: A → A, g : A → A are two bijections.
Then, fog : A → A

(i) Injectivity of fog:
Let x and y be two elements of the domain (A), such that
$(f o g) (x) = (f o g) (y) \Rightarrow f (g (x)) = f (g (y)) \Rightarrow g (x) = g (y) (As, f is one-one) \Rightarrow x = y (As, g is one-one)$
So, fog is an injection.

(ii) Surjectivity of fog:
Let z be an element in the co-domain of fog (A).
$Now, z∈A (co-domain of f) and f is a surjection. So, z = f (y), where y∈A (domain of f) . . . (1) Now, y∈A (co-domain of g) and g is a surjection. So, y = g (x), where x∈A (domain of g) . . . (2) From (1) and (2), z = f (y) = f (g (x)) = (f o g) (x), where x \in A (domain of f o g)$
So, fog is a surjection.

Page No 2.72:

Question 1:

Let $A = \{x \in R : - 1 \leq x \leq 1\} = B and C = \{x \in R : x \geq 0\} and$
let $S = \{(x, y) \in A \times B : x^{2} + y^{2} = 1\} and S_{0} = \{(x, y) \in A \times C : x^{2} + y^{2} = 1\}$ . Then,
(a) S defines a function from A to B
(b) S₀ defines a function from A to C
(c) S₀ defines a function from A to B
(d) S defines a function from A to C

Answer:

(a) S defines a function from A to B

$Let x \in A \Rightarrow - 1 \leq x \leq 1 Now, x^{2} + y^{2} = 1 \Rightarrow y^{2} = 1 - x^{2} \Rightarrow y = \pm \sqrt{1 - x^{2}} \Rightarrow - 1 \leq y \leq 1 ∴ y \in B Thus, S defines a function from A to B .$

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Question 2:

$f : R \to R given by f (x) = x + \sqrt{x^{2}} is$
(a) injective
(b) surjective
(c) bijective
(d) None of these

Answer:

$f (x) = x + \sqrt{x^{2}} = x \pm x = 0 or 2 x ⇒ Each element of the domain has 2 images.$
$\Rightarrow$ f is not a function.
So, the answer is (d).

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Question 3:

If $f : A \to B given by 3^{f (x)} + 2^{- x} = 4$ is a bijection, then
(a) $A = \{x \in R : - 1 < x < \infty\}, B = \{x \in R : 2 < x < 4\}$
(b) $A = \{x \in R : - 3 < x < \infty\}, B = \{x \in R : 2 < x < 4\}$
(c) $A = \{x \in R : - 2 < x < \infty\}, B = \{x \in R : 2 < x < 4\}$
(d) None of these

Answer:

(d) None of these

$f : A \to B 3^{f (x)} + 2^{- x} = 4 \Rightarrow 3^{f (x)} = 4 - 2^{- x} Taking \log on both the sides, f (x) \log 3 = \log (4 - 2^{- x}) \Rightarrow f (x) = \frac{\log (4 - 2^{- x})}{\log 3} Logaritmic function will only be defined if 4 - 2^{- x} > 0 \Rightarrow 4 > 2^{- x} \Rightarrow 2^{2} > 2^{- x} \Rightarrow 2 > - x \Rightarrow - 2 < x \Rightarrow x \in (- 2, \infty) That means A = \{x \in R : - 2 < x < \infty\} As we know that, f (x) = \frac{\log (4 - 2^{- x})}{\log 3} We take x = 0 \in (- 2, \infty) \Rightarrow f (x) = 1 which does not belong to any of the options .$

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Question 4:

The function f : R → R defined by $f (x) = 2^{x} + 2^{|x|} is$
(a) one-one and onto
(b) many-one and onto
(c) one-one and into
(d) many-one and into

Answer:

(d) many-one and into

Graph for the given function is as follows.

A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y.
That means it is many one function.

From the given graph we can see that the range is $[2, \infty)$
and R is the co-domain of the given function.
Hence, Co-domain $\neq$ Range
Therefore, the given function is into.

Page No 2.72:

Question 5:

Let the function $f : R - \{- b\} \to R - \{1\}$ be defined by

$f (x) = \frac{x + a}{x + b}, a \neq b . T hen,$

(a) f is one-one but not onto
(b) f is onto but not one-one
(c) f is both one-one and onto
(d) None of these

Answer:

(c) f is both one-one and onto

Injectivity:
Let x and y be two elements in the domain R- {-b}, such that
$f (x) = f (y) \Rightarrow \frac{x + a}{x + b} = \frac{y + a}{y + b} \Rightarrow (x + a) (y + b) = (x + b) (y + a) \Rightarrow x y + b x + a y + a b = x y + a x + b y + a b \Rightarrow b x + a y = a x + b y \Rightarrow (a - b) x = (a - b) y \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain of f, i.e. R-{1}, such that f (x)=y

$f (x) = y \Rightarrow \frac{x + a}{x + b} = y \Rightarrow x + a = y x + y b \Rightarrow x - y x = y b - a \Rightarrow x (1 - y) = y b - a \Rightarrow x = \frac{y b - a}{1 - y} \in R - \{- b\}$
So, f is onto.

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Question 6:

The function $f : A \to B$ defined by $f (x) = - x^{2} + 6 x - 8$ is a bijection if
(a) $A = (- \infty, 3] and B = (- \infty, 1]$
(b) $A = [- 3, \infty) and B = (- \infty, 1]$
(c) $A = (- \infty, 3] and B = [1, \infty)$
(d) $A = [3, \infty) and B = [1, \infty)$

Answer:

(a) $A = (- \infty, 3] and B = (- \infty, 1]$

$f (x) = - x^{2} + 6 x - 8, is a polynomial function And the domain of polynomial function is real number . ∴ x \in R$

$f (x) = - x^{2} + 6 x - 8 = - (x^{2} - 6 x + 8) = - (x^{2} - 6 x + 9 - 1) = - {(x - 3)}^{2} + 1 Maximum value of - {(x - 3)}^{2} woud be 0 ∴ Maximum value of - {(x - 3)}^{2} + 1 woud be 1 ∴ f (x) \in (- \infty, 1]$

$We can see from the given graph that function is symmetrical about x = 3 & the given function is bijective . So, x would be either (- \infty, 3] or [3, \infty) The correct option which satisfy A and B both is :$
$A = (- \infty, 3] and B = (- \infty, 1]$

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Question 7:

Let $A = \{x \in R : - 1 \leq x \leq 1\} = B$ . Then, the mapping $f : A \to B given by f (x) = x |x|$ is
(a) injective but not surjective
(b) surjective but not injective
(c) bijective
(d) none of these

Answer:

Injectivity:
Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that
$f (x) = f (y) \Rightarrow x |x| = y |y| \Rightarrow x (x) = y (y) \Rightarrow x^{2} = y^{2} \Rightarrow x = y$

Case-2: Let x and y be two negative numbers, such that
$f (x) = f (y) \Rightarrow x |x| = y |y| \Rightarrow x (- x) = y (- y) \Rightarrow - x^{2} = - y^{2} \Rightarrow x^{2} = y^{2} \Rightarrow x = y$

Case-3: Let x be positive and y be negative.
$Then, x \neq y \Rightarrow f (x) = x |x| is positive and f (y) = y |y| is negative \Rightarrow f (x) \neq f (y) So, x \neq y \Rightarrow f (x) \neq f (y)$
From the 3 cases, we can conclude that f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
$Case-1: Let y >0. Then, 0< y ≤1 \Rightarrow y = f (x) = x |x| > 0 \Rightarrow x > 0 \Rightarrow |x| = x f (x) = y \Rightarrow x |x| = y \Rightarrow x (x) = y \Rightarrow x^{2} = y \Rightarrow x = \sqrt{y} \in A (We do not get ± because x >0) Case-2: Let y <0. Then, -1≤ y <0 \Rightarrow y = f (x) = x |x| < 0 \Rightarrow x < 0 \Rightarrow |x| = - x f (x) = y \Rightarrow x |x| = y \Rightarrow x (- x) = y \Rightarrow - x^{2} = y \Rightarrow x^{2} = - y \Rightarrow x = - \sqrt{- y} \in A (We do not get ± because x >0)$
$\Rightarrow$ f is onto.
$\Rightarrow$ f is a bijection.
So, the answer is (c).

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Question 8:

Let $f : R \to R$ be given by $f (x) = [x^{2}] + [x + 1] - 3$ where [x] denotes the greatest integer less than or equal to x. Then, f(x) is
(a) many-one and onto
(b) many-one and into
(c) one-one and into
(d) one-one and onto

Answer:

(b) many-one and into

$f : R \to R$

$f (x) = [x^{2}] + [x + 1] - 3$

It is many one function because in this case for two different values of x
we would get the same value of f(x) .
$For x = 1.1, 1.2 \in R f (1.1) = [{(1.1)}^{2}] + [1.1 + 1] - 3 = [1.21] + [2.1] - 3 = 1 + 2 - 3 = 0 f (1.1) = [{(1.2)}^{2}] + [1.2 + 1] - 3 = [1.44] + [2.2] - 3 = 1 + 2 - 3 = 0$

It is into function because for the given domain we would only get the integral values of
f(x).
but R is the codomain of the given function.
That means , Codomain $\neq$ Range
Hence, the given function is into function.

Therefore, f(x) is many one and into

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Question 9:

Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M → R defined by f(A) = |A| for every A ∈ M, is
(a) one-one and onto
(b) neither one-one nor onto
(c) one-one but-not onto
(d) onto but not one-one

Answer:

$M = \{A = [\begin{matrix} a & b \\ c & d \end{matrix}] : a, b, c, d \in R\} f : M \to R is given by f (A) = |A|$

Injectivity:
$f ([\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]) = |\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}| = 0 and f ([\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]) = |\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}| = 0 \Rightarrow f ([\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]) = f ([\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]) = 0$
So, f is not one-one.

Surjectivity:
Let y be an element of the co-domain, such that
$f (A) = - y, A = [\begin{matrix} a & b \\ c & d \end{matrix}] \Rightarrow |\begin{matrix} a & b \\ c & d \end{matrix}| = y \Rightarrow a d - b c = y \Rightarrow a, b, c, d \in R \Rightarrow A = [\begin{matrix} a & b \\ c & d \end{matrix}] \in M$
$\Rightarrow$ f is onto.
So, the answer is (d).

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Question 10:

The function $f : [0, \infty) \to R given by f (x) = \frac{x}{x + 1} is$
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) onto but not one-one

Answer:

Injectivity:
Let x and y be two elements in the domain, such that
$f (x) = f (y) \Rightarrow \frac{x}{x + 1} = \frac{y}{y + 1} \Rightarrow x y + x = x y + y \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that
$y = f (x) \Rightarrow y = \frac{x}{x + 1} \Rightarrow x y + y = x \Rightarrow x (y - 1) = - y \Rightarrow x = \frac{- y}{y - 1} Range of f = R - \{1\} \neq co domain (R)$
$\Rightarrow$ f is not onto.
So, the answer is (b).

Page No 2.73:

Question 11:

The range of the function $f (x) =^{7 - x} P_{x - 3}$ is
(a) {1, 2, 3, 4, 5}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3, 4}
(d) {1, 2, 3}

Answer:

We know that
$7 - x > 0; x - 3 \geq 0 and 7 - x \geq x - 3 \Rightarrow x < 7; x \geq 3 and 2 x \leq 10 \Rightarrow x < 7; x \geq 3 and x \leq 5 So, x = \{3, 4, 5\} Range of f = \{{}^{(7 - 3)}P_{(3 - 3)}, {}^{(7 - 4)}P_{(4 - 3)}, {}^{(7 - 5)}P (_{5 - 3})\} = \{4 P_{0}, 3 P_{1}, 2 P_{2}\} = \{1, 3, 2\} = \{1, 2, 3\}$
So, the answer is (d).

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Question 12:

A function f from the set of natural numbers to integers defined by

$f (n) = \{\begin{cases} \frac{n - 1}{2}, when n is odd \\ - \frac{n}{2}, when n is even \end{cases} is$

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto both

Answer:

(d) one-one and onto both

Injectivity:

Let x and y be any two elements in the domain (N).
$Case-1: Both x and y are even. Let f (x) = f (y) \Rightarrow \frac{- x}{2} = \frac{- y}{2} \Rightarrow - x = - y \Rightarrow x = y Case-2: B oth x and y are odd. Let f (x) = f (y) \Rightarrow \frac{x - 1}{2} = \frac{y - 1}{2} \Rightarrow x - 1 = y - 1 \Rightarrow x = y Case- 3 : Let x be even and y be odd. Then, f (x) = \frac{- x}{2} and f (y) = \frac{y - 1}{2} Then, clearly x \neq y \Rightarrow f (x) \neq f (y) From all the cases, f is one-one.$

Surjectivity:

$Co-domain of f = Z = \{. . ., - 3, - 2, - 1, 0, 1, 2, 3, . . . .\} Range of f = \{. . ., \frac{- 3 - 1}{2}, \frac{- (- 2)}{2}, \frac{- 1 - 1}{2}, \frac{0}{2}, \frac{1 - 1}{2}, \frac{- 2}{2}, \frac{3 - 1}{2}, . . .\} \Rightarrow Range of f = \{. . ., - 2, 1, - 1, 0, 0, - 1, 1, . . .\} \Rightarrow Range of f = \{. . ., - 2, - 1, 0, 1, 2, . . . .\} \Rightarrow Co-domain of f = Range of f$
$\Rightarrow$ f is onto.

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Question 13:

Let f be an injective map with domain {x, y, z} and range {1, 2, 3}, such that exactly one of the following statements is correct and the remaining are false.

$f (x) = 1, f (y) \neq 1, f (z) \neq 2 .$

The value of $f^{- 1} (1)$ is
(a) x
(b) y
(c) z
(d) none of these

Answer:

$Case-1: Let f (x) = 1 be true. Then, f (y) ≠1 and f (z) \neq 2 are false. So, f (y) = 1 and f (z) = 2 ⇒ f (x) = 1, f (y) = 1 \Rightarrow x and y have the same images. This contradicts the fact that f is one-one. Case-2: Let f (y) \neq1 be true. Then, f (x) = 1 and f (z) \neq 2 are false. So, f (x) \neq1 and f (z) = 2 ⇒ f (x) \neq 1, f (y) \neq 1 and f (z) = 2 ⇒There is no pre-image for 1. This contradicts the fact that range is \{1, 2, 3\} . Case-3: Let f (z) \neq 2 be true. Then, f (x) = 1 and f (y) \neq 1 are false. So, f (x) \neq1 and f (y) = 1 ⇒ f (x) = 2, f (y) = 1 and f (z) = 3 \Rightarrow f (y) = 1 \Rightarrow f^{- 1} (1) = y$
So, the answer is (b).

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Question 14:

Which of the following functions form Z to itself are bijections?
(a) $f (x) = x^{3}$
(b) $f (x) = x + 2$
(c) $f (x) = 2 x + 1$
(d) $f (x) = x^{2} + x$

Answer:

$(a) f is not onto because for y = 3∈Co-domain(Z), there is no value of x∈Domain(Z) x^{3} = 3 \Rightarrow x = \sqrt[3]{3} \notin Z ⇒ f is not onto. So, f is not a bijection.$

(b) Injectivity:
Let x and y be two elements of the domain (Z), such that
$x + 2 = y + 2 \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that
$y = f (x) \Rightarrow y = x + 2 \Rightarrow x = y - 2 \in Z (Domain)$
$\Rightarrow$ f is onto.
So, f is a bijection.

$(c) f (x) = 2 x + 1 is not onto because if we take 4 ∈ Z (co domain), then 4 = f (x) \Rightarrow 4 = 2 x + 1 \Rightarrow 2 x = 3 \Rightarrow x = \frac{3}{2} \notin Z So, f is not a bijection. (d) f (0) = 0^{2} + 0 = 0 and f (- 1) = {(- 1)}^{2} + (- 1) = 1 - 1 = 0 ⇒0 and -1 have the same image. ⇒ f is not one-one. So, f is not a bijection.$

So, the answer is (b).

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Question 15:

Which of the following functions from $A = \{x : - 1 \leq x \leq 1\}$ to itself are bijections?
(a) $f (x) = \frac{x}{2}$
(b) $g (x) = \sin (\frac{π x}{2})$
(c) $h (x) = | x |$
(d) $k (x) = x^{2}$

Answer:

$(a) Range of f = [\frac{- 1}{2}, \frac{1}{2}] ≠ A So, f is not a bijection. (b) Range = [\sin (\frac{- π}{2}), \sin (\frac{π}{2})] = [- 1, 1] = A So, g is a bijection. (c) h (- 1) = |- 1| = 1 and h (1) = |1| = 1 ⇒-1 and 1 have the same images So, h is not a bijection. (d) k (- 1) = {(- 1)}^{2} = 1 and k (1) = {(1)}^{2} = 1 ⇒-1 and 1 have the same images So, k is not a bijection.$
So, the answer is (b).

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Question 16:

Let $A = \{x : - 1 \leq x \leq 1\} a n d f : A \to A such that f (x) = x | x |$ , then f is
(a) a bijection
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective

Answer:

Injectivity:
Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that
$f (x) = f (y) \Rightarrow x |x| = y |y| \Rightarrow x (x) = y (y) \Rightarrow x^{2} = y^{2} \Rightarrow x = y$

Case-2: Let x and y be two negative numbers, such that
$f (x) = f (y) \Rightarrow x |x| = y |y| \Rightarrow x (- x) = y (- y) \Rightarrow - x^{2} = - y^{2} \Rightarrow x^{2} = y^{2} \Rightarrow x = y$

Case-3: Let x be positive and y be negative.
$Then, x \neq y \Rightarrow f (x) = x |x| is positive and f (y) = y |y| is negative \Rightarrow f (x) \neq f (y) So, x \neq y \Rightarrow f (x) \neq f (y)$
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
$Case-1: Let y >0. Then, 0< y ≤1 y = f (x) = x |x| > 0 \Rightarrow x > 0 \Rightarrow |x| = x \Rightarrow f (x) = y \Rightarrow x |x| = y \Rightarrow x (x) = y \Rightarrow x^{2} = y \Rightarrow x = \sqrt{y} \in A (We do not get ±, as x >0) Case-2: Let y <0 . Then, -1≤ y <0 y = f (x) = x |x| < 0 \Rightarrow x < 0 \Rightarrow |x| = - x \Rightarrow f (x) = y \Rightarrow x |x| = y \Rightarrow x (- x) = y \Rightarrow - x^{2} = y \Rightarrow x^{2} = - y \Rightarrow x = - \sqrt{- y} \in A (We do not get ±, as x>0)$
$\Rightarrow$ f is onto
$\Rightarrow$ f is a bijection.
So, the answer is (a).

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Question 17:

If the function $f : R \to A given by f (x) = \frac{x^{2}}{x^{2} + 1}$ is a surjection, then A =
(a) R
(b) [0, 1]
(c) [0, 1)
(d) [0, 1)

Answer:

$As f is surjective, r ange of f = co - domain of f \Rightarrow A = range of f ∵ f (x) = \frac{x^{2}}{x^{2} + 1}, y = \frac{x^{2}}{x^{2} + 1} \Rightarrow y (x^{2} + 1) = x^{2} \Rightarrow (y - 1) x^{2} + y = 0 \Rightarrow x^{2} = \frac{- y}{(y - 1)} \Rightarrow x = \sqrt{\frac{y}{(1 - y)}} \Rightarrow \frac{y}{(1 - y)} \geq 0 \Rightarrow y \in [0, 1) \Rightarrow Range of f = [0, 1) \Rightarrow A = [0, 1)$
So, the answer is (d).

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Question 18:

If a function $f : [2, \infty) \to B defined by f (x) = x^{2} - 4 x + 5$ is a bijection, then B =
(a) R
(b) [1, ∞)
(c) [4, ∞)
(d) [5, ∞)

Answer:

Since f is a bijection, co-domain of f = range of f
$\Rightarrow$ B = range of f
$Given : f (x) = x^{2} - 4 x + 5 Let f (x) = y \Rightarrow y = x^{2} - 4 x + 5 \Rightarrow x^{2} - 4 x + (5 - y) = 0 ∵ Discrimant, D = b^{2} - 4 a c \geq 0, {(- 4)}^{2} - 4 \times 1 \times (5 - y) \geq 0 \Rightarrow 16 - 20 + 4 y \geq 0 \Rightarrow 4 y \geq 4 \Rightarrow y \geq 1 \Rightarrow y \in [1, \infty) \Rightarrow Range of f = [1, \infty) \Rightarrow B = [1, \infty)$
So, the answer is (b).

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Question 19:

The function $f : R \to R$ defined by
$f (x) = (x - 1) (x - 2) (x - 3)$ is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto

Answer:

$f (x) = (x - 1) (x - 2) (x - 3)$

Injectivity:
$f (1) = (1 - 1) (1 - 2) (1 - 3) = 0 f (2) = (2 - 1) (2 - 2) (2 - 3) = 0 f (3) = (3 - 1) (3 - 2) (3 - 3) = 0 ⇒ f (1) = f (2) = f (3) = 0$
So, f is not one-one.

Surjectivity:
Let y be an element in the co domain R, such that
$y = f (x) \Rightarrow y = (x - 1) (x - 2) (x - 3) Since y \in R and x \in R, f is onto.$

So, the answer is (b).

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Question 20:

The function $f : [- 1 / 2, 1 / 2, 1 / 2] \to [- π / 2, π / 2]$ , defined by $f (x) = \sin^{- 1} (3 x - 4 x^{3})$ , is
(a) bijection
(b) injection but not a surjection
(c) surjection but not an injection
(d) neither an injection nor a surjection

Answer:

$f (x) = \sin^{- 1} (3 x - 4 x^{3}) \Rightarrow f (x) = 3 \sin^{- 1} x$

Injectivity:
Let x and y be two elements in the domain $[\frac{- 1}{2}, \frac{1}{2}]$ , such that
$f (x) = f (y) \Rightarrow 3 \sin^{- 1} x = 3 \sin^{- 1} y \Rightarrow \sin^{- 1} x = \sin^{- 1} y \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain, such that
$f (x) = y \Rightarrow 3 \sin^{- 1} (x) = y \Rightarrow \sin^{- 1} (x) = \frac{y}{3} \Rightarrow x = \sin \frac{y}{3} \in [\frac{- 1}{2}, \frac{1}{2}]$
$\Rightarrow$ f is onto.
$\Rightarrow$ f is a bijection.
So, the answer is (a).

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Question 21:

Let $f : R \to R$ be a function defined by $f (x) = \frac{e^{| x |} - e^{- x}}{e^{x} + e^{- x}} . Then,$
(a) f is a bijection
(b) f is an injection only
(c) f is surjection on only
(d) f is neither an injection nor a surjection

Answer:

(d) f is neither an injection nor a surjection

$f : R \to R$

$f (x) = \frac{e^{| x |} - e^{- x}}{e^{x} + e^{- x}} For x = - 2 and - 3 \in R f (- 2) = \frac{e^{|- 2|} - e^{2}}{e^{- 2} + e^{2}} = \frac{e^{2} - e^{2}}{e^{- 2} + e^{2}} = 0 & f (- 3) = \frac{e^{|- 3|} - e^{3}}{e^{- 3} + e^{3}} = \frac{e^{3} - e^{3}}{e^{- 3} + e^{3}} = 0 Hence, for different values of x we are getting same values of f (x) That means, the given function is many one .$
Therefore, this function is not injective.

$For x < 0 f (x) = 0 For x > 0 f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} = \frac{e^{x} + e^{- x}}{e^{x} + e^{- x}} - \frac{2 e^{- x}}{e^{x} + e^{- x}} = 1 - \frac{2 e^{- x}}{e^{x} + e^{- x}} The value of \frac{2 e^{- x}}{e^{x} + e^{- x}} is always positive . Therefore, the value of f (x) is always less than 1 Numbers more than 1 are not included in the range but they are included in codomain . As the codomain is R . ∴ Codomain \neq Range Hence, the given function is not onto .$
Therefore, this function is not surjective .

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Question 22:

Let $f : R - \{n\} \to R$ be a function defined by

$f (x) = \frac{x - m}{x - n}, where m \neq n .$ Then,

(a) f is one-one onto
(b) f is one-one into
(c) f is many one onto
(d) f is many one into

Answer:

Injectivity:
Let x and y be two elements in the domain R-{n}, such that
$f (x) = f (y) \Rightarrow \frac{x - m}{x - n} = \frac{y - m}{y - n} \Rightarrow (x - m) (y - n) = (x - n) (y - m) \Rightarrow x y - n x - m y + m n = x y - m x - n y + m n \Rightarrow (m - n) x = (m - n) y \Rightarrow x = y$
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that
$f (x) = y \Rightarrow \frac{x - m}{x - n} = y \Rightarrow x - m = x y - n y \Rightarrow n y - m = x y - x \Rightarrow n y - m = x (y - 1) \Rightarrow x = \frac{n y - m}{y - 1}, which is not defined for y =1 So, 1 ∈ R (co domain) has no pre image in R - \{n\}$
$\Rightarrow$ f is not onto.
Thus, the answer is (b).

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Question 23:

Let $f : R \to R$ be a function defined by $f (x) = \frac{x^{2} - 8}{x^{2} + 2}$ . Then, f is
(a) one-one but not onto
(b) one-one and onto
(c) onto but not one-one
(d) neither one-one nor onto

Answer:

Injectivity:
Let x and y be two elements in the domain (R), such that
$f (x) = f (y) \Rightarrow \frac{x^{2} - 8}{x^{2} + 2} = \frac{y^{2} - 8}{y^{2} + 2} \Rightarrow (x^{2} - 8) (y^{2} + 2) = (x^{2} + 2) (y^{2} - 8) \Rightarrow x^{2} y^{2} + 2 x^{2} - 8 y^{2} - 16 = x^{2} y^{2} - 8 x^{2} + 2 y^{2} - 16 \Rightarrow 10 x^{2} = 10 y^{2} \Rightarrow x^{2} = y^{2} \Rightarrow x = \pm y$
So, f is not one-one.

Surjectivity:
$f (- 1) = \frac{{(- 1)}^{2} - 8}{{(- 1)}^{2} + 2} = \frac{1 - 8}{1 + 2} = \frac{- 7}{3} and f (1) = \frac{{(1)}^{2} - 8}{{(1)}^{2} + 2} = \frac{1 - 8}{1 + 2} = \frac{- 7}{3} \Rightarrow f (- 1) = f (1) = \frac{- 7}{3}$
$\Rightarrow$ f is not onto.
The correct answer is (d).

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Question 24:

$f : R \to R$ is defined by $f (x) = \frac{e^{x^{2}} - e^{- x^{2}}}{e^{x^{2} + e^{- x^{2}}}} is$
(a) one-one but not onto
(b) many-one but onto
(c) one-one and onto
(d) neither one-one nor onto

Answer:

(d) neither one-one nor onto

$We have, f (x) = \frac{e^{x^{2}} - e^{- x^{2}}}{e^{x^{2} + e^{- x^{2}}}} Here, - 2, 2 \in R Now, 2 \neq - 2 But, f (2) = f (- 2) Therefore, function is not one - one . And, The minimum value of the function is 0 and maximum value is 1 That is range of the function is [0, 1] but the co - domain of the function is given R . Therefore, function is not onto . ∴ function is neither one - one nor onto .$

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Question 25:

The function $f : R \to R, f (x) = x^{2}$ is
(a) injective but not surjective
(b) surjective but not injective
(c) injective as well as surjective
(d) neither injective nor surjective

Answer:

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y). Then,

$x^{2} = y^{2} \Rightarrow x = \pm y$

So, f is not one-one.

Surjectivity:
$As f (- 1) = {(- 1)}^{2} = 1 and f (1) = 1^{2} = 1, f (- 1) = f (1)$
So, both -1 and 1 have the same images.
$\Rightarrow$ f is not onto.
So, the answer is (d).

x2+x+1=y2+y+1(x2−y2)+(x−y)=0(x+y)(x−y)+(x−y)=0(x−y)(x+y+1)=0x−y=0 (x+y+1) cannot be zero because x and y are natural numbers)x=y

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Question 26:

A function f from the set of natural numbers to the set of integers defined by

$f (n) \{\begin{cases} \frac{n - 1}{2}, & when n is odd \\ - \frac{n}{2}, & when n is even \end{cases}$ is

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto

Answer:

Injectivity:
Let x and y be any two elements in the domain (N).
$Case-1: Both x and y are even. Let f (x) = f (y) \Rightarrow \frac{- x}{2} = \frac{- y}{2} \Rightarrow - x = - y \Rightarrow x = y Case-2: B oth x and y are odd. Let f (x) = f (y) \Rightarrow \frac{x - 1}{2} = \frac{y - 1}{2} \Rightarrow x - 1 = y - 1 \Rightarrow x = y Case- 3 : Let x be even and y be odd. Then, f (x) = \frac{- x}{2} and f (y) = \frac{y - 1}{2} Then, clearly x \neq y \Rightarrow f (x) \neq f (y) From all the cases, f is one-one.$

Surjectivity:
$Co-domain of f = Z = \{. . ., - 3, - 2, - 1, 0, 1, 2, 3, . . . .\} Range of f = \{. . ., \frac{- 3 - 1}{2}, \frac{- (- 2)}{2}, \frac{- 1 - 1}{2}, \frac{0}{2}, \frac{1 - 1}{2}, \frac{- 2}{2}, \frac{3 - 1}{2}, . . .\} \Rightarrow Range of f = \{. . ., - 2, 1, - 1, 0, 0, - 1, 1, . . .\} \Rightarrow Range of f = \{. . ., - 2, - 1, 0, 1, 2, . . . .\} \Rightarrow Co-domain of f = Range of f$
$\Rightarrow$ f is onto.
So, the answer is (d).

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Question 27:

Which of the following functions from $A = \{x \in R : - 1 \leq x \leq 1\}$ to itself are bijections?

(a) $f (x) = | x |$
(b) $f (x) = \sin \frac{π x}{2}$
(c) $f (x) = \sin \frac{π x}{4}$
(d) None of these

Answer:

(b) $f (x) = \sin \frac{π x}{2}$

It is clear that f(x) is one-one.

$Range of f = [\sin \frac{π (- 1)}{2}, \sin \frac{π (1)}{2}] = [\sin \frac{- π}{2}, \sin \frac{π}{2}] = [- 1, 1] = A = Co domain of f$
⇒ f is onto.
So, f is a bijection.

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Question 28:

Let $f : Z \to Z$ be given by $f (x) = \{\begin{cases} \frac{x}{2}, if x is even \\ 0, if x is odd \end{cases}$
Then, f is
(a) onto but not one-one
(b) one-one but not onto
(c) one-one and onto
(d) neither one-one nor onto

Answer:

Injectivity:
Let x and y be two elements in the domain (Z), such that
$f (x) = f (y) Case-1: Let both x and y be even. Then, f (x) = f (y) \Rightarrow \frac{x}{2} = \frac{y}{2} \Rightarrow x = y Case-2: Let both x and y be odd. Then, f (x) = f (y) \Rightarrow 0 = 0 Here, we cannot determine whether x = y .$
So, f is not one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that
$Co-domain of f = Z = {0, \pm 1, \pm 2, \pm 3, \pm 4, . . .} Range of f = \{0, 0, \frac{\pm 2}{2}, 0, \frac{\pm 4}{2}, . . .\} = \{0, \pm 1, \pm 2, . . .\} \Rightarrow Co-domain of f = Range of f$
$\Rightarrow$ f is onto.
So, the answer is (a).

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Question 29:

The function $f : R \to R$ defined by $f (x) = 6^{x} + 6^{| x |}$ is
(a) one-one and onto
(b) many one and onto
(c) one-one and into
(d) many one and into

Answer:

(d) many one and into

Graph of the given function is as follows :

A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y .
That means it is many one function .

From the given graph we can see that the range is $[2, \infty)$
and R is the codomain of the given function .
Hence, Codomain $\neq$ Range
Therefore, the given function is into .

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Question 30:

Let $f (x) = x^{2} and g (x) = 2^{x}$ . Then, the solution set of the equation $f o g (x) = g o f (x)$ is
(a) R
(b) {0}
(c) {0, 2}
(d) none of these

Answer:

$Since (f o g) (x) = (g o f) (x), f (g (x)) = g (f (x)) \Rightarrow f (2^{x}) = g (x^{2}) \Rightarrow {(2^{x})}^{2} = 2^{x^{2}} \Rightarrow 2^{2 x} = 2^{x^{2}} \Rightarrow x^{2} = 2 x \Rightarrow x^{2} - 2 x = 0 \Rightarrow x (x - 2) = 0 \Rightarrow x = 0, 2 \Rightarrow x \in \{0, 2\}$
So, the answer is (c).

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Question 31:

If $f : R \to R is given by f (x) = 3 x - 5, then f^{- 1} (x)$
(a) is given by $\frac{1}{3 x - 5}$
(b) is given by $\frac{x + 5}{3}$
(c) does not exist because f is not one-one
(d) does not exist because f is not onto

Answer:

Clearly, f is a bijection.
So, f^-1 exists.
$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow 3 y - 5 = x \Rightarrow 3 y = x + 5 \Rightarrow y = \frac{x + 5}{3} \Rightarrow f^{- 1} (x) = \frac{x + 5}{3} [from (1)]$
So, the answer is (b).

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Question 32:

If $g (f (x)) = |\sin x| and f (g (x)) = {(\sin \sqrt{x})}^{2}, then$
(a) $f (x) = \sin^{2} x, g (x) = \sqrt{x}$
(b) $f (x) = \sin x, g (x) = | x |$
(c) $f (x) = x^{2}, g (x) = \sin \sqrt{x}$
(d) f and g cannot be determined.

Answer:

If we solve it by the trial-and-error method, we can see that (a) satisfies the given condition.
From (a):
$f (x) = \sin^{2} x and g (x) = \sqrt{x} \Rightarrow f (g (x)) = f (\sqrt{x}) = \sin^{2} \sqrt{x} = {(\sin \sqrt{x})}^{2}$
So, the answer is (a).

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Question 33:

The inverse of the function $f : R \to \{x \in R : x < 1\}$ given by $f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ is
(a) $\frac{1}{2} \log \frac{1 + x}{1 - x}$

(b) $\frac{1}{2} \log \frac{2 + x}{2 - x}$

(c) $\frac{1}{2} \log \frac{1 - x}{1 + x}$

(d) none of these

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow \frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} = x \Rightarrow \frac{e^{- y} (e^{2 y} - 1)}{e^{- y} (e^{2 y} + 1)} = x \Rightarrow (e^{2 y} - 1) = x (e^{2 y} + 1) \Rightarrow e^{2 y} - 1 = x e^{2 y} + x \Rightarrow e^{2 y} (1 - x) = x + 1 \Rightarrow e^{2 y} = \frac{1 + x}{1 - x} \Rightarrow 2 y = \log_{e} (\frac{1 + x}{1 - x}) \Rightarrow y = \frac{1}{2} l {og}_{e} (\frac{1 + x}{1 - x}) \Rightarrow f^{- 1} (x) = \frac{1}{2} l {og}_{e} (\frac{1 + x}{1 - x}) [from (1)]$
So, the answer is (a).

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Question 34:

Let $A = \{x \in R : x \geq 1\}$ . The inverse of the function, $f : A \to A$ given by $f (x) = 2^{x (x - 1)}, is$

(a) ${(\frac{1}{2})}^{x (x - 1)}$

(b) $\frac{1}{2} \{1 + \sqrt{1 + 4 \log_{2} x}\}$

(c) $\frac{1}{2} \{1 - \sqrt{1 + 4 \log_{2} x}\}$

(d) not defined

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow 2^{y (y - 1)} = x \Rightarrow 2^{y^{2} - y} = x \Rightarrow y^{2} - y = \log_{2} x \Rightarrow y^{2} - y + \frac{1}{4} = \log_{2} x + \frac{1}{4} \Rightarrow {(y - \frac{1}{2})}^{2} = \frac{4 \log_{2} x + 1}{4} \Rightarrow y - \frac{1}{2} = \pm \frac{\sqrt{4 \log_{2} x + 1}}{2} \Rightarrow y = \frac{1}{2} \pm \frac{\sqrt{4 \log_{2} x + 1}}{2} \Rightarrow y = \frac{1}{2} + \frac{\sqrt{4 \log_{2} x + 1}}{2} (∵ y ≥1) So, f^{- 1} (x) = \frac{1}{2} (1 + \sqrt{1 + 4 \log_{2} x}) [from (1)]$
So, the answer is (b).

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Question 35:

Let $A = \{x \in R : x \leq 1\} and f : A \to A$ be defined as $f (x) = x (2 - x)$ . Then, $f^{- 1} (x)$ is
(a) $1 + \sqrt{1 - x}$
(b) $1 - \sqrt{1 - x}$
(c) $\sqrt{1 - x}$
(d) $1 \pm \sqrt{1 - x}$

Answer:

$Let y be the element in the codomain R such that f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x and y \leq1 \Rightarrow y (2 - y) = x \Rightarrow 2 y - y^{2} = x \Rightarrow y^{2} - 2 y + x = 0 \Rightarrow y^{2} - 2 y = - x \Rightarrow y^{2} - 2 y + 1 = 1 - x \Rightarrow {(y - 1)}^{2} = 1 - x \Rightarrow y - 1 = \pm \sqrt{1 - x} \Rightarrow y = 1 \pm \sqrt{1 - x} \Rightarrow y = 1 - \sqrt{1 - x} (∵ y ≤1)$
The correct answer is (b).

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Question 36:

Let $f (x) = \frac{1}{1 - x} . Then, \{f o (f o f)\} (x)$
(a) $x for all x \in R$
(b) $x for all x \in R - \{1\}$
(c) $x for all x \in R - \{0, 1\}$
(d) none of these

Answer:

$Domain of f : 1 - x \neq 0 \Rightarrow x \neq 1 Domain of f = R - \{1\} Range of f : y = \frac{1}{1 - x} \Rightarrow 1 - x = \frac{1}{y} \Rightarrow x = 1 - \frac{1}{y} \Rightarrow x = \frac{y - 1}{y} \Rightarrow y \neq 0 Range of f = R - \{0\} So, f : R - \{1\} \to R - \{0\} and f : R - \{1\} \to R - \{0\} Range of f is not a subset of the domain of f . Domain (f o f) = \{x : x \in domain of f and f (x) \in domain of f\} Domain (f o f) = \{x : x \in R - \{1\} and \frac{1}{1 - x} \in R - \{1\}\} Domain (f o f) = \{x : x \neq 1 and \frac{1}{1 - x} \neq 1\} Domain (f o f) = \{x : x \neq 1 and 1 - x \neq 1\} Domain (f o f) = \{x : x \neq 1 and x \neq 0\} Domain (f o f) = R - \{0, 1\} (f o f) (x) = f (f (x)) = f (\frac{1}{1 - x}) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{1 - x}{1 - x - 1} = \frac{1 - x}{- x} = \frac{x - 1}{x} For range of f o f, x \neq 0 Now, f o f : R - \{0, 1\} \to R - \{0\} and f : R - \{1\} \to R - \{0\} Range of f o f is not a subset of domain of f . Domain (f o (f o f)) = \{x : x \in domain of f o f and (f o f) (x) \in domain of f\} Domain (f o (f o f)) = \{x : x \in R - \{0, 1\} and \frac{x - 1}{x} \in R - \{1\}\} Domain (f o (f o f)) = \{x : x \neq 0, 1 and \frac{x - 1}{x} \neq 1\} Domain (f o (f o f)) = \{x : x \neq 0, 1 and x -1 \neq x\} Domain (f o (f o f)) = \{x : x \neq 0, 1 and x \in R\} Domain (f o (f o f)) = R - \{0, 1\} (f o (f o f)) (x) = f ((f o f) (x)) = f (\frac{x - 1}{x}) = \frac{1}{1 - \frac{x - 1}{x}} = \frac{x}{x - x + 1} = x So, (f o (f o f)) (x) = x, where x \neq 0, 1$
So, the answer is (c).

Page No 2.75:

Question 37:

If the function $f : R \to R$ be such that $f (x) = x - [x]$ , where [x] denotes the greatest integer less than or equal to x, then $f^{- 1} (x)$ is

(a) $\frac{1}{x - [x]}$

(b) [x] − x

(c) not defined

(d) none of these

Answer:

f(x) = x - [x]
We know that the range of f is [0, 1).
Co-domain of f = R
As range of f $\neq$ Co-domain of f, f is not onto.
$\Rightarrow$ f is not a bijective function.
So, f^-1 does not exist.
Thus, the answer is (c).

Page No 2.75:

Question 38:

If $F : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + \frac{1}{x}, then f^{- 1} (x)$ equals

(a) $\frac{x + \sqrt{x^{2} - 4}}{2}$

(b) $\frac{x}{1 + x^{2}}$

(c) $\frac{x - \sqrt{x^{2} - 4}}{2}$

(d) $1 + \sqrt{x^{2} - 4}$

Answer:

$Let f^{- 1} (x) = y \Rightarrow f (y) = x \Rightarrow y + \frac{1}{y} = x \Rightarrow y^{2} + 1 = x y \Rightarrow y^{2} - x y + 1 = 0 \Rightarrow y^{2} - 2 \times y \times \frac{x}{2} + {(\frac{x}{2})}^{2} - {(\frac{x}{2})}^{2} + 1 = 0 \Rightarrow y^{2} - 2 \times y \times \frac{x}{2} + {(\frac{x}{2})}^{2} = \frac{x^{2} - 1}{4} \Rightarrow {(y - \frac{x}{2})}^{2} = \frac{x^{2} - 1}{4} \Rightarrow y - \frac{x}{2} = \frac{\sqrt{x^{2} - 4}}{2} \Rightarrow y = \frac{x}{2} + \frac{\sqrt{x^{2} - 4}}{2} \Rightarrow y = \frac{x + \sqrt{x^{2} - 4}}{2} \Rightarrow f^{- 1} (x) = \frac{x + \sqrt{x^{2} - 4}}{2}$

So, the answer is (a).

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Question 39:

Let $g (x) = 1 + x - [x] and f (x) = \{\begin{cases} - 1, & x < 0 \\ 0, & x = 0, \\ 1, & x > 0 \end{cases}$ , where [x] denotes the greatest integer less than or equal to x. Then for all $x, f (g (x))$ is equal to
(a) x
(b) 1
(c) f(x)
(d) g(x)

Answer:

(b) 1

$When, - 1 < x < 0 Then, g (x) = 1 + x - [x] = 1 + x - (- 1) = 2 + x ∴ f (g (x)) = 1 When, x = 0 Then, g (x) = 1 + x - [x] = 1 + x - 0 = 1 + x ∴ f (g (x)) = 1 When, x > 1 Then, g (x) = 1 + x - [x] = 1 + x - 1 = x ∴ f (g (x)) = 1$

Therefore, for each interval f(g(x))=1

Page No 2.75:

Question 40:

Let $f (x) = \frac{α x}{x + 1}, x \neq - 1$ . Then, for what value of α is $f (f (x)) = x ?$
(a) $\sqrt{2}$
(b) $- \sqrt{2}$
(c) 1
(d) −1

Answer:

(d) −1

$f (f (x)) = x \Rightarrow f (\frac{α x}{x + 1}) = x \Rightarrow \frac{α (\frac{α x}{x + 1})}{(\frac{α x}{x + 1}) + 1} = x \Rightarrow \frac{α^{2} x}{α x + x + 1} = x \Rightarrow α^{2} x = α x^{2} + x^{2} + x \Rightarrow α^{2} x - α x^{2} - x^{2} - x = 0 \Rightarrow α^{2} x - α x^{2} - (x^{2} + x) = 0 Solving for the α we get, α = \frac{- (- x^{2}) \pm \sqrt{{(- x^{2})}^{2} - 4 \times x \times [- (x^{2} + x)]}}{2 x} = \frac{x^{2} \pm \sqrt{x^{4} + 4 x^{3} + 4 x^{2}}}{2 x} = x + 1, - 1 Here, - 1 is independent of x, ∴ for, α = - 1, f (f (x)) = x$

Page No 2.76:

Question 41:

The distinct linear functions that map [−1, 1] onto [0, 2] are
(a) $f (x) = x + 1, g (x) = - x + 1$
(b) $f (x) = x - 1, g (x) = x + 1$
(c) $f (x) = - x - 1, g (x) = x - 1$
(d) None of these

Answer:

Let us substitute the end-points of the intervals in the given functions. Here, domain = [-1, 1] and range =[0, 2]
By substituting -1 or 1 in each option, we get:

Option (a):
$f (- 1) = - 1 + 1 = 0 and f (1) = 1 + 1 = 2 g (- 1) = 1 + 1 = 2 and g (1) = - 1 + 1 = 0$
So, option (a) is correct.

Option (b):
$f (- 1) = - 1 - 1 = - 2 and f (1) = 1 - 1 = 0 g (- 1) = - 1 + 1 = 0 and g (1) = 1 + 1 = 2$
Here, f(-1) gives -2 $\notin [0, 2]$
So, (b) is not correct.

Similarly, we can see that (c) is also not correct.

Page No 2.76:

Question 42:

Let $f : [2, \infty) \to X$ be defined by $f (x) = 4 x - x^{2}$ . Then, f is invertible if X =
(a) $[2, \infty)$
(b) $(- \infty, 2]$
(c) $(- \infty, 4]$
(d) $[4, \infty)$

Answer:

Since f is invertible, range of f = co domain of f = X
So, we need to find the range of f to find X.
For finding the range, let
$f (x) = y \Rightarrow 4 x - x^{2} = y \Rightarrow x^{2} - 4 x = - y \Rightarrow x^{2} - 4 x + 4 = 4 - y \Rightarrow {(x - 2)}^{2} = 4 - y \Rightarrow x - 2 = \pm \sqrt{4 - y} \Rightarrow x = 2 \pm \sqrt{4 - y} This is defined only when 4 - y \geq 0 \Rightarrow y \leq 4 X = Range of f = (- \infty, 4]$
So, the answer is (c).

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Question 43:

If $f : R \to (- 1, 1)$ is defined by $f (x) = \frac{- x | x |}{1 + x^{2}}, then f^{- 1} (x)$ equals
(a) $\sqrt{\frac{|x|}{1 - |x|}}$

(b) $S g n (x) \sqrt{\frac{|x|}{1 - |x|}}$

(c) $- \sqrt{\frac{x}{1 - x}}$

(d) None of these

Answer:

(b) $- Sgn (x) \sqrt{\frac{|x|}{1 - |x|}}$

$We have, f (x) = \frac{- x | x |}{1 + x^{2}} x \in (- 1, 1) Case - (I) When, x < 0, Then, |x| = - x And f (x) > 0 Now, f (x) = \frac{- x (- x)}{1 + x^{2}} \Rightarrow y = \frac{x^{2}}{1 + x^{2}} \Rightarrow \frac{y}{1} = \frac{x^{2}}{1 + x^{2}} \Rightarrow \frac{y + 1}{y - 1} = \frac{x^{2} + 1 + x^{2}}{x^{2} - 1 - x^{2}} [Using Componendo and dividendo] \Rightarrow \frac{y + 1}{y - 1} = \frac{2 x^{2} + 1}{- 1} \Rightarrow - \frac{y + 1}{y - 1} = 2 x^{2} + 1 \Rightarrow \frac{2 y}{1 - y} = 2 x^{2} \Rightarrow \frac{y}{1 - y} = x^{2} \Rightarrow x = - \sqrt{\frac{y}{1 - y}} (As x < 0) \Rightarrow x = - \sqrt{\frac{|y|}{1 - |y|}} [As y > 0] To find the inverse interchanging x and y we get, f^{- 1} (x) = - \sqrt{\frac{|x|}{1 - |x|}} . . . (i) Case - (II) When, x > 0, Then, |x| = x And f (x) < 0 Now, f (x) = \frac{- x (x)}{1 + x^{2}} \Rightarrow y = \frac{- x^{2}}{1 + x^{2}} \Rightarrow \frac{y}{1} = \frac{- x^{2}}{1 + x^{2}} \Rightarrow \frac{y + 1}{y - 1} = \frac{- x^{2} + 1 + x^{2}}{- x^{2} - 1 - x^{2}} [Using Componendo and dividendo] \Rightarrow \frac{y + 1}{y - 1} = \frac{1}{- 2 x^{2} - 1} \Rightarrow \frac{1 + y}{1 - y} = \frac{1}{2 x^{2} + 1} \Rightarrow \frac{1 - y}{1 + y} = 2 x^{2} + 1 \Rightarrow \frac{- 2 y}{1 + y} = 2 x^{2} \Rightarrow x^{2} = \frac{- y}{1 + y} \Rightarrow x = \sqrt{\frac{- y}{1 + y}} (As x > 0) \Rightarrow x = \sqrt{\frac{|y|}{1 - |y|}} [As y < 0] To find the inverse interchanging x and y we get, f^{- 1} (x) = \sqrt{\frac{|x|}{1 - |x|}} . . . (ii) Case - (III) When, x = 0, Then, f (x) = 0 Hence, f^{- 1} (x) = 0 . . . (iii) Combinig equation (i), (ii) and (iii) we get, f^{- 1} (x) = - Sgn (x) \sqrt{\frac{|x|}{1 - |x|}}$

Page No 2.76:

Question 44:

Let [x] denote the greatest integer less than or equal to x. If $f (x) = \sin^{- 1} x, g (x) = [x^{2}] and h (x) = 2 x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}$ , then
(a) $f o g o h (x) = \frac{π}{2}$
(b) $f o g o h (x) = π$
(c) $h o f o g = h o g o f$
(d) $h o f o g \neq h o g o f$

Answer:

(c) $h o f o g = h o g o f$

$We have, g (x) = [x^{2}] = 0 (A s \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}} ∴ \frac{1}{4} \leq x^{2} \leq \frac{1}{2}) f o g (x) = f (g (x)) = \sin^{- 1} (0) = 0 h o f o g (x) = h (f (g (x))) = 2 \times 0 = 0$

$And f (x) = \sin^{- 1} x Now, f o r, x \in [\frac{1}{2}, \frac{1}{\sqrt{2}}] f (x) \in [\frac{π}{6}, \frac{π}{4}] f (x) \in [0.52, 0.78] g o f (x) = 0 (As, f (x) \in [0.52, 0.78]) = 0 h o g o f (x) = h (g (f (x))) = 2 \times 0 = 0$

$∴ h o f o g = h o g o f = 0$

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Question 45:

If $g (x) = x^{2} + x - 2 and \frac{1}{2} g o f (x) = 2 x^{2} - 5 x + 2$ , then f(x) is equal to
(a) $2 x - 3$
(b) $2 x + 3$
(c) $2 x^{2} + 3 x + 1$
(d) 2 $x^{2} - 3 x - 1$

Answer:

We will solve this problem by the trial-and-error method.
Let us check option (a) first.
$If f (x) = 2 x - 3 \frac{1}{2} (g o f) (x) = g (f (x)) = \frac{1}{2} g (2 x - 3) = \frac{1}{2} [{(2 x - 3)}^{2} + (2 x - 3) - 2] = \frac{1}{2} [4 x^{2} + 9 - 12 x + 2 x - 3 - 2] = \frac{1}{2} [4 x^{2} - 10 x + 4] = 2 x^{2} - 5 x + 2$
The given condition is satisfied by (a).
So, the answer is (a).

Page No 2.76:

Question 46:

If $f (x) = \sin^{2} x$ and the composite function $g (f (x)) = |\sin x|$ , then g(x) is equal to
(a) $\sqrt{x - 1}$
(b) $\sqrt{x}$
(c) $\sqrt{x + 1}$
(d) $- \sqrt{x}$

Answer:

(b)

$If we take g (x) = \sqrt{x}, then g (f (x)) = g (\sin^{2} x) = \sqrt{\sin^{2} x} = \pm \sin x = |\sin x|$

So, the answer is (b).

Page No 2.76:

Question 47:

If $f : R \to R$ is given by $f (x) = x^{3} + 3, then f^{- 1} (x)$ is equal to
(a) $x^{1 / 3} - 3$
(b) $x^{1 / 3} + 3$
(c) ${(x - 3)}^{1 / 3}$
(d) $x + 3^{1 / 3}$

Answer:

(c)

$Let f^{- 1} (x) = y f (y) = x \Rightarrow y^{3} + 3 = x \Rightarrow y^{3} = x - 3 \Rightarrow y = \sqrt[3]{x - 3} \Rightarrow y = {(x - 3)}^{\frac{1}{3}}$
So, the answer is (c).

Page No 2.76:

Question 48:

Let $f (x) = x^{3}$ be a function with domain {0, 1, 2, 3}. Then domain of $f^{- 1}$ is
(a) {3, 2, 1, 0}
(b) {0, −1, −2, −3}
(c) {0, 1, 8, 27}
(d) {0, −1, −8, −27}

Answer:

(c) {0, 1, 8, 27}

f (x) = x^{3} Domain = \{0, 1, 2, 3\} Range = \{0^{3}, 1^{3}, 2^{3}, 3^{3}\} = \{0, 1, 8, 27\} So, f = \{(0, 0), (1, 1), (2, 8), (3, 27)\} f^{- 1} = \{(0, 0), (1, 1), (8, 2), (27, 3)\} Domain of f^{- 1} = \{0, 1, 8, 27\}

Page No 2.76:

Question 49:

Let $f : R \to R$ be given by $f (x) = x^{2} - 3$ . Then, $f^{- 1}$ is given by
(a) $\sqrt{x + 3}$
(b) $\sqrt{x} + 3$
(c) $x + \sqrt{3}$
(d) None of these

Answer:

(d)

$Let f^{- 1} (x) = y f (y) = x y^{2} - 3 = x y^{2} = x + 3 y = \pm \sqrt{x + 3}$
So, the answer is (d).

Page No 2.76:

Question 50:

Mark the correct alternative in the following question:

Let f : R $\to$ R be given by f(x) = tanx. Then, f ^{$-$ 1}(1) is

(a) $\frac{π}{4}$ (b) $\{n π + \frac{π}{4} : n \in Z\}$ (c) does not exist (d) none of these

Answer:

$We have, f : R \to R is given by f (x) = \tan x \Rightarrow f^{- 1} (x) = \tan^{- 1} x ∴ f^{- 1} (1) = \tan^{- 1} 1 = \{n π + \frac{π}{4} : n \in Z\}$

Hence, the correct alternative is option (b).

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Question 51:

Mark the correct alternative in the following question:

Let f : R $\to$ R be defined as f(x) = $\{\begin{matrix} 2 x, if x > 3 \\ x^{2}, if 1 < x \leq 3 \\ 3 x, if x \leq 1 \end{matrix}$

Then, find f( $-$ 1) + f(2) + f(4)

(a) 9 (b) 14 (c) 5 (d) none of these

Answer:

$We have, f (x) = \{\begin{matrix} 2 x, if x > 3 \\ x^{2}, if 1 < x \leq 3 \\ 3 x, if x \leq 1 \end{matrix} Now, f (- 1) + f (2) + f (4) = 3 (- 1) + 2^{2} + 2 (4) = - 3 + 4 + 8 = 9$

Hence, the correct alternative is option (a).

Page No 2.76:

Question 52:

Mark the correct alternative in the following question:

Let A = {1, 2, ... , n} and B = {a, b}. Then the number of subjections from A into B is

(a) ⁿP₂ (b) 2ⁿ $-$ 2 (c) 2ⁿ $-$ 1 (d) ⁿC₂

Answer:

As, the number of surjections from A to B is equal to the number of functions from A to B minus the number of functions from A to B whose images are proper subsets of B.

And, the number of functions from a set with n number of elements into a set with m number of elements = mⁿ

So, the number of subjections from A into B where A = {1, 2, ... , n} and B = {a, b} is 2ⁿ $-$ 2. (As, two functions can be many-one into functions)

Hence, the correct alternative is option (b).

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Question 53:

Mark the correct alternative in the following question:

If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is

(a) 720 (b) 120 (c) 0 (d) none of these

Answer:

$As, the number of bijection from A into B can only be possible when provided n (A) \geq n (B) But here n (A) < n (B) So, the number of bijection i . e . one - one and onto mappings from A to B = 0$

Hence, the correct alternative is option (c).

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Question 54:

Mark the correct alternative in the following question:

If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is

(a) ¹⁰C₇ (b) ¹⁰C₇ $\times$ 7! (c) 7¹⁰ (d)10⁷

Answer:

As, the number of one-one functions from A to B with m and n elements, respectively = ⁿP_m = ⁿC_m $\times$ m!

So, the number of one-one functions from A to B with 7 and 10 elements, respectively = ¹⁰P₇ = ¹⁰C₇ $\times$ 7!

Hence, the correct alternative is option (b).

Page No 2.76:

Question 55:

Mark the correct alternative in the following question:

Let f : R $-$ $\{\frac{3}{5}\}$ $\to$ R be defined by f(x) = $\frac{3 x + 2}{5 x - 3}$ . Then,

(a) f ^{$-$ l}(x) = f(x) (b) f ^{$-$ 1}(x) = $-$ f(x) (c) fof(x) = $-$ x (d) f ^{$-$ 1}(x) = $\frac{1}{19}$ f(x)

Answer:

We have,

f : R $-$ $\{\frac{3}{5}\}$ $\to$ R is defined by f(x) = $\frac{3 x + 2}{5 x - 3}$

$f o f (x) = f (f (x)) = f (\frac{3 x + 2}{5 x - 3}) = \frac{3 (\frac{3 x + 2}{5 x - 3}) + 2}{5 (\frac{3 x + 2}{5 x - 3}) - 3} = \frac{(\frac{9 x + 6}{5 x - 3}) + 2}{(\frac{15 x + 10}{5 x - 3}) - 3} = \frac{(\frac{9 x + 6 + 10 x - 6}{5 x - 3})}{(\frac{15 x + 10 - 15 x + 9}{5 x - 3})} = \frac{19 x}{19} = x Let y = \frac{3 x + 2}{5 x - 3} \Rightarrow 5 x y - 3 y = 3 x + 2 \Rightarrow 5 x y - 3 x = 3 y + 2 \Rightarrow x (5 y - 3) = 3 y + 2 \Rightarrow x = \frac{3 y + 2}{5 y - 3} \Rightarrow f^{- 1} (y) = \frac{3 y + 2}{5 y - 3} So, f^{- 1} (x) = \frac{3 x + 2}{5 x - 3} = f (x)$

Hence, the correct alternative is option (a).

Page No 2.77:

Question 56:

Let f : R → R be defined by $f (x) = \frac{1}{x} .$ Then, f is
(a) one-one
(b) onto
(e) bijective
(d) not defined

Answer:

Given: The function f : R → R be defined by $f (x) = \frac{1}{x} .$

To check f is one-one:

$Let f (x_{1}) = f (x_{2}) \Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}} \Rightarrow x_{1} = x_{2} Hence, f is one - one . To check f is onto : Since, y = \frac{1}{x} \Rightarrow x = \frac{1}{y} \Rightarrow y \in R - \{0\} \neq R There is no pre - image of y = 0 . Hence, f is not onto .$

Hence, the correct option is (a).

Page No 2.77:

Question 57:

Let f : R → R be defined by f(x) = 3x² – 5 and g : R → R by $g (x) = \frac{x}{x^{2} + 1} .$ Then (gof) (x) is
(a) $\frac{3 x^{2} - 5}{9 x^{4} - 30 x^{2} + 26}$

(b) $\frac{3 x^{2} - 5}{9 x^{4} - 6 x^{2} + 26}$

(c) $\frac{3 x^{2}}{x^{4} + 2 x^{2} - 4}$

(d) $\frac{3 x^{2}}{9 x^{4} + 30 x^{2} - 2}$

Answer:

Given: f(x) = 3x² – 5 and $g (x) = \frac{x}{x^{2} + 1}$

$(g o f) (x) = g (f (x)) = g (3 x^{2} - 5) = \frac{3 x^{2} - 5}{{(3 x^{2} - 5)}^{2} + 1} = \frac{3 x^{2} - 5}{{(3 x^{2})}^{2} + 5^{2} - 2 (3 x^{2}) (5) + 1} = \frac{3 x^{2} - 5}{9 x^{4} + 25 - 30 x^{2} + 1} = \frac{3 x^{2} - 5}{9 x^{4} - 30 x^{2} + 26}$

Hence, the correct option is (a).

Page No 2.77:

Question 58:

Which of the following functions from Z to Z are bijections?
(a) f(x) = x³
(b) f(x) = x + 2
(c) f(x) = 2x + 1
(d) f(x) = x² + 1

Answer:

Given: f : Z → Z

(a) f(x) = x³

It is one-one but not onto.

Thus, it is not bijective.

(b) f(x) = x + 2

It is one-one and onto.

Thus, it is bijective.

(c) f(x) = 2x + 1

It is one-one but not onto.

Thus, it is not bijective.

(d) f(x) = x² + 1

It is neither one-one nor onto.

Thus, it is not bijective.

Hence, the correct option is (b).

Page No 2.77:

Question 59:

Let f : A → B and g : B → C be the bijective functions. Then, (gof)^–1 =
(a) f^–1o g^–1
(b) fog
(c) g^–1of^–1
(d) gof

Answer:

Given: f : A → B and g : B → C be the bijective functions

$Since, f : A \to B Thus, f^{- 1} : B \to A . . . (1) Since, g : B \to C Thus, g^{- 1} : C \to B . . . (2) From (1) and (2), we get f^{- 1} o g^{- 1} : C \to A . . . (3) Also, g o f : A \to C \Rightarrow {(g o f)}^{- 1} : C \to A . . . (4) Therefore, {(g o f)}^{- 1} = f^{- 1} o g^{- 1}$

Hence, the correct option is (a).

Page No 2.77:

Question 60:

Let f : N → R be the function defined by $f (x) = \frac{2 x - 1}{2}$ and g : Q → R be another function defined by g(x) = x + 2. Then, (gof) (3/2) is
(a) 1

(b) 2

(c) $\frac{7}{2}$

(d) none of these

Answer:

Given: $f (x) = \frac{2 x - 1}{2}$ and g(x) = x + 2

$(g o f) (x) = g (f (x)) = g (\frac{2 x - 1}{2}) = \frac{2 x - 1}{2} + 2 = \frac{2 x - 1 + 4}{2} = \frac{2 x + 3}{2} (g o f) (\frac{3}{2}) = \frac{2 (\frac{3}{2}) + 3}{2} = \frac{3 + 3}{2} = \frac{6}{2} = 3$

Hence, the correct option is (d).

Page No 2.77:

Question 1:

The total number of functions from the set A = (1, 2, 3, 4 to the set B = a, b, c) is _________.

Answer:

Given: $f : A \to B where A = \{1, 2, 3, 4\} and B = \{a, b, c\}$

Number of elements in A = 4
Number of elements in B = 3
Each element of A have 3 options to form an image.

Thus, Number of functions that can be formed = 3 × 3 × 3 × 3 = 81

Hence, the total number of functions from the set A = {1, 2, 3, 4} to the set B = {a, b, c} is 81.

Page No 2.77:

Question 2:

The total number of one-one functions from the set A = {a, b, c} to the set B = {x, y, z, t} is _________.

Answer:

Given: $f : A \to B where A = \{a, b, c\} and B = \{x, y, z, t\}$

Number of elements in A = 3
Number of elements in B = 4

To form a one-one function,

Element a ∈ A have 4 options to form an image.
Element b ∈ A have 3 options to form an image.
Element c ∈ A have 2 options to form an image.

Thus, Number of one-one functions that can be formed = 4 × 3 × 2 = 24

Hence, the total number of one-one functions from the set A = {a, b, c} to the set B = {x, y, z, t} is 24.

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Question 3:

The total number of onto functions from the set A = (1, 2, 3, 4, 5) to the set B = {x, y} is _________.

Answer:

Given: $f : A \to B where A = \{1, 2, 3, 4, 5\} and B = \{x, y\}$

Number of elements in A = 5
Number of elements in B = 2

Each Element of A have 2 options to form an image.

Thus, Total number of functions that can be formed = 2 × 2 × 2 × 2 × 2 = 32

Number of functions having only one image i.e., {x} = 1
Number of functions having only one image i.e., {y} = 1

Thus, Number of onto functions that can be formed = 32 − 1 − 1 = 30

Hence, the total number of onto functions from the set A = {1, 2, 3, 4, 5} to the set B = {x, y} is 30.

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Question 4:

The domain of the real function $f (x) = \sqrt{16 - x^{2}}$ is _________.

Answer:

Given: $f (x) = \sqrt{16 - x^{2}}$

To find the domain, we find the real values of x for which the function is defined.

$16 - x^{2} \geq 0 \Rightarrow 16 \geq x^{2} \Rightarrow x^{2} \leq 16 \Rightarrow x \leq 4 and x \geq - 4 \Rightarrow - 4 \leq x \leq 4 \Rightarrow x \in [- 4, 4]$

Hence, the domain of the real function $f (x) = \sqrt{16 - x^{2}}$ is [−4, 4].

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Question 5:

The domain of the real function $f (x) = \frac{x}{\sqrt{9 - x^{2}}}$ is ___________.

Answer:

Given: $f (x) = \frac{x}{\sqrt{9 - x^{2}}}$

To find the domain, we find the real values of x for which the function is defined.

$x \in R and 9 - x^{2} > 0 \Rightarrow x \in R and 9 > x^{2} \Rightarrow x \in R and x^{2} < 9 \Rightarrow x \in R and - 3 < x < 3 \Rightarrow - 3 < x < 3 \Rightarrow x \in (- 3, 3)$

Hence, the domain of the real function $f (x) = \frac{x}{\sqrt{9 - x^{2}}}$ is (−3, 3).

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Question 6:

The range of the function f : R → R given by $f (x) = x + \sqrt{x^{2}}$ is _________.

Answer:

Given: $f (x) = x + \sqrt{x^{2}}$

$f (x) = x + \sqrt{x^{2}} = x + |x| = \{\begin{cases} x + x, x \geq 0 \\ x - x, x < 0 \end{cases} = \{\begin{cases} 2 x, x \geq 0 \\ 0, x < 0 \end{cases}$

To find the range, we find the real values of y obtained.

$y = 2 x when x \geq 0 \Rightarrow x = \frac{y}{2} \geq 0 \Rightarrow y \geq 0 \Rightarrow y \in [0, \infty) . . . (1) y = 0 when x < 0 . . . (2) Thus, from (1) and (2), y \in [0, \infty)$

Hence, the range of the function f : R → R given by $f (x) = x + \sqrt{x^{2}}$ is $[0, \infty)$ .

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Question 7:

The range of the function f : R –{–2) → R given by $f (x) = \frac{x + 2}{|x + 2|}$ is _________.

Answer:

Given: $f (x) = \frac{x + 2}{|x + 2|}$

$f (x) = \frac{x + 2}{|x + 2|} = \{\begin{cases} \frac{x + 2}{x + 2}, x + 2 \geq 0 \\ \frac{x + 2}{- (x + 2)}, x + 2 < 0 \end{cases} = \{\begin{cases} 1, x + 2 \geq 0 \\ - 1, x + 2 < 0 \end{cases}$

To find the range, we find the real values of y obtained.

Hence, the range of the function f : R –{–2) → R given by $f (x) = \frac{x + 2}{|x + 2|}$ is {–1, 1}.

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Question 8:

If f : C → C is defined by f(x) = 8x³, then f^–1(8) = . _________.

Answer:

Given: f(x) = 8x³

$f (x) = 8 x^{3} \Rightarrow y = 8 x^{3} \Rightarrow x^{3} = \frac{y}{8} \Rightarrow x = {(\frac{y}{8})}^{\frac{1}{3}} Thus, f^{- 1} (x) = {(\frac{x}{8})}^{\frac{1}{3}} f^{- 1} (8) = {(\frac{8}{8})}^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1, ω, ω^{2} (∵ f : C \to C) where, ω is the cube root of unity .$

Hence, if f : C → C is defined by f(x) = 8x³, then f⁻¹(8) = $1, ω, ω^{2}$ .

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Question 9:

If f : R → R is defined by f(x) = 8x³ then, f^–1(8) = _________.

Answer:

Given: f(x) = 8x³

$f (x) = 8 x^{3} \Rightarrow y = 8 x^{3} \Rightarrow x^{3} = \frac{y}{8} \Rightarrow x = {(\frac{y}{8})}^{\frac{1}{3}} Thus, f^{- 1} (x) = {(\frac{x}{8})}^{\frac{1}{3}} f^{- 1} (8) = {(\frac{8}{8})}^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1 (∵ f : R \to R)$

Hence, if f : R → R is defined by f(x) = 8x³ then f⁻¹(8) = 1.

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Question 10:

If f : R – {0} → R – {0} is defined as $f (x) = \frac{2}{3 x},$ then f^–1(x) = ___________.

Answer:

Given: A function f : R – {0} → R – {0} is defined as $f (x) = \frac{2}{3 x}$

$f (x) = \frac{2}{3 x} \Rightarrow y = \frac{2}{3 x} \Rightarrow 3 x = \frac{2}{y} \Rightarrow x = \frac{2}{3 y} Thus, f^{- 1} (x) = \frac{2}{3 x}$

Hence, if f : R – {0} → R – {0} is defined as $f (x) = \frac{2}{3 x},$ then f⁻¹(x) = $\frac{2}{3 x}$ .

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Question 11:

If f : R → R is defined by f(x) = 6 – (x – 9)³, then f^–1(x) = ___________.

Answer:

Given: A function f : R → R is defined by f(x) = 6 – (x – 9)³

$f (x) = 6 - {(x - 9)}^{3} \Rightarrow y = 6 - {(x - 9)}^{3} \Rightarrow {(x - 9)}^{3} = 6 - y \Rightarrow x - 9 = {(6 - y)}^{\frac{1}{3}} \Rightarrow x = 9 + {(6 - y)}^{\frac{1}{3}} Thus, f^{- 1} (x) = 9 + {(6 - x)}^{\frac{1}{3}}$

Hence, if f : R → R is defined by f(x) = 6 – (x – 9)³, then f⁻¹(x) = $9 + {(6 - x)}^{\frac{1}{3}}$ .

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Question 12:

Let A = {1, 2, 3, 4} and f : A → A be given by f = {(1, 4), (2, 3), (3, 2), (4, 1)}. Then f^–1 = ___________.

Answer:

Given: A function f : A → A be given by f = {(1, 4), (2, 3), (3, 2), (4, 1)}

$f = \{(1, 4), (2, 3), (3, 2), (4, 1)\} \Rightarrow f^{- 1} = \{(4, 1), (3, 2), (2, 3), (1, 4)\} Thus, f^{- 1} = \{(4, 1), (3, 2), (2, 3), (1, 4)\}$

Hence, $f^{- 1} = \{(4, 1), (3, 2), (2, 3), (1, 4)\} .$

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Question 13:

If f : R → R be defined by f(x) = (2 – x⁵)^1/5, then fof(x) = ___________.

Answer:

Given: f(x) = (2 – x⁵)^1/5

$f o f (x) = f (f (x)) = f ({(2 - x^{5})}^{\frac{1}{5}}) = {[2 - {({(2 - x^{5})}^{\frac{1}{5}})}^{5}]}^{\frac{1}{5}} = {[2 - {(2 - x^{5})}^{\frac{1}{5} \times 5}]}^{\frac{1}{5}} = {[2 - 2 + x^{5}]}^{\frac{1}{5}} = {[x^{5}]}^{\frac{1}{5}} = {x^{5 \times}}^{\frac{1}{5}} = x$

Hence, if f : R → R be defined by f(x) = (2 – x⁵)^1/5, then fof(x) = x.

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Question 14:

Let A = {1, 2, 3, 4, 5, ..., 10} and f : A → A be an invertible function. Then, $\sum_{r = 1}^{10} (f^{- 1} o f) (r) =$ ___________.

Answer:

Given: f : A → A is an invertible function, where A = {1, 2, 3, 4, 5, ..., 10}

$Since, f is invertible Therefore, f^{- 1} o f (x) = x . . . (1) Now, \sum_{r = 1}^{10} (f^{- 1} o f) (r) = f^{- 1} o f (1) + f^{- 1} o f (2) + f^{- 1} o f (3) + . . . . + f^{- 1} o f (10) = 1 + 2 + 3 + . . . . + 10 (From (1)) = \frac{10 (10 + 1)}{2} (∵ 1 + 2 + 3 + . . . + n = \frac{n (n + 1)}{2}) = 5 (11) = 55$

Hence, $\sum_{r = 1}^{10} (f^{- 1} o f) (r) =$ 55.

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Question 15:

Let A = {1, 2, 3, 4, 5, 6) and B = (2, 4, 6, 8, 10, 12). If f : A → B is given by f(x) = 2x, then f^–1 as set of ordered pairs, is ___________.

Answer:

Given: A function f : A → B defined as f(x) = 2x, where A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10, 12}

$Since, f (x) = 2 x Therefore, f = \{(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12)\} Hence, f^{- 1} = \{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)\}$

Hence, if f : A → B is given by f(x) = 2x, then f^–1 as set of ordered pairs, is $\{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)\} .$

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Question 16:

Let f = {(0, –1), (–1, 3), (2, 3), (3, 5)} be a function from Z to Z defined by f(x) = ax + b. Then, (a, b) = ___________.

Answer:

Given: f  = {(0, –1), (–1, –3), (2, 3), (3, 5)} is a function from Z to Z defined by f(x) = ax + b

f  = {(0, –1), (–1, –3), (2, 3), (3, 5)} defined by f(x) = ax + b

$f (0) = - 1 \Rightarrow a (0) + b = - 1 \Rightarrow 0 + b = - 1 \Rightarrow b = - 1 . . . (1) f (2) = 3 \Rightarrow a (2) + b = 3 \Rightarrow 2 a + b = 3 \Rightarrow 2 a - 1 = 3 (From (1)) \Rightarrow 2 a = 3 + 1 \Rightarrow 2 a = 4 \Rightarrow a = 2 . . . (2) Thus, a = 2 and b = - 1$

Hence, (a, b) = (2, –1).

Disclaimer: The function f must be equal to f  =  {(0, –1), (–1, –3), (2, 3), (3, 5)}.

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Question 17:

Let f : R → R and g : R → R be functions defined by f(x) = 5 – x² and g(x) = 3x – 4. Then the value of fog (–1) is ___________.

Answer:

Given: f(x) = 5 – x² and g(x) = 3x – 4

$f o g (- 1) = f (g (- 1)) = f (3 (- 1) - 4) = f (- 3 - 4) = f (- 7) = 5 - {(- 7)}^{2} = 5 - 49 = - 44$

Hence, the value of fog (–1) is –44.

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Question 18:

Let f be the greatest integer function defined as f(x) = [x] and g be the modules function defined as g(x) = |x|, then the value of $g o f (- \frac{5}{4})$ is ___________.

Answer:

Given: f(x) = [x] and g(x) = |x|

$g o f (- \frac{5}{4}) = g (f (- \frac{5}{4})) = g ([- \frac{5}{4}]) = g (- 2) = |- 2| = 2$

Hence, the value of $g o f (- \frac{5}{4})$ is 2.

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Question 19:

If f(x) = cos [e] x + cos [–e] x, then f(π) = ___________.

Answer:

Given: f(x) = cos[e]x + cos[–e]x

$f (x) = \cos [e] x + \cos [- e] x = \cos 2 x + \cos (- 3 x) (∵ e = 2.718 approx) = \cos 2 x + \cos 3 x \Rightarrow f (π) = \cos 2 π + \cos 3 π = 1 - 1 = 0$

Hence, f(π) = 0.

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Question 20:

Let A = {1, 2, 3} and B = {a, b} be two sets. Then the number of constant functions from A to B is ___________.

Answer:

Given: Sets A = {1, 2, 3} and B = {a, b}

Two constant functions can be formed from A to B.
i.e., one is f(x) = a and other is f(x) = b

Hence, the number of constant functions from A to B is 2.

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Question 21:

If f(x) = cos [π²] x + cos [–π²] x, then $f (\frac{π}{2}) =$ ______________.

Answer:

Given: f(x) = cos [π²] x + cos [–π²] x

$f (x) = \cos [π^{2}] x + \cos [- π^{2}] x = \cos 9 x + \cos (- 10 x) (∵ π^{2} = 9.85 approx) = \cos 9 x + \cos 10 x \Rightarrow f (\frac{π}{2}) = \cos 9 (\frac{π}{2}) + \cos 10 (\frac{π}{2}) = \cos \frac{9 π}{2} + \cos 5 π = 0 - 1 = - 1$

Hence, $f (\frac{π}{2}) =$ –1.

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Question 22:

The number of onto functions from A = {a, b, c} to B = {1, 2, 3, 4} is __________.

Answer:

Given: A function from A = {a, b, c} to B = {1, 2, 3, 4}

If a function from A to B is onto, then number of elements of A ≥ number of elements of B.

But here, number of elements of A < number of elements of B

Thus, no onto function exist from A = {a, b, c} to B = {1, 2, 3, 4}.

Hence, the number of onto functions from A = {a, b, c} to B = {1, 2, 3, 4} is 0.

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Question 23:

If f(0, ∞) → R is given by f(x) = log₁₀ x, then f^–1(x) = ___________.

Answer:

Given: f(x) = log₁₀x

$f (x) = \log_{10} x \Rightarrow y = \log_{10} x \Rightarrow x = 10^{y} Thus, f^{- 1} (y) = 10^{y} .$

Hence, f⁻¹(x) = 10^x.

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Question 24:

If f : R⁺ → R is defined as f(x) = log₃x, then f^–1(x) = ______________.

Answer:

Given: f(x) = log₃x

$f (x) = \log_{3} x \Rightarrow y = \log_{3} x \Rightarrow x = 3^{y} Thus, f^{- 1} (y) = 3^{y} .$

Hence, f⁻¹(x) = 3^x.

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Question 25:

If f : R → R, g : R → R are defined by f(x) = 5x – 3, g(x) = x² + 3, then (gof^–1) (3) = _______________.

Answer:

Given: f(x) = 5x – 3 and g(x) = x² + 3

$f (x) = 5 x - 3 \Rightarrow y = 5 x - 3 \Rightarrow 5 x = y + 3 \Rightarrow x = \frac{y + 3}{5} Thus, f^{- 1} (y) = \frac{y + 3}{5} . Now, g o f^{- 1} (3) = g (f^{- 1} (3)) = g (\frac{3 + 3}{5}) = g (\frac{6}{5}) = {(\frac{6}{5})}^{2} + 3 = \frac{36}{25} + 3 = \frac{36 + 75}{25} = \frac{111}{25}$

Hence, gof⁻¹(3) = $\frac{111}{25}$ .

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Question 26:

If f : R → R is given by f(x) = 2x + |x|, then f(2x) + f(–x) + 4x = _______________.

Answer:

Given: f(x) = 2x + |x|

$f (x) = 2 x + |x| f (2 x) = 2 (2 x) + |2 x| \Rightarrow f (2 x) = 4 x + 2 |x| . . . (1) f (- x) = 2 (- x) + |- x| \Rightarrow f (- x) = - 2 x + |x| . . . (2) Now, f (2 x) + f (- x) + 4 x = 4 x + 2 |x| - 2 x + |x| + 4 x = 6 x + 3 |x| = 3 (2 x + |x|) = 3 f (x)$

Hence, f(2x) + f(–x) + 4x = 3f(x).

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Question 27:

If $f (x) = \frac{1 - x}{1 + x},$ then fof(cos 2θ) = ______________.

Answer:

Given: $f (x) = \frac{1 - x}{1 + x}$

$f o f (\cos 2 θ) = f (f (\cos 2 θ)) = f (\frac{1 - \cos 2 θ}{1 + \cos 2 θ}) = f (\frac{2 \sin^{2} θ}{2 \cos^{2} θ}) = f (\tan^{2} θ) = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = \cos 2 θ$

Hence, fof(cos 2θ) = $\cos 2 θ$ .

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Question 28:

Let $f (x) = \frac{x}{x - 1} and \frac{f (α)}{f (α + 1)} = f (α^{k}),$ then k = ______________.

Answer:

Given: $f (x) = \frac{x}{x - 1} and \frac{f (α)}{f (α + 1)} = f (α^{k})$

$\frac{f (a)}{f (a + 1)} = \frac{\frac{a}{a - 1}}{\frac{a + 1}{a + 1 - 1}} = \frac{\frac{a}{a - 1}}{\frac{a + 1}{a}} = \frac{a^{2}}{(a - 1) (a + 1)} = \frac{a^{2}}{a^{2} - 1} . . . (1) It is given that, \frac{f (a)}{f (a + 1)} = f (a^{k}) \Rightarrow \frac{a^{2}}{a^{2} - 1} = \frac{a^{k}}{a^{k} - 1} \Rightarrow k = 2$

Hence, k = 2.

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Question 29:

If f (f(x)) = x + 1 for all x ∈ R and if $f (0) = \frac{1}{2},$ then f(1) = ____________.

Answer:

Given: f (f(x)) = x + 1 for all x ∈ R and $f (0) = \frac{1}{2}$

$f (f (x)) = x + 1 \Rightarrow f (f (0)) = 0 + 1 \Rightarrow f (\frac{1}{2}) = 1 (∵ f (0) = \frac{1}{2}) . . . . (1) Now, f (f (\frac{1}{2})) = \frac{1}{2} + 1 \Rightarrow f (1) = \frac{1 + 2}{2} (∵ f (\frac{1}{2}) = 1) \Rightarrow f (1) = \frac{3}{2}$

Hence, f(1) = $\frac{3}{2}$ .

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Question 30:

If f(x) = 3x + 10 and g(x) = x² –1, then (fog)^–1 is equal to ___________.

Answer:

Given: f(x) = 3x + 10 and g(x) = x² –1

$f o g (x) = f (g (x)) = f (x^{2} - 1) = 3 (x^{2} - 1) + 10 = 3 x^{2} - 3 + 10 = 3 x^{2} + 7 Thus, f o g (x) = 3 x^{2} + 7 \Rightarrow y = 3 x^{2} + 7 \Rightarrow 3 x^{2} = y - 7 \Rightarrow x^{2} = \frac{y - 7}{3} \Rightarrow x = \pm \sqrt{\frac{y - 7}{3}} f o g^{- 1} (x) = \pm \sqrt{\frac{x - 7}{3}}$

Hence, (fog)^–1 is equal to $\pm \sqrt{\frac{x - 7}{3}}$ .

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Question 31:

Let f = {(1, 2), (3, 5), (4, 1)) and g = {(2, 3), (5, 1), (1, 3)}. Then, gof = __________ and fog = __________.

Answer:

Given: f = {(1, 2), (3, 5), (4, 1)) and g = {(2, 3), (5, 1), (1, 3)}

$f o g (x) = f (g (x)) f o g (1) = f (g (1)) = f (3) = 5 f o g (2) = f (g (2)) = f (3) = 5 f o g (5) = f (g (5)) = f (1) = 2 Hence, f o g = \{(1, 5), (2, 5), (5, 2)\} g o f (x) = g (f (x)) g o f (1) = g (f (1)) = g (2) = 3 g o f (3) = g (f (3)) = g (5) = 1 g o f (4) = g (f (4)) = g (1) = 3 Hence, g o f = \{(1, 3), (3, 1), (4, 3)\} .$

Hence, gof = $\{(1, 3), (3, 1), (4, 3)\}$ and fog = $\{(1, 5), (2, 5), (5, 2)\}$ .

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Question 1:

Which one of the following graphs represents a function?

Answer:

In graph (b), 0 has more than one image, whereas every value of x in graph (a) has a unique image.
Thus, graph (a) represents a function.
So, the answer is (a).

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Question 2:

Which of the following graphs represents a one-one function?

Answer:

In the graph of (b), different elements on the x-axis have different images on the y-axis.

But in (a), the graph cuts the x-axis at 3 points, which means that 3 points on the x-axis have the same image as 0 and hence, it is not one-one.

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Question 3:

If A = {1, 2, 3} and B = {a, b}, write the total number of functions from A to B.

Answer:

Formula:
If set A has m elements and set B has n elements, then the number of functions from A to B is $n^{m}$ .
Given:
A = {1, 2, 3} and B = {a, b}
$\Rightarrow n (A)$ = 3 and $n (B)$ = 2
$∴$ Number of functions from A to B = 2³ = 8

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Question 4:

If A = {a, b, c} and B = {−2, −1, 0, 1, 2}, write the total number of one-one functions from A to B.

Answer:

Let $f : A \to B$ be a one-one function.
$Then, f (a) can take 5 values, f (b) can take 4 values and f (c) can take 3 values.$

Then, the number of one-one functions = 5 $\times$ 4 $\times$ 3 = 60

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Question 5:

Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {a, b, c}.

Answer:

A has 4 elements and B has 3 elements.
Also, one-one function is only possible from A to B if $n (A) \leq n (B)$ .
But, here $n (A) > n (B)$ .
So, the number of one-one functions from A to B is 0.

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Question 6:

If f : R → R is defined by f(x) = x², write f⁻¹ (25).

Answer:

$Let f^{- 1} (25) = x . . . (1) \Rightarrow f (x) = 25 \Rightarrow x^{2} = 25 \Rightarrow x^{2} - 25 = 0 \Rightarrow (x - 5) (x + 5) = 0 \Rightarrow x = \pm 5 \Rightarrow f^{- 1} (25) = \{- 5, 5\} [from (1)]$

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Question 7:

If f : C → C is defined by f(x) = x², write f⁻¹ (−4). Here, C denotes the set of all complex numbers.

Answer:

$Let f^{- 1} (- 4) = x . . . (1) \Rightarrow f (x) = - 4 \Rightarrow x^{2} = - 4 \Rightarrow x^{2} + 4 = 0 \Rightarrow (x + 2 i) (x - 2 i) = 0 [u sing the identity : a^{2} + b^{2} = (a - i b) (a + i b)] \Rightarrow x = \pm 2 i [as x \in C] \Rightarrow f^{- 1} (25) = \{- 2 i, 2 i\} [f rom (1)]$

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Question 8:

If f : R → R is given by f(x) = x³, write f⁻¹ (1).

Answer:

$Let f^{- 1} (1) = x . . . (1) \Rightarrow f (x) = 1 \Rightarrow x^{3} = 1 \Rightarrow x^{3} - 1 = 0 \Rightarrow (x - 1) (x^{2} + x + 1) = 0 [u sing the identity : a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})] \Rightarrow x = 1 (as x \in R) \Rightarrow f^{- 1} (1) = \{1\} [from (1)]$

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Question 9:

Let C denote the set of all complex numbers. A function f : C → C is defined by f(x) = x³. Write f⁻¹ (1).

Answer:

$Let f^{- 1} (1) = x . . . (1) \Rightarrow f (x) = 1 \Rightarrow x^{3} = 1 \Rightarrow x^{3} - 1 = 0 \Rightarrow (x - 1) (x^{2} + x + 1) = 0 [Using identity : a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})] \Rightarrow (x - 1) (x - ω) (x - ω^{2}) = 0, where ω = \frac{1 \pm i \sqrt{3}}{2} \Rightarrow x = 1, ω or ω^{2} (as x \in C) \Rightarrow f^{- 1} (1) = \{1, ω, ω^{2}\} [from (1)]$

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Question 10:

Let f be a function from C (set of all complex numbers) to itself given by f(x) = x³. Write f⁻¹ (−1).

Answer:

$Let f^{- 1} (- 1) = x . . . (1) \Rightarrow f (x) = - 1 \Rightarrow x^{3} = - 1 \Rightarrow x^{3} + 1 = 0 \Rightarrow (x + 1) (x^{2} - x + 1) = 0 [u sing the identity : a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})] \Rightarrow (x + 1) (x + ω) (x + ω^{2}) = 0, where ω = \frac{1 \pm i \sqrt{3}}{2} \Rightarrow x = - 1, - ω, - ω^{2} (as x \in C) \Rightarrow f^{- 1} (- 1) = \{- 1, - ω, - ω^{2}\} [from (1)]$

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Question 11:

If f : R → R be defined by f(x) = x⁴, write f⁻¹ (1).

Answer:

$Let f^{- 1} (1) = x . . . (1) \Rightarrow f (x) = 1 \Rightarrow x^{4} = 1 \Rightarrow x^{4} - 1 = 0 \Rightarrow (x^{2} - 1) (x^{2} + 1) = 0 [u sing identity : a^{2} - b^{2} = (a - b) (a + b)] \Rightarrow (x - 1) (x + 1) (x^{2} + 1) = 0 [u sing identity : a^{2} - b^{2} = (a - b) (a + b)] \Rightarrow x = \pm 1 [as x \in R] \Rightarrow f^{- 1} (1) = \{- 1, 1\} [from (1)]$

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Question 12:

If f : C → C is defined by f(x) = x⁴, write f⁻¹ (1).

Answer:

$Let f^{- 1} (1) = x . . . (1) \Rightarrow f (x) = 1 \Rightarrow x^{4} = 1 \Rightarrow x^{4} - 1 = 0 \Rightarrow (x^{2} - 1) (x^{2} + 1) = 0 [u sing identity : a^{2} - b^{2} = (a - b) (a + b)] \Rightarrow (x - 1) (x + 1) (x - i) (x + i) = 0, w here i = \sqrt{- 1} [u sing identity : a^{2} - b^{2} = (a - b) (a + b)] \Rightarrow x = \pm 1, \pm i \Rightarrow f^{- 1} (1) = \{- 1, 1, i, - i\} [from (1)]$

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Question 13:

If f : R → R is defined by f(x) = x², find f⁻¹ (−25).

Answer:

$Let f^{- 1} (- 25) = x \Rightarrow f (x) = - 25 \Rightarrow x^{2} = - 25 We cannot find x∈R, such that x^{2} = - 25 ({as x}^{2} ≥0 for all x∈R) So, f^{- 1} (- 25) = ϕ$

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Question 14:

If f : C → C is defined by f(x) = (x − 2)³, write f⁻¹ (−1).

Answer:

$Let f^{- 1} (- 1) = x . . . (1) \Rightarrow f (x) = - 1 \Rightarrow {(x - 2)}^{3} = - 1 \Rightarrow x - 2 = - 1 or - ω or - ω^{2} (as the roots of {(- 1)}^{\frac{1}{3}} are - 1, - ω and - ω^{2}, where ω = \frac{1 \pm i \sqrt{3}}{2}) \Rightarrow x = - 1 + 2 or 2 - ω or 2 - ω^{2} = 1, 2 - ω, 2 - ω \Rightarrow f^{- 1} (- 1) = \{1, 2 - ω, 2 - ω^{2}\} [from (1)]$

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Question 15:

If f : R → R is defined by f(x) = 10 x − 7, then write f⁻¹ (x).

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow 10 y - 7 = x \Rightarrow 10 y = x + 7 \Rightarrow y = \frac{x + 7}{10} \Rightarrow f^{- 1} (x) = \frac{x + 7}{10} (From (1))$

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Question 16:

Let $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ be a function defined by f(x) = cos [x]. Write range (f).

Answer:

$Domain = (\frac{- π}{2}, \frac{π}{2}) = (- 1.57, 1.57) (as π= \frac{22}{7}) So, \cos [x] = \cos (- 2) = \cos 2 \forall x \in (- 1.57, 0) Also, \cos 0 = 1 for x = 0 And \cos [x] = \cos 1 \forall x \in (0, 1.57) ∴ Range= \{1, \cos 1, \cos 2\}$

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Question 17:

If f : R → R defined by f(x) = 3x − 4 is invertible, then write f⁻¹ (x).

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow 3 y - 4 = x \Rightarrow 3 y = x + 4 \Rightarrow y = \frac{x + 4}{3} \Rightarrow f^{- 1} (x) = \frac{x + 4}{3} [from (1)]$

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Question 18:

If f : R → R, g : R → are given by f(x) = (x + 1)² and g(x) = x² + 1, then write the value of fog (−3).

Answer:

$(f o g) (- 3) = f (g (- 3)) = f ({(- 3)}^{2} + 1) = f (10) = {(10 + 1)}^{2} = 121$

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Question 19:

Let A = {x ∈ R : −4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by $f (x) = \frac{|x|}{x}$ . Write the range of f.

Answer:

$∵ f (x) = \frac{|x|}{x} = \frac{\pm x}{x} = \pm 1 \forall x \in A, range of f = \{- 1, 1\} .$

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Question 20:

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to$ A be defined by f(x) = sin x. If f is a bijection, write set A.

Answer:

$∵$ f is a bijection,
co-domain of f = range of f
As $- 1 \leq \sin x \leq 1$ ,
$- 1 \leq y \leq 1$
So, A = [-1, 1]

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Question 21:

Let f : R → R⁺ be defined by f(x) = a^x, a > 0 and a ≠ 1. Write f⁻¹ (x).

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow a^{y} = x \Rightarrow y = \log_{a} x \Rightarrow f^{- 1} (x) = \log_{a} x [from (1)]$

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Question 22:

Let f : R − {−1} → R − {1} be given by $f (x) = \frac{x}{x + 1} . Write f^{- 1} (x)$ .

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow \frac{y}{y + 1} = x \Rightarrow y = x y + x \Rightarrow y - x y = x \Rightarrow y (1 - x) = x \Rightarrow y = \frac{x}{1 - x} \Rightarrow f^{- 1} (x) = \frac{x}{1 - x} [from (1)]$

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Question 23:

Let $f : R - \{- \frac{3}{5}\} \to R$ be a function defined as $f (x) = \frac{2 x}{5 x + 3}$ .

Write $f^{- 1} : Range of f \to R - \{- \frac{3}{5}\}$ .

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow \frac{2 y}{5 y + 3} = x \Rightarrow 2 y = 5 x y + 3 x \Rightarrow 2 y - 5 x y = 3 x \Rightarrow y (2 - 5 x) = 3 x \Rightarrow y = \frac{3 x}{2 - 5 x} \Rightarrow f^{- 1} (x) = \frac{3 x}{2 - 5 x} [from (1)]$

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Question 24:

Let f : R → R, g : R → R be two functions defined by f(x) = x² + x + 1 and g(x) = 1 − x². Write fog (−2).

Answer:

$(f o g) (- 2) = f (g (- 2)) = f (1 - {(- 2)}^{2}) = f (- 3) = {(- 3)}^{2} + (- 3) + 1 = 9 - 3 + 1 = 7$

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Question 25:

Let f : R → R be defined as $f (x) = \frac{2 x - 3}{4} . Write f o f^{- 1} (1) .$

Answer:

$Let f^{- 1} (x) = y . . . (1) \Rightarrow f (y) = x \Rightarrow \frac{2 y - 3}{4} = x \Rightarrow 2 y - 3 = 4 x \Rightarrow 2 y = 4 x + 3 \Rightarrow y = \frac{4 x + 3}{2} \Rightarrow f^{- 1} (x) = \frac{4 x + 3}{2} [from (1)] \Rightarrow f^{- 1} (x) = \frac{4 x + 3}{2} ∴ (f o f^{- 1}) (1) = f (\frac{4 (1) + 3}{2}) = f (\frac{7}{2}) = \frac{2 (\frac{7}{2}) - 3}{4} = \frac{7 - 3}{4} = \frac{4}{4} = 1$

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Question 26:

Let f be an invertible real function. Write

$(f^{- 1} o f) (1) + (f^{- 1} o f) (2) + . . . + (f^{- 1} o f) (100) .$

Answer:

Given that f is an invertible real function.
$f^{- 1} o f = I, where I is an identity function. So, (f^{- 1} o f) (1) + (f^{- 1} o f) (2) + . . . + (f^{- 1} o f) (100) = I (1) + I (2) + . . . + I (100) = 1 + 2 + . . . + 100 (As I (x) = x, \forall x \in R) = \frac{100 (100 + 1)}{2} [Sum of first n natural numbers= \frac{n (n + 1)}{2}] = 5050$

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Question 27:

Let A = {1, 2, 3, 4} and B = {a, b} be two sets. Write the total number of onto functions from A to B.

Answer:

Formula:

When two sets A and B have m and n elements respectively, then the number of onto functions from A to B is
$\{\begin{cases} \sum_{r = 1}^{n} {(- 1)}^{r} n C_{r} r^{m}, if m \geq n \\ o, if m < n \end{cases}$

Here, number of elements in A = 4 = m
Number of elements in B = 2 = n
So, m > n
Number of onto functions
$= \sum_{r = 1}^{2} {(- 1)}^{r} 2 C_{r} r^{4} = {(- 1)}^{1} 2 C_{1} 1^{4} + {(- 1)}^{2} 2 C_{2} 2^{4} = - 2 + 16 = 14$

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Question 28:

Write the domain of the real function $f (x) = \sqrt{x - [x]} .$

Answer:

[x] is the greatest integral function.
$So, 0≤ x - [x] < 1 \Rightarrow \sqrt{x - [x]} exists for every x \in R . ⇒Domain = R$

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Question 29:

Write the domain of the real function $f (x) = \sqrt{[x] - x} .$

Answer:

[x] is the greatest integer function.
$[x] \leq x, \forall x \in R \Rightarrow [x] - x \leq 0, \forall x \in R \Rightarrow \sqrt{[x] - x} does not exist for any x \in R . Domain = ϕ$

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Question 30:

Write the domain of the real function $f (x) = \frac{1}{\sqrt{| x | - x}} .$

Answer:

$Case-1: When x > 0 |x| = x \Rightarrow \frac{1}{\sqrt{|x| - x}} = \frac{1}{\sqrt{x - x}} = \frac{1}{0} = \infty Case-2: When x < 0 |x| = - x \Rightarrow \frac{1}{\sqrt{|x| - x}} = \frac{1}{\sqrt{- x - x}} = \frac{1}{\sqrt{- 2 x}} (exists because when x<0, -2x>0) ⇒ f (x) is defined when x < 0 So, domain = (- \infty, 0)$

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Question 31:

Write whether f : R → R, given by $f (x) = x + \sqrt{x^{2}}$ , is one-one, many-one, onto or into.

Answer:

$f (x) = x + \sqrt{x^{2}} = x \pm x = 0 o r 2 x So, each element x in the domain may contain 2 images. For example, f (0) = 0 + \sqrt{0^{2}} = 0 f (- 1) = - 1 + \sqrt{{(- 1)}^{2}} = - 1 + \sqrt{1} = - 1 + 1 = 0 Here, the image of 0 and -1 is 0.$
Hence, f is may-one.

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Question 32:

If f(x) = x + 7 and g(x) = x − 7, x ∈ R, write fog (7).

Answer:

$(f o g) (7) = f (g (7)) = f (7 - 7) = f (0) = 0 + 7 = 7$

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Question 33:

What is the range of the function $f (x) = \frac{|x - 1|}{x - 1} ?$

Answer:

$f (x) = \frac{|x - 1|}{x - 1} = \frac{\pm (x - 1)}{x - 1} = \pm 1 Range of f = \{- 1, 1\}$

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Question 34:

If f : R → R be defined by f(x) = (3 − x³)1/3, then find fof (x).

Answer:

$(f o f) (x) = f (f (x)) = f ({(3 - x^{3})}^{\frac{1}{3}}) = {[3 - {({(3 - x^{3})}^{\frac{1}{3}})}^{3}]}^{\frac{1}{3}} = {[3 - (3 - x^{3})]}^{\frac{1}{3}} = {(x^{3})}^{\frac{1}{3}} = x$

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Question 35:

If f : R → R is defined by f(x) = 3x + 2, find f (f (x)).

Answer:

$f (f (x)) = f (3 x + 2) = 3 (3 x + 2) + 2 = 9 x + 6 + 2 = 9 x + 8$

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Question 36:

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. State whether f is one-one or not.

Answer:

f = {(1, 4), (2, 5), (3, 6)}

Here, different elements of the domain have different images in the co-domain.
So, f is one-one.

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Question 37:

If f : {5, 6} $\to$ {2, 3} and g : {2, 3} $\to$ {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, then find fog. [NCERT EXEMPLAR]

Answer:

We have,
f : {5, 6} $\to$ {2, 3} and g : {2, 3} $\to$ {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}

As,
fog(2) = f(g(2)) = f(5) = 2,
fog(3) = f(g(3)) = f(6) = 3,

So,
fog : {2, 3} $\to$ {2, 3} is defined as
fog = {(2, 2), (3, 3)}

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Question 38:

Let f : R $\to$ R be the function defined by f(x) = 4x $-$ 3 for all x $\in$ R. Then write f ^{$-$ 1}. [NCERT EXEMPLAR]

Answer:

We have,
f : R $\to$ R is the function defined by f(x) = 4x $-$ 3 for all x $\in$ R

$Let f (x) = y . Then, y = 4 x - 3 \Rightarrow 4 x = y + 3 \Rightarrow x = \frac{y + 3}{4} So, f^{- 1} (y) = \frac{y + 3}{4} or, f^{- 1} (x) = \frac{x + 3}{4}$

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Question 39:

Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)} [NCERT EXEMPLAR]

Answer:

As, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}

So, f is a function.

Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}

So, g is not a function.

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Question 40:

Write the domain of the real function f defined by f(x) = $\sqrt{25 - x^{2}}$ . [NCERT EXEMPLAR]

Answer:

$We have, f (x) = \sqrt{25 - x^{2}} The function is defined only when 25 - x^{2} \geq 0 \Rightarrow x^{2} - 25 \leq 0 \Rightarrow (x + 5) (x - 5) \leq 0 \Rightarrow x \in [- 5, 5] So, the domain of the given function is [- 5, 5] .$

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Question 41:

Let A = {a, b, c, d} and f : A $\to$ A be given by f = {(a, b), (b, d), (c, a), (d, c)}. Write f ^{$-$ 1}. [NCERT EXEMPLAR]

Answer:

We have,

A = {a, b, c, d} and f : A $\to$ A be given by f = {(a, b), (b, d), (c, a), (d, c)}

Since, the elements of a function when interchanged gives inverse function.

So, f ^{$-$ 1} = {(b, a), (d, b), (a, c), (c, d)}

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Question 42:

Let f, g : R $\to$ R be defined by f(x) = 2x + l and g(x) = x² $-$ 2 for all x $\in$ R, respectively. Then, find gof. [NCERT EXEMPLAR]

Answer:

We have,

f, g : R $\to$ R are defined by f(x) = 2x + l and g(x) = x² $-$ 2 for all x $\in$ R, respectively

$Now, g o f (x) = g (f (x)) = g (2 x + 1) = {(2 x + 1)}^{2} - 2 = 4 x^{2} + 4 x + 1 - 2 = 4 x^{2} + 4 x - 1$

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Question 43:

If the mapping f : {1, 3, 4} $\to$ {1, 2, 5} and g : {1, 2, 5} $\to$ {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, then write fog. [NCERT EXEMPLAR]

Answer:

We have,

f : {1, 3, 4} $\to$ {1, 2, 5} and g : {1, 2, 5} $\to$ {1, 3}, are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, respectively

As,

$f o g (2) = f (g (2)) = f (3) = 5, f o g (5) = f (g (5)) = f (1) = 2, f o g (1) = f (g (1)) = f (3) = 5, So, f o g : \{1, 2, 5\} \to \{1, 2, 5\} is given by f o g = \{(2, 5), (5, 2), (1, 5)\}$

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Question 44:

If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = $α x + β$ , then find the values of $α$ and $β$ . [NCERT EXEMPLAR]

Answer:

We have,

A function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = $α x + β$

$As, g (1) = 1 and g (2) = 3 So, α (1) + β = 1 \Rightarrow α + β = 1 . . . . . (i) and α (2) + β = 3 \Rightarrow 2 α + β = 3 . . . . . (ii) (ii) - (i), we get 2 α - α = 2 \Rightarrow α = 2 Substituting α = 2 in (i), we get 2 + β = 1 \Rightarrow β = - 1$

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Question 45:

If f(x) = 4 $-$ (x $-$ 7)³, then write f ^{$-$ 1}(x). [NCERT EXEMPLAR]

Answer:

$We have, f (x) = 4 - {(x - 7)}^{3} Let y = 4 - {(x - 7)}^{3} \Rightarrow {(x - 7)}^{3} = 4 - y \Rightarrow x - 7 = \sqrt[3]{4 - y} \Rightarrow x = 7 + \sqrt[3]{4 - y} \Rightarrow f^{- 1} (y) = 7 + \sqrt[3]{4 - y} ∴ f^{- 1} (x) = 7 + \sqrt[3]{4 - x}$

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