Question 40:

Prove that $f (x) = \{\begin{cases} \frac{x - |x|}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases}$ is discontinuous at x = 0

Answer:

The given function can be rewritten as

$f (x) = \{\begin{matrix} \frac{x - x}{x}, when x > 0 \\ \frac{x + x}{x}, when x < 0 \\ 2, when x = 0 \end{matrix}$

$\Rightarrow$ $f (x) = \{\begin{matrix} 0, when x > 0 \\ 2, when x < 0 \\ 2, when x = 0 \end{matrix}$

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} 2 = 2$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} 0 = 0$

∴ $\lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x)$

Thus, f(x) is discontinuous at x = 0.

Page No 8.21:

Question 41:

If $f (x) = \{\begin{cases} 2 x^{2} + k, & if x \geq 0 \\ - 2 x^{2} + k, & if x < 0 \end{cases}$ , then what should be the value of
k so that f(x) is continuous at x = 0.

Answer:

The given function can be rewritten as

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

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} - 2 {(- h)}^{2} + k = k$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (2 h^{2} + k) = k$

If $f (x)$ is continuous at $x = 0$ , then

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \Rightarrow \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = k$

∴ k can be any real number.

Page No 8.21:

Question 42:

For what value of λ is the function
$f (x) = \{\begin{cases} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{cases}$
continuous at x = 0? What about continuity at x = ± 1?

Answer:

The given function f is

If f is continuous at x = 0, then

Therefore, there is no value of λ for which f(x) is continuous at x = 0.

At x = 1,

f (1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1

Discuss the continuity of the function $f (x) = \{\begin{array}{l} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{array} .$

Answer:

Given: $f (x) = \{\begin{cases} \frac{x}{|x|}, x \neq 0 \\ 0, x = 0 \end{cases}$

$|x| = \{\begin{cases} x, x \geq 0 \\ - x, x < 0 \end{cases} \Rightarrow f (x) = \{\begin{matrix} 1, x > 0 \\ - 1, x < 0 \\ 0, x = 0 \end{matrix}$

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} (- 1) = - 1$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (1) = 1$

∴ $\lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x)$

Thus, $f (x)$ is discontinuous at x = 0.

Page No 8.34:

Question 3:

Find the points of discontinuity, if any, of the following functions:
(i) $f (x) = \{\begin{cases} x^{3} - x^{2} + 2 x - 2, & if x \neq 1 \\ 4, & if x = 1 \end{cases}$

(ii) $f (x) = \{\begin{cases} \frac{x^{4} - 16}{x - 2}, & if x \neq 2 \\ 16, & if x = 2 \end{cases}$

(iii) $f (x) = \{\begin{cases} \frac{\sin x}{x}, & if x < 0 \\ 2 x + 3, & x \geq 0 \end{cases}$

(iv) $f (x) = \{\begin{cases} \frac{\sin 3 x}{x}, & if x \neq 0 \\ 4, & if x = 0 \end{cases}$

(v) $f (x) = \{\begin{cases} \frac{\sin x}{x} + \cos x, & if x \neq 0 \\ 5, & if x = 0 \end{cases}$

(vi) $f (x) = \{\begin{cases} \frac{x^{4} + x^{3} + 2 x^{2}}{\tan^{- 1} x}, & if x \neq 0 \\ 10, & if x = 0 \end{cases}$

(vii) $f (x) = \{\begin{cases} \frac{e^{x} - 1}{\log_{e} (1 + 2 x)}, & if x \neq 0 \\ 7, & if x = 0 \end{cases}$

(viii) $f (x) = \{\begin{cases} |x - 3|, & if x \geq 1 \\ \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, & if x < 1 \end{cases}$

(ix) $f (x) = \{\begin{matrix} |x| + 3, & if x \leq - 3 \\ - 2 x, & if - 3 < x < 3 \\ 6 x + 2, & if x > 3 \end{matrix}$

(x) $f (x) = \{\begin{cases} x^{10} - 1, & if x \leq 1 \\ x^{2}, & if x > 1 \end{cases}$

(xi) $f (x) = \{\begin{matrix} 2 x, & if & x < 0 \\ 0, & if & 0 \leq x \leq 1 \\ 4 x, & if & x > 1 \end{matrix}$

(xii) $f (x) = \{\begin{cases} \sin x - \cos x, & if x \neq 0 \\ - 1, & if x = 0 \end{cases}$

(xiii) $f (x) = \{\begin{matrix} - 2, & if & x \leq - 1 \\ 2 x, & if & - 1 < x < 1 \\ 2, & if & x \geq 1 \end{matrix}$

Answer:

(i)

When x $\neq$ 1, then
$f (x) = x^{3} - x^{2} + 2 x - 2$

We know that a polynomial function is everywhere continuous.
So, $f (x) = x^{3} - x^{2} + 2 x - 2$ is continuous at each x $\neq$ 1.

At x = 1, we have

(LHL at x = 1) = $\lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} ({(1 - h)}^{3} - {(1 - h)}^{2} + 2 (1 - h) - 2) = 1 - 1 + 2 - 2 = 0$

(RHL at x = 1) = $\lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} ({(1 + h)}^{3} - {(1 + h)}^{2} + 2 (1 + h) - 2) = 1 - 1 + 2 - 2 = 0$

Also, $f (1) = 4$
∴ $\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) \neq f (1)$

Thus, $f (x)$ is discontinuous at x = 1.

Hence, the only point of discontinuity for $f (x)$ is x = 1.

(ii)
Given: $f (x) = \{\begin{cases} \frac{x^{4} - 16}{x - 2}, if x \neq 2 \\ 16, if x = 2 \end{cases}$

When x $\neq$ 2, then
$f (x) = \frac{x^{4} - 16}{x - 2} = \frac{x^{4} - 2^{4}}{x - 2} = \frac{(x^{2} + 4) (x - 2) (x + 2)}{x - 2} = (x^{2} + 4) (x + 2)$

We know that a polynomial function is everywhere continuous.
Therefore, the functions $(x^{2} + 4) and (x + 2)$ are everywhere continuous.
So, the product function $(x^{2} + 4) (x + 2)$ is everywhere continuous.
Thus, f(x) is continuous at every x $\neq$ 2.

At x = 2, we have

(LHL at x = 2) = $\lim_{x \to 2^{-}} f (x) = \lim_{h \to 0} f (2 - h) = \lim_{h \to 0} [{(2 - h)}^{2} + 4] (2 - h + 2) = 8 (4) = 32$

(RHL at x = 2) = $\lim_{x \to 2^{+}} f (x) = \lim_{h \to 0} f (2 + h) = \lim_{h \to 0} [{(2 + h)}^{2} + 4] (2 + h + 2) = 8 (4) = 32$

Also, $f (2) = 16$

∴ $\lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{+}} f (x) \neq f (2)$

Thus, $f (x)$ is discontinuous at x = 2.

Hence, the only point of discontinuity for $f (x)$ is x = 2.

(iii)

When x < 0, then
$f (x) = \frac{\sin x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\sin x}{x}$ is continuous at each x < 0.

When x > 0, then
$f (x) = 2 x + 3$ , which is a polynomial function.
Therefore, $f (x)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given: $f (x) = \{\begin{cases} \frac{\sin x}{x}, if x < 0 \\ 2 x + 3, if x \geq 0 \end{cases}$

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} (\frac{\sin (- h)}{- h}) = \lim_{h \to 0} (\frac{\sin (h)}{h}) = 1$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (2 h + 3) = 3$

∴ $\lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x)$

Thus, $f (x)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f (x)$ is x = 0.

(iv)

When x $\neq$ 0, then
$f (x) = \frac{\sin 3 x}{x}$

We know that sin 3x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\sin 3 x}{x}$ is continuous at each x $\neq$ 0.

Let us consider the point x = 0.
Given: $f (x) = \{\begin{cases} \frac{\sin 3 x}{x}, if x \neq 0 \\ 4, if x = 0 \end{cases}$

We have

(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} (\frac{\sin (- 3 h)}{- h}) = \lim_{h \to 0} (\frac{3 \sin (3 h)}{3 h}) = 3$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (\frac{\sin (3 h)}{h}) = \lim_{h \to 0} (\frac{3 \sin (3 h)}{3 h}) = 3$
Also, $f (0) = 4$

∴ $\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) \neq f (0)$
Thus, $f (x)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f (x)$ is x = 0.

(v)

When x $\neq$ 0, then
$f (x) = \frac{\sin x}{x} + \cos x$

We know that sin x as well as the identity function x both are everywhere continuous.
So, the quotient function $\frac{\sin x}{x}$ is continuous at each x $\neq$ 0.
Also, cos x is everywhere continuous.
Therefore, $\frac{\sin x}{x} + \cos x$ is continuous at each x $\neq$ 0.

Let us consider the point x = 0.
Given: $f (x) = \{\begin{cases} \frac{\sin x}{x} + \cos x, if x \neq 0 \\ 5, if x = 0 \end{cases}$

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} (\frac{\sin (- h)}{- h} + \cos (- h)) = \lim_{h \to 0} (\frac{\sin (- h)}{- h}) + \lim_{h \to 0} \cos (- h) = 1 + 1 = 2$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (\frac{\sin (h)}{h} + \cos (h)) = \lim_{h \to 0} (\frac{\sin (h)}{h}) + \lim_{h \to 0} \cos (h) = 1 + 1 = 2$
Also, $f (0) = 5$

∴ $\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) \neq f (0)$
Thus, $f (x)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f (x)$ is x = 0.

(vi)

When x $\neq$ 0, then
$f (x) = \frac{x^{4} + x^{3} + 2 x^{2}}{\tan^{- 1} x}$

We know that $x^{4} + x^{3} + 2 x^{2}$ is a polynomial function which is everywhere continuous.
Also, $\tan^{- 1} x$ is everywhere continuous.
So, the quotient function $\frac{x^{4} + x^{3} + 2 x^{2}}{\tan^{- 1} x}$ is continuous at each x $\neq$ 0.

Let us consider the point x = 0.

Given: $f (x) = \{\begin{cases} \frac{x^{4} + x^{3} + 2 x^{2}}{\tan^{- 1} x}, if x \neq 0 \\ 10, if x = 0 \end{cases}$

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} (\frac{{(- h)}^{4} + {(- h)}^{3} + 2 {(- h)}^{2}}{\tan^{- 1} (- h)}) = \lim_{h \to 0} (\frac{{(h)}^{3} - {(h)}^{2} + 2 (h)}{- \frac{\tan^{- 1} (h)}{h}}) = \frac{0}{(- 1)} = 0$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (\frac{{(h)}^{4} + {(h)}^{3} + 2 {(h)}^{2}}{\tan^{- 1} (h)}) = \lim_{h \to 0} (\frac{{(h)}^{3} + {(h)}^{2} + 2 (h)}{\frac{\tan^{- 1} (h)}{h}}) = \frac{0}{1} = 0$
Also, $f (0) = 10$

∴ $\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) \neq f (0)$
Thus, $f (x)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f (x)$ is x = 0.

(vii)
Given: $f (x) = \{\begin{cases} \frac{e^{x} - 1}{\log_{e} (1 + 2 x)}, if x \neq 0 \\ 7, if x = 0 \end{cases}$

We have
$\lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{e^{x} - 1}{\log_{e} (1 + 2 x)} = \lim_{x \to 0} \frac{(\frac{e^{x} - 1}{x})}{(\frac{2 \log_{e} (1 + 2 x)}{2 x})} = \frac{1}{2} \times \frac{\lim_{x \to 0} (\frac{e^{x} - 1}{x})}{\lim_{x \to 0} (\frac{\log_{e} (1 + 2 x)}{2 x})} = \frac{1}{2}$
It is given that $f (0) = 7$

⇒ $\lim_{x \to 0} f (x) \neq f (0)$

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.

(viii)

When x > 1, then
$f (x) = |x - 3|$

Since modulus function is a continuous function, $f (x)$ is continuous for each x > 1.

When x < 1, then
$f (x) = \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}$

Since, $x^{2} & 3 x$ are continuous being polynomial functions, $\frac{x^{2}}{4} & \frac{3 x}{2}$ will also be continuous.
Also, $\frac{13}{4}$ is continuous being a polynomial function.

$\Rightarrow \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}$ is continuous for each $x < 1$ .

$\Rightarrow f (x)$ is continuous for each x < 1.

At x = 1, we have
(LHL at x=1) = $\lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} [\frac{{(1 - h)}^{2}}{4} - \frac{3 (1 - h)}{2} + \frac{13}{4}] = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = 2$

(RHL at x=1) = $\lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} [|1 + h - 3|] = |- 2| = 2$
Also, $f (1) = |1 - 3| = |- 2| = 2$

Thus, $\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1)$

Hence, $f (x)$ is continuous at x= 1.

Thus, the given function is nowhere discontinuous.

(ix)

At $x \leq - 3$ , we have
$f (x) = |x| + 3$

Since modulus function and constant function are continuous, $f (x) = |x| + 3$ is continuous for each $x \leq - 3$ .

At $- 3 < x < 3$ , we have
$f (x) = - 2 x$
Since polynomial function is continuous and constant function is continuous, $f (x) = - 2 x$ is continuous for each $- 3 < x < 3$ .

At $x > 3$ , we have
$f (x) = 6 x + 2$

Since polynomial function is continuous and constant function is continuous, $f (x) = 6 x + 2$ is continuous for each $x > 3$ .

Now, we check the continuity of the function at the point $x = 3$ .

We have
(LHL at x=3) = $\lim_{x \to 3^{-}} f (x) = \lim_{h \to 0} f (3 - h) = \lim_{h \to 0} - 2 (3 - h) = - 6$

(RHL at x=3) = $\lim_{x \to 3^{+}} f (x) = \lim_{h \to 0} f (3 + h) = \lim_{h \to 0} 6 (3 + h) + 2 = 20$
$\Rightarrow \lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x)$

Hence, the only point of discontinuity of the given function is $x = 3$

(x)
Given:

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

(xi) The given function is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 0

Case II:

The left hand limit of f at x = 0 is,

The right hand limit of f at x = 0 is,

Therefore, f is continuous at x = 0

Case III:

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

Therefore, f is continuous at all points x, such that x > 1

Hence, f is not continuous only at x = 1

(xii)

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

Therefore, f is continuous at all points x, such that x ≠ 0

Case II:

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

(xiii)
The given function f is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < −1

Case II:

The left hand limit of f at x = −1 is,

The right hand limit of f at x = −1 is,

Therefore, f is continuous at x = −1

Case III:

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

Therefore, f is continuous at x = 2

Discuss the continuity of f(x) = sin | x |.

Answer:

This function f is defined for every real number and f can be written as the composition of two functions as,

f = h o g, where

$[∵ h o g (x) = h (g (x)) = h (|x|) = \sin |x|]$

It has to be proved first that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

Therefore, g is continuous at all points x < 0

Case II:

Therefore, g is continuous at all points x > 0

Case III:

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = sin x

It is evident that h (x) = sin x is defined for every real number.

Let c be a real number.
Put x = c + k

If x → c, then k → 0

h (c) = sin c

So, h is a continuous function.

$∴ f (x) = h o g (x) = h (g (x)) = h (|x|) = \sin |x|$ is a continuous function.

Page No 8.37:

Question 11:

Prove that
$f (x) = \{\begin{cases} \frac{\sin x}{x}, & x < 0 \\ x + 1, & x \geq 0 \end{cases}$
is everywhere continuous.

Answer:

When x < 0, we have
$f (x) = \frac{\sin x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\sin x}{x}$ is continuous at each x < 0.

When x > 0, we have
$f (x) = x + 1$ , which is a polynomial function.
Therefore, $f (x)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given: $f (x) = \{\begin{cases} \frac{\sin x}{x}, x < 0 \\ x + 1, x \geq 0 \end{cases}$

We have
(LHL at x = 0) = $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} f (- h) = \lim_{h \to 0} (\frac{\sin (- h)}{- h}) = \lim_{h \to 0} (\frac{\sin (h)}{h}) = 1$

(RHL at x = 0) = $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} f (h) = \lim_{h \to 0} (h + 1) = 1$
Also,
$f (0) = 0 + 1 = 1$

∴ $\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0)$

Thus, $f (x)$ is continuous at x = 0.

Hence, $f (x)$ is everywhere continuous.

Page No 8.37:

Question 12:

Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Answer:

Given:

It is evident that g is defined at all integral points.

Let $n \in Z$ .

Then,

The left hand limit of f at x = n is,

The right hand limit of f at x = n is,

It is observed that the left and right hand limits of f at x = n do not coincide.
i.e. $\lim_{x \to n^{-}} g (x) \neq \lim_{x \to n^{+}} g (x)$

So, f is not continuous at x = n, $n \in Z$

Hence, g is discontinuous at all integral points.

Page No 8.37:

Question 13:

Discuss the continuity of the following functions:
(i) f(x) = sin x + cos x
(ii) f(x) = sin x − cos x
(iii) f(x) = sin x cos x

Answer:

It is known that if g and h are two continuous functions, then $g + h, g - h and g \times h$ are also continuous.

It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x → c, then h → 0

So, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x → c, then h → 0

h (c) = cos c

So, h is a continuous function.

Therefore, it can be concluded that

(i) f (x) = g (x) + h (x) = sin x + cos x is a continuous function.

(ii) f (x) = g (x) − h (x) = sin x − cos x is a continuous function.

(iii) f (x) = g (x) $\times$ h (x) = sin x cos x is a continuous function.

Page No 8.37:

Question 14:

Show that f (x) = cos x² is a continuous function.

Answer:

Given: f (x) = cos (x²)

This function f is defined for every real number and f can be written as the composition of two functions as

f = g o h, where g (x) = cos x and h (x) = x²

It has to be first proved that g (x) = cos x and h (x) = x² are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g (c) = cos c

So, g (x) = cos x is a continuous function.

Now,
h (x) = x²

Clearly, h is defined for every real number.

Let k be a real number, then h (k) = k²

So, h is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (f o g) is continuous at x = c.

Therefore, is a continuous function.

Page No 8.37:

Question 15:

Show that f (x) = | cos x | is a continuous function.

Answer:

The given function is

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where

It has to be first proved that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number.
Put x = c + h

If x → c, then h → 0

h (c) = cos c

So, h (x) = cos x is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then (f o g) is continuous at x = c.

Therefore, is a continuous function.

Page No 8.37:

Question 16:

Find all the points of discontinuity of f defined by f (x) = | x | − | x + 1 |.

Answer:

Given:

The two functions, g and h, are defined as

Then, f = g − h

The continuity of g and h is examined first.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Clearly, h is defined for every real number.

Let c be a real number.

Case I:

So, h is continuous at all points x < −1.

Case II:

So, h is continuous at all points x > −1.

Case III:

So, h is continuous at x = −1

From the above three observations, it can be concluded that h is continuous at all points of the real line.

So, g and h are continuous functions.

Thus, f = g − h is also a continuous function.

Therefore, f has no point of discontinuity.

Page No 8.37:

Question 17:

Determine if $f (x) = \{\begin{cases} x^{2} \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is a continuous function?

Answer:

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

So, f is continuous at all points x ≠ 0

Case II:

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (x^{2} \sin \frac{1}{x}) = \lim_{x \to 0} (x^{2} \sin \frac{1}{x}) It is known that - 1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 . \Rightarrow - x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \Rightarrow \lim_{x \to 0} (- x^{2}) \leq \lim_{x \to 0} (x^{2} \sin \frac{1}{x}) \leq \lim_{x \to 0} x^{2} \Rightarrow 0 \leq \lim_{x \to 0} (x^{2} \sin \frac{1}{x}) \leq 0 \Rightarrow \lim_{x \to 0} (x^{2} \sin \frac{1}{x}) = 0 \Rightarrow \lim_{x \to 0^{-}} f (x) = 0 Similarly, \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (x^{2} \sin \frac{1}{x}) = \lim_{x \to 0} (x^{2} \sin \frac{1}{x}) = 0$

If $f (x) = \{\begin{matrix} \frac{\sin (a + 1) x + \sin x}{x}, & x < 0 \\ c, & x = 0 \\ \frac{\sqrt{x + b x^{2}} - \sqrt{x}}{b x \sqrt{x}}, & x > 0 \end{matrix}$ is continuous at x = 0, then
(a) a = $- \frac{3}{2}$ , b = 0, c = $\frac{1}{2}$
(b) a = $- \frac{3}{2}$ , b = 1, c = $- \frac{1}{2}$
(c) a = $- \frac{3}{2}$ , b ∈ R − {0}, c = $\frac{1}{2}$
(d) none of these

Question 38:

The points of discontinuity of the function
$f (x) = \{\begin{matrix} 2 \sqrt{x}, & 0 \leq x \leq 1 \\ 4 - 2 x, & 1 < x < \frac{5}{2} \\ 2 x - 7, & \frac{5}{2} \leq x \leq 4 \end{matrix} is (are)$

(a) x = 1, $x = \frac{5}{2}$

(b) $x = \frac{5}{2}$

(c) $x = 1, \frac{5}{2}, 4$

(d) x = 0, 4

Answer:

$(b x = \frac{5}{2} = 5If0 \leq x \leq 1, thenf (x) = 2 \sqrt{x}.Sincef (x) = 2 \sqrt{x}is a polynomial function, it is continuous.Thus,f (x)is continuous for every0 \leq x \leq 1.If 1 < x < \frac{5}{2}, then f (x) = 4 - 2 x . Since 2 x is a polynomial function and 4 is a constant function, both of them are continuous. So, their difference will also be continuous. Thus, f (x) is continuous for every 1 < x < \frac{5}{2} . If \frac{5}{2} \leq x \leq 4, then f (x) = 2 x - 7 . Since 2 x is a polynomial function and 7 is continuous function, their difference will also be continuous. Thus, f (x) is continuous for every \frac{5}{2} \leq x \leq 4 . Now, Consider the point x = 1 . Here, \lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} (2 (\sqrt{1 - h})) = 2 \lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} (4 - 2 (1 + h)) = 2 Also, f (1) = 2 \sqrt{1} = 2 \Rightarrow \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) Thus, f (x) is continuous at x = 1 . Now, Consider the point x = \frac{5}{2} . Here, \lim_{x \to {\frac{5}{2}}^{-}} f (x) = \lim_{h \to 0} f (\frac{5}{2} - h) = \lim_{h \to 0} (4 - 2 (\frac{5}{2} - h)) = - 1 \lim_{x \to {\frac{5}{2}}^{+}} f (x) = \lim_{h \to 0} f (\frac{5}{2} + h) = \lim_{h \to 0} (2 (\frac{5}{2} - h) - 7) = - 2 \Rightarrow \lim_{x \to {\frac{5}{2}}^{+}} f (x) \neq \lim_{x \to {\frac{5}{2}}^{-}} f (x) Thus, f (x) is discontinuous at x = \frac{5}{2} .Page No 8.46:Question 39: If f (x) = \{\begin{matrix} \frac{1 - \sin^{2} x}{3 \cos^{2} x}, & x < \frac{π}{2} \\ a, & x = \frac{π}{2} \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > \frac{π}{2} \end{matrix} . Then, f (x) is continuous at x = \frac{π}{2}, if (a) a = \frac{1}{3}, b = 2 (b) a = \frac{1}{3}, b = \frac{8}{3} (c) a = \frac{2}{3}, b = \frac{8}{3} (d) none of theseAnswer: (b) a = \frac{1}{3}, b = \frac{8}{3} Given: f (x) = \{\begin{matrix} \frac{1 - \sin^{2} x}{3 \cos^{2} x}, if x < \frac{π}{2} \\ a, if x = \frac{π}{2} \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, if x > \frac{π}{2} \end{matrix} We have (LHL at x = \frac{π}{2}) = \lim_{x \to {\frac{π}{2}}^{-}} f (x) = \lim_{h \to 0} f (\frac{π}{2} - h) = \lim_{h \to 0} (\frac{1 - \sin^{2} (\frac{π}{2} - h)}{3 \cos^{2} (\frac{π}{2} - h)}) = \lim_{h \to 0} (\frac{1 - \cos^{2} h}{3 \sin^{2} h}) = \frac{1}{3} \lim_{h \to 0} (\frac{\sin^{2} h}{\sin^{2} h}) = \frac{1}{3} (RHL at x = \frac{π}{2}) = \lim_{x \to {\frac{π}{2}}^{+}} f (x) = \lim_{h \to 0} f (\frac{π}{2} + h) = \lim_{h \to 0} (\frac{b [1 - \sin (\frac{π}{2} + h)]}{{[π - 2 (\frac{π}{2} + h)]}^{2}}) = \lim_{h \to 0} (\frac{b (1 - \cos h)}{{[- 2 h]}^{2}}) = \lim_{h \to 0} (\frac{2 b \sin^{2} \frac{h}{2}}{4 h^{2}}) = \lim_{h \to 0} (\frac{2 b \sin^{2} \frac{h}{2}}{16 \frac{h^{2}}{4}}) = \frac{b}{8} \lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} = \frac{b}{8} \times 1 = \frac{b}{8} Also, f (\frac{π}{2}) = a If f (x) is continuous at x = \frac{π}{2}, then \lim_{x \to {\frac{π}{2}}^{-}} f (x) = \lim_{x \to {\frac{π}{2}}^{+}} f (x) = f (\frac{π}{2}) \Rightarrow \frac{1}{3} = \frac{b}{8} = a \Rightarrow a = \frac{1}{3} and b = \frac{8}{3}Page No 8.46:Question 40: The points of discontinuity of the function f (x) = \{\begin{matrix} \frac{1}{5} (2 x^{2} + 3), & x \leq 1 \\ 6 - 5 x, & 1 < x < 3 \\ x - 3, & x \geq 3 \end{matrix} is (are) (a) x = 1 (b) x = 3 (c) x = 1, 3 (d) none of theseAnswer: (b) x = 3 If x \leq 1, then f (x) = \frac{1}{5} (2 x^{2} + 3) . Since 2 x^{2} + 3 is a polynomial function and \frac{1}{5} is a constant function, both of them are continuous. So, their product will also be continuous. Thus, f (x) is continuous at x \leq 1 . If 1 < x < 3, then f (x) = 6 - 5 x . Since 5 x is a polynomial function and 6 is a constant function, both of them are continuous. So, their difference will also be continuous. Thus, f (x) is continuous for every 1 < x < 3 . If x \geq 3, then f (x) = x - 3 . Since x - 3 is a polynomial function, it is continuous. So, f (x) is continuous for every x \geq 3 . Now, Consider the point x = 1 . Here, \lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} (\frac{1}{5} [2 {(1 - h)}^{2} + 3]) = 1 \lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} (6 - 5 (1 + h)) = 1 Also, f (1) = \frac{1}{5} (2 {(1)}^{2} + 3) = 1 Thus, \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) Hence, f (x) is continuous at x = 1 . Now, Consider the point x = 3 . Here, \lim_{x \to 3^{-}} f (x) = \lim_{h \to 0} f (3 - h) = \lim_{h \to 0} (6 - 5 (3 - h)) = - 9 \lim_{x \to 3^{+}} f (x) = \lim_{h \to 0} f (3 + h) = \lim_{h \to 0} ((3 + h) - 3) = 0 Also, f (1) = \frac{1}{5} (2 {(1)}^{2} + 3) = 1 Thus, \lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x) Hence, f (x) is discontinuous at x = 3 . So, the only point of discontinuity of f (x) is x = 3 .Page No 8.46:Question 41: The value of a for which the function f (x) = \{\begin{cases} 5 x - 4, & if 0 < x \leq 1 \\ 4 x^{2} + 3 a x, & if 1 < x < 2 \end{cases} is continuous at every point of its domain, is (a) \frac{13}{3} (b) 1 (c) 0 (d) −1Answer: (d) −1 Given: f (x) = \{\begin{cases} 5 x - 4, if 0 < x \leq 1 \\ 4 x^{2} + 3 a x, if 1 < x < 2 \end{cases} If f (x) is continuous in its domain, then it will be continuous at x = 1 . Now, \lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} [5 (1 - h) - 4] = 5 - 4 = 1 \lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} [4 {(1 + h)}^{2} + 3 a (1 + h)] = 4 + 3 a Since f(x) is continuous at x = 1, \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) \Rightarrow 4 + 3 a = 1 \Rightarrow 3 a = - 3 \Rightarrow a = - 1Page No 8.46:Question 42: If f (x) = \{\begin{cases} \frac{\sin (\cos x) - \cos x}{{(π - 2 x)}^{2}}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{cases} is continuous at x = π/2, then k is equal to (a) 0 (b) \frac{1}{2} (c) 1 (d) −1Answer: (a) 0 Given: f (x) = \{\begin{cases} \frac{\sin (\cos x) - \cos x}{{(π - 2 x)}^{2}}, x \neq \frac{π}{2} \\ k, x = \frac{π}{2} \end{cases} If f (x) is continuous at x = \frac{π}{2}, then \lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \Rightarrow \lim_{x \to \frac{π}{2}} \frac{\sin (\cos x) - \cos x}{{(π - 2 x)}^{2}} = k Now, \frac{π}{2} - x = y \Rightarrow π - 2 x = 2 y Also, x \to \frac{π}{2}, y \to 0 \Rightarrow \lim_{y \to 0} \frac{\sin (\cos (\frac{π}{2} - y)) - \cos (\frac{π}{2} - y)}{4 y^{2}} = k \Rightarrow \lim_{y \to 0} \frac{\sin (\sin y) - \sin (y)}{4 y^{2}} = k \Rightarrow \lim_{y \to 0} \frac{2 \sin (\frac{\sin y - y}{2}) \cos (\frac{\sin y + y}{2})}{4 y^{2}} = k [∵ \sin C - \sin D = 2 s in (\frac{C - D}{2}) \cos (\frac{C + D}{2})] \Rightarrow \frac{1}{2} \lim_{y \to 0} \frac{\sin (\frac{\sin y - y}{2})}{y} \frac{\cos (\frac{\sin y + y}{2})}{y} = k \Rightarrow \frac{1}{2} \lim_{y \to 0} \frac{(\frac{\sin y - y}{2}) \sin (\frac{\sin y - y}{2})}{y (\frac{\sin y - y}{2})} \frac{\cos (\frac{\sin y + y}{2})}{y} = k \Rightarrow \frac{1}{2} \lim_{y \to 0} (\frac{(\frac{\sin y - y}{2})}{y}) (\frac{\sin (\frac{\sin y - y}{2})}{(\frac{\sin y - y}{2})}) (\frac{\cos (\frac{\sin y + y}{2})}{y}) = k \Rightarrow \frac{1}{2} \lim_{y \to 0} (\frac{(\frac{\sin y - y}{2})}{y}) \lim_{y \to 0} (\frac{\sin (\frac{\sin y - y}{2})}{(\frac{\sin y - y}{2})}) \lim_{y \to 0} (\frac{\cos (\frac{\sin y + y}{2})}{y}) = k \Rightarrow \frac{1}{4} \lim_{y \to 0} (\frac{\sin y}{y} - 1) \lim_{y \to 0} (\frac{\sin (\frac{\sin y - y}{2})}{(\frac{\sin y - y}{2})}) \lim_{y \to 0} (\frac{\cos (\frac{\sin y + y}{2})}{y}) = k \Rightarrow \frac{1}{4} \times 0 \times 1 \times \lim_{y \to 0} (\frac{\cos (\frac{\sin y + y}{2})}{y}) = k \Rightarrow 0 = k Page No 8.47:Question 43: If f (x) = 2 x and g (x) = \frac{x^{2}}{2} + 1, then which of the following can be a discontinuous function (a) f (x) + g (x) (b) f (x) – g (x) (c) f (x) g (x) (d) \frac{g (x)}{f (x)}Answer: f (x) = 2 x and g (x) = \frac{x^{2}}{2} + 1 are polynomial functions. We know polynomial functions are continuous for all values of x . So, f (x) = 2 x and g (x) = \frac{x^{2}}{2} + 1 are continuous functions. Also, sum, difference and product of continuous functions is continuous functions. ∴ f (x) + g (x), f (x) − g (x) and f (x) g (x) are continuous functions. Now, \frac{g (x)}{f (x)} is continuous if f (x) ≠ 0. But, f (x) = 2 x = 0 when x = 0. So, \frac{g (x)}{f (x)} is discontinuous at x = 0. Thus, \frac{g (x)}{f (x)} can be a discontinuous function. Hence, the correct answer is option (d).Page No 8.47:Question 44: The function f (x) = cot x is discontinuous on the set (a) {x : x = n π, n ∈ Z} (b) {x : x = 2 n π, n ∈ Z} (C) \{x : x = (2 n + 1) \frac{π}{2}, n \in Z\} (d) \{x : x = \frac{n π}{2}, n \in Z\}Answer: f (x) = cot x = \frac{\cos x}{\sin x} Now, f (x) is discontinuous when sin x = 0. sin x = 0 ⇒ x = n π, n ∈ Z So, f (x) = cot x is discontinuous on the set {x : x = n π, n ∈ Z} Hence, the correct answer is option (a).Page No 8.47:Question 45: If f (x) = x^{2} \sin \frac{1}{x}, where x ≠ 0, then the value of the function f at x = 0, so that the function is continuous at x = 0, is (a) 0 (b) –1 (c) 1 (d) noneAnswer: The given function is f (x) = x^{2} \sin \frac{1}{x}, where x ≠ 0. Now, f (x) is continuous at x = 0. ∴ f (0) = \lim_{x \to 0} f (x) \Rightarrow f (0) = \lim_{x \to 0} x^{2} \sin \frac{1}{x} \Rightarrow f (0) = \lim_{x \to 0} x^{2} \times \lim_{x \to 0} \sin \frac{1}{x} ⇒ f (0) = 0 × a finite value between − 1 and 1 (- 1 \leq \sin \frac{1}{x} \leq 1) ⇒ f (0) = 0 Thus, the value of the function f at x = 0 so that the function is continuous at x = 0 is 0. Hence, the correct answer is option (a).Page No 8.47:Question 46: The number of points at which the function f (x) = \frac{1}{x - [x]} is not continuous is (a) 1 (b) 2 (c) 3 (d) none of theseAnswer: The function f (x) = \frac{1}{x - [x]} is discontinuous when x − [x] = 0. x − [x] = 0 ⇒ x = [x] ⇒ x is an integer So, f (x) is not continuous for all x ∈ Z. Thus, the function f (x) = \frac{1}{x - [x]} is not continuous at infinitely many points. Hence, the correct answer is option (d).Page No 8.47:Question 47: The function f (x) = [x] is continuous at (a) 4 (b) –2 (c) 1 (d) 1.5Answer: The graph of f (x) = [x] is shown below. It can be seen that, the function f (x) = [x] is discontinuous at all integral values of x . It is continuous at all points except the integer points. Thus, the function f (x) = [x] is continuous at x = 1.5 and discontinuous at x = 4, x = –2 and x = 1. Hence, the correct answer is option (d).Page No 8.47:Question 48: The function f (x) = \{\begin{array}{lc} \frac{\sin x}{x} + \cos x, & if x \neq 0 \\ k, & if x = 0 \end{array} is continuous at x = 0, then the value of k is (a) 3 (b) 2 (c) 1 (d) 1.5Answer: It is given that, the function f (x) = \{\begin{array}{lc} \frac{\sin x}{x} + \cos x, & if x \neq 0 \\ k, & if x = 0 \end{array} is continuous at x = 0. ∴ f (0) = \lim_{x \to 0} f (x) \Rightarrow k = \lim_{x \to 0} (\frac{\sin x}{x} + \cos x) \Rightarrow k = \lim_{x \to 0} \frac{\sin x}{x} + \lim_{x \to 0} \cos x \Rightarrow k = 1 + 1 = 2 Thus, the value of k is 2. Hence, the correct answer is option (b).Page No 8.47:Question 1: If f (x) = \{\begin{matrix} \frac{x^{3} - a^{3}}{x - a}, & x \neq a \\ b, & x = a \end{matrix} is continuous at x = a, then b = ______________.Answer: It is given that, the function f (x) = \{\begin{matrix} \frac{x^{3} - a^{3}}{x - a}, & x \neq a \\ b, & x = a \end{matrix} is continuous at x = a . ∴ f (a) = \lim_{x \to a} f (x) \Rightarrow b = \lim_{x \to a} \frac{x^{3} - a^{3}}{x - a} \Rightarrow b = 3 a^{2} (\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}) If f (x) = \{\begin{matrix} \frac{x^{3} - a^{3}}{x - a}, & x \neq a \\ b, & x = a \end{matrix} is continuous at x = a, then b = 3 a^{2} .Page No 8.47:Question 2: If the function f (x) = \{\begin{matrix} \frac{\sin^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x = 0 \end{matrix} is continuous at x = 0, then a = _____________.Answer: The function f (x) = \{\begin{matrix} \frac{\sin^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x = 0 \end{matrix} is continuous at x = 0. ∴ \lim_{x \to 0} f (x) = f (0) \Rightarrow \lim_{x \to 0} \frac{\sin^{2} a x}{x^{2}} = 1 \Rightarrow a^{2} {(\lim_{x \to 0} \frac{\sin a x}{a x})}^{2} = 1 \Rightarrow a^{2} \times {(1)}^{2} = 1 (\lim_{x \to 0} \frac{\sin x}{x} = 1) \Rightarrow a^{2} = 1 \Rightarrow a = \pm 1 Thus, the value of a is ±1. If the function f (x) = \{\begin{matrix} \frac{\sin^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x = 0 \end{matrix} is continuous at x = 0, then a =___±1___.Page No 8.47:Question 3: If the function f (x) = \{\begin{matrix} a x^{2} - b, & 0 \leq x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \leq 2 \end{matrix} is continuous at x = 1, then a – b = _____________.Answer: The function f (x) = \{\begin{matrix} a x^{2} - b, & 0 \leq x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \leq 2 \end{matrix} is continuous at x = 1. ∴ f (1) = \lim_{x \to 1} f (x) \Rightarrow f (1) = \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) .....(1) Now, f (1) = 2 ....(2) \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1} (a x^{2} - b) = a \times {(1)}^{2} - b = a - b .....(3) \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} (x + 1) = 1 + 1 = 2 .....(4) From (1), (2), (3) and (4), we have 2 = a − b = 2 ∴ a − b = 2 Thus, the value of a − b is 2. If the function f (x) = \{\begin{matrix} a x^{2} - b, & 0 \leq x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \leq 2 \end{matrix} is continuous at x = 1, then a – b =____2____.Page No 8.48:Question 4: If f (x) = \{\begin{matrix} x + k, & x < 3 \\ 4, & x = 3 \\ 3 x - 5, & x > 3 \end{matrix} is continuous at x = 3, then k = _____________.Answer: The function f (x) = \{\begin{matrix} x + k, & x < 3 \\ 4, & x = 3 \\ 3 x - 5, & x > 3 \end{matrix} is continuous at x = 3. ∴ f (3) = \lim_{x \to 3} f (x) \Rightarrow f (3) = \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) .....(1) Now, f (3) = 4 .....(2) \lim_{x \to 3^{-}} f (x) = \lim_{x \to 3} (x + k) = 3 + k .....(3) \lim_{x \to 3^{+}} f (x) = \lim_{x \to 3} (3 x - 5) = 3 \times 3 - 5 = 9 - 5 = 4 .....(4) From (1), (2), (3) and (4), we have 4 = 3 + k = 4 ⇒ 3 + k = 4 ⇒ k = 4 − 3 = 1 Thus, the value of k is 1. If f (x) = \{\begin{matrix} x + k, & x < 3 \\ 4, & x = 3 \\ 3 x - 5, & x > 3 \end{matrix} is continuous at x = 3, then k =___1___.Page No 8.48:Question 5: If f (x) = \{\begin{matrix} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{matrix} . Then f (x) is continuous at x = 4, then a + b = _____________.Answer: The function f (x) = \{\begin{matrix} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{matrix} is continuous at x = 4. ∴ f (4) = \lim_{x \to 4} f (x) \Rightarrow f (4) = \lim_{x \to 4^{-}} f (x) = \lim_{x \to 4^{+}} f (x) .....(1) |x - 4| = \{\begin{cases} - (x - 4), & x < 4 \\ x - 4, & x \geq 4 \end{cases} Now, f (4) = a + b ..... (2) \lim_{x \to 4^{-}} f (x) = \lim_{x \to 4} [\frac{x - 4}{- (x - 4)} + a] = \lim_{x \to 4} \frac{x - 4}{- (x - 4)} + \lim_{x \to 4} a = - 1 + a .....(3) \lim_{x \to 4^{+}} f (x) = \lim_{x \to 4} [\frac{x - 4}{(x - 4)} + b] = \lim_{x \to 4} \frac{x - 4}{(x - 4)} + \lim_{x \to 4} b = 1 + b .....(4) From (1), (2), (3) and (4), we get a + b = −1 + a = 1 + b So, a + b = −1 + a ⇒ b = −1 Also, a + b = 1 + b ⇒ a = 1 ∴ a + b = 1 + (−1) = 0 Thus, the value of a + b is 0. If f (x) = \{\begin{matrix} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{matrix} . Then f (x) is continuous at x = 4, then a + b =___0___.Page No 8.48:Question 6: If f : R → R defined by f (x) = \{\begin{matrix} \frac{\cos 3 x - \cos x}{x^{2}}, & x \neq 0 \\ λ, & x = 0 \end{matrix} is continuous at x = 0, then λ = _____________.Answer: The function f (x) = \{\begin{matrix} \frac{\cos 3 x - \cos x}{x^{2}}, & x \neq 0 \\ λ, & x = 0 \end{matrix} is continuous at x = 0. ∴ f (0) = \lim_{x \to 0} f (x) \Rightarrow λ = \lim_{x \to 0} \frac{\cos 3 x - \cos x}{x^{2}} \Rightarrow λ = \lim_{x \to 0} \frac{- 2 \sin (\frac{3 x + x}{2}) \sin (\frac{3 x - x}{2})}{x^{2}} \Rightarrow λ = \lim_{x \to 0} \frac{- 2 \sin 2 x \sin x}{x^{2}} \Rightarrow λ = - 2 \times \lim_{x \to 0} \frac{\sin 2 x}{x} \times \lim_{x \to 0} \frac{\sin x}{x} \Rightarrow λ = - 2 \times 2 \lim_{x \to 0} \frac{\sin 2 x}{2 x} \times \lim_{x \to 0} \frac{\sin x}{x} \Rightarrow λ = - 2 \times (2 \times 1) \times 1 (\lim_{x \to 0} \frac{\sin x}{x} = 1) \Rightarrow λ = - 4 Thus, the value of λ is −4. If f : R → R defined by f (x) = \{\begin{matrix} \frac{\cos 3 x - \cos x}{x^{2}}, & x \neq 0 \\ λ, & x = 0 \end{matrix} is continuous at x = 0, then λ =____−4____.Page No 8.48:Question 7: If f (x) = \{\begin{matrix} \frac{1 - \sin x}{π - 2 x}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{matrix} is continuous at x = \frac{π}{2}, then k = _______________.Answer: The function f (x) = \{\begin{matrix} \frac{1 - \sin x}{π - 2 x}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{matrix} is continuous at x = \frac{π}{2} . ∴ f (\frac{π}{2}) = \lim_{x \to \frac{π}{2}} f (x) \Rightarrow k = \lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{π - 2 x} Put x = \frac{π}{2} - h When x \to \frac{π}{2}, h \to 0 ∴ k = \lim_{h \to 0} \frac{1 - \sin (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} \Rightarrow k = \lim_{h \to 0} \frac{1 - \cos h}{2 h} \Rightarrow k = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{2 h} \Rightarrow k = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{h} \times \lim_{h \to 0} \sin \frac{h}{2} \Rightarrow k = \lim_{h \to 0} \frac{\sin \frac{h}{2}}{2 \times \frac{h}{2}} \times \lim_{h \to 0} \sin \frac{h}{2} \Rightarrow k = \frac{1}{2} \times \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim_{h \to 0} \sin \frac{h}{2} \Rightarrow k = \frac{1}{2} \times 1 \times 0 \Rightarrow k = 0 Thus, the value of k is 0. If f (x) = \{\begin{matrix} \frac{1 - \sin x}{π - 2 x}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{matrix} is continuous at x = \frac{π}{2}, then k =____0____.Page No 8.48:Question 8: If f (x) = \frac{2 - \sqrt{x + 4}}{\sin 2 x}, x \neq 0, is continuous at x = 0, then f (0) = ______________.Answer: The function f (x) = \frac{2 - \sqrt{x + 4}}{\sin 2 x}, x \neq 0, is continuous at x = 0. ∴ f (0) = \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{2 - \sqrt{x + 4}}{\sin 2 x} = \lim_{x \to 0} \frac{2 - \sqrt{x + 4}}{\sin 2 x} \times \frac{2 + \sqrt{x + 4}}{2 + \sqrt{x + 4}} = \lim_{x \to 0} \frac{4 - (x + 4)}{\sin 2 x (2 + \sqrt{x + 4})} = \lim_{x \to 0} \frac{- x}{\sin 2 x (2 + \sqrt{x + 4})} = - \frac{1}{2} \times \lim_{x \to 0} \frac{2 x}{\sin 2 x (2 + \sqrt{x + 4})} = - \frac{1}{2} \times \frac{1}{\lim_{x \to 0} (\frac{\sin 2 x}{2 x})} \times \frac{1}{\lim_{x \to 0} (2 + \sqrt{x + 4})} = - \frac{1}{2} \times \frac{1}{1} \times \frac{1}{(2 + \sqrt{0 + 4})} (\lim_{x \to 0} \frac{\sin x}{x} = 1) = - \frac{1}{2} \times \frac{1}{4} = - \frac{1}{8} Thus, the value of f (0) is - \frac{1}{8} . If f (x) = \frac{2 - \sqrt{x + 4}}{\sin 2 x}, x \neq 0, is continuous at x = 0, then f (0) = - \frac{1}{8} .Page No 8.48:Question 9: If f (x) = \{\begin{matrix} \frac{x^{2} - 9}{x - 3}, & x \neq 3 \\ 2 x + k, & x = 3 \end{matrix} is continuous at x = 3, then k = ____________.Answer: The function f (x) = \{\begin{matrix} \frac{x^{2} - 9}{x - 3}, & x \neq 3 \\ 2 x + k, & x = 3 \end{matrix} is continuous at x = 3. ∴ f (3) = \lim_{x \to 3} f (x) \Rightarrow 2 \times 3 + k = \lim_{x \to 3} \frac{x^{2} - 9}{x - 3} \Rightarrow 6 + k = \lim_{x \to 3} \frac{(x - 3) (x + 3)}{x - 3} \Rightarrow 6 + k = \lim_{x \to 3} (x + 3) \Rightarrow 6 + k = 3 + 3 = 6 \Rightarrow k = 0 Thus, the value of k is 0. If f (x) = \{\begin{matrix} \frac{x^{2} - 9}{x - 3}, & x \neq 3 \\ 2 x + k, & x = 3 \end{matrix} is continuous at x = 3, then k =___0___.Page No 8.48:Question 10: If the function f (x) = \{\begin{matrix} \frac{x^{2} - 1}{x - 1}, & x \neq 1 \\ k, & x = 1 \end{matrix} is given to be continuous at x = 1, then the value of k is ____________.Answer: The function f (x) = \{\begin{matrix} \frac{x^{2} - 1}{x - 1}, & x \neq 1 \\ k, & x = 1 \end{matrix} is continuous at x = 1. ∴ f (1) = \lim_{x \to 1} f (x) \Rightarrow k = \lim_{x \to 1} \frac{x^{2} - 1}{x - 1} \Rightarrow k = \lim_{x \to 1} \frac{(x - 1) (x + 1)}{x - 1} \Rightarrow k = \lim_{x \to 1} (x + 1) \Rightarrow k = 1 + 1 = 2 Thus, the value of k is 2. If the function f (x) = \{\begin{matrix} \frac{x^{2} - 1}{x - 1}, & x \neq 1 \\ k, & x = 1 \end{matrix} is given to be continuous at x = 1, then the value of k is___2___.Page No 8.48:Question 11: The set of points where f (x) = x – [x] is discontinuous is _____________.Answer: The graph of f (x) = x – [x] is shown below. It can be seen that, the function f (x) = x – [x] is discontinuous at all integral values of x . So, the set of point where where f (x) = x – [x] is discontinuous is Z i.e. the set of integers. The set of points where f (x) = x – [x] is discontinuous is_____the set of integers i.e. Z______.Page No 8.48:Question 12: Let f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] . If f (x) is continuous in [0, \frac{π}{2}], then f (\frac{π}{4}) =_________.Answer: The given function is f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] . It is given that, the function f (x) is continuous in [0, \frac{π}{2}] . So, the function is continuous at x = \frac{π}{4} . ∴ f (\frac{π}{4}) = \lim_{x \to \frac{π}{4}} f (x) = \lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{4 x - π} Put x = \frac{π}{4} + h When x \to \frac{π}{4}, h \to 0 So, f (\frac{π}{4}) = \lim_{h \to 0} \frac{1 - \tan (\frac{π}{4} + h)}{4 (\frac{π}{4} + h) - π} = \lim_{h \to 0} \frac{1 - \frac{\tan \frac{π}{4} + \tan h}{1 - \tan \frac{π}{4} \tan h}}{4 h} = \lim_{h \to 0} \frac{1 - \tan h - 1 - \tan h}{4 h (1 - \tan h)} (\tan \frac{π}{4} = 1) = \lim_{h \to 0} \frac{- 2 \tan h}{4 h (1 - \tan h)} = - \frac{1}{2} \times \lim_{h \to 0} \frac{\tan h}{h} \times \frac{1}{\lim_{h \to 0} (1 - \tan h)} = - \frac{1}{2} \times 1 \times \frac{1}{(1 - 0)} (\lim_{x \to 0} \frac{\tan x}{x} = 1) = - \frac{1}{2} Thus, the value of f (\frac{π}{4}) is - \frac{1}{2} . Let f (x) = \frac{1 - \tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{2}] . If f (x) is continuous in [0, \frac{π}{2}], then f (\frac{π}{4}) = - \frac{1}{2} .Page No 8.48:Question 13: If f (x) = x \sin (\frac{π}{4}) is everywhere continuous, then f (0) = ____________.Answer: f (x) = x \sin (\frac{π}{4}) = x \times \frac{1}{\sqrt{2}} = \frac{x}{\sqrt{2}} Thus, f (x) is a polynomial function which is continuous everywhere. So, f (x) is continuous at x = 0. ∴ f (0) = \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times 0 = 0 If f (x) = x \sin (\frac{π}{4}) is everywhere continuous, then f (0) =____0____.Page No 8.48:Question 14: The set of points at which the function f (x) = \frac{1}{\log |x|} is not continuous, is ___________.Answer: The given function f (x) = \frac{1}{\log |x|} is discontinuous when \log |x| = 0 . Also, x ≠ 0 (log 0 is not defined) Now, \log |x| = 0 \Rightarrow |x| = 1 \Rightarrow x = \pm 1 Thus, the given function is not continuous at x = 0, x = −1 and x = 1. Hence, the set of points at which the given function is not continuous is {−1, 0, 1}. The set of points at which the function f (x) = \frac{1}{\log |x|} is not continuous, is___{−1, 0, 1}___.Page No 8.48:Question 15: If f (x) = \{\begin{matrix} a x + 1, & if x \geq 1 \\ x + 2, & if x < 1 \end{matrix} is continuous, then ' a' should be equal to __________.Answer: It is given that, the function f (x) = \{\begin{matrix} a x + 1, & if x \geq 1 \\ x + 2, & if x < 1 \end{matrix} is continuous. So, the function f (x) is continuous at x = 1. ∴ f (1) = \lim_{x \to 1} f (x) \Rightarrow f (1) = \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) .....(1) Now, f (1) = a × 1 + 1 = a + 1 .....(2) \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} (a x + 1) = a \times 1 + 1 = a + 1 .....(3) \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1} (x + 2) = 1 + 2 = 3 .....(4) From (1), (2), (3) and (4), we have a + 1 = 3 = a + 1 ⇒ a = 3 − 1 = 2 Thus, the value of a is 2. If f (x) = \{\begin{matrix} a x + 1, & if x \geq 1 \\ x + 2, & if x < 1 \end{matrix} is continuous, then ' a' should be equal to___2___.Page No 8.48:Question 16: If f (x) is continuous at x = a and \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = k, then k is equal to _____________.Answer: It is given that, f (x) is continuous at x = a . ∴ f (a) = \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) .....(1) Also, \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = k .....(2) From (1) and (2), we have f (a) = k Thus, the value of k is f (a). If f (x) is continuous at x = a and \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = k, then k is equal to f (a) .Page No 8.48:Question 17: If f (x) = \{\begin{matrix} \frac{\sin 3 x}{x}, & if x \neq 0 \\ \frac{k}{2}, & if x = 0 \end{matrix} is continuous at x = 0, then k is equal to _____________.Answer: The given function f (x) = \{\begin{matrix} \frac{\sin 3 x}{x}, & if x \neq 0 \\ \frac{k}{2}, & if x = 0 \end{matrix} is continuous at x = 0. ∴ f (0) = \lim_{x \to 0} f (x) \Rightarrow \frac{k}{2} = \lim_{x \to 0} \frac{\sin 3 x}{x} \Rightarrow \frac{k}{2} = 3 \lim_{x \to 0} \frac{\sin 3 x}{3 x} \Rightarrow \frac{k}{2} = 3 \times 1 = 3 (\lim_{x \to 0} \frac{\sin x}{x} = 1) \Rightarrow k = 6 Thus, the value of k is 6. If f (x) = \{\begin{matrix} \frac{\sin 3 x}{x}, & if x \neq 0 \\ \frac{k}{2}, & if x = 0 \end{matrix} is continuous at x = 0, then k is equal to____6____.Page No 8.48:Question 18: The set of points of discontinuity of f (x) = tan x is ___________.Answer: The given function is f (x) = \tan x . f (x) = \tan x = \frac{\sin x}{\cos x} The function f (x) is discontinuous when \cos x = 0 . \cos x = 0 \Rightarrow x = (2 n + 1) \frac{π}{2}, n \in Z Thus, the set of point of discontinuity of f (x) = \tan x is \{(2 n + 1) \frac{π}{2} : n \in Z\} . The set of points of discontinuity of f (x) = tan x is \{(2 n + 1) \frac{π}{2} : n \in Z\} .Page No 8.48:Question 19: The set of points of discontinuity of f (x) = [x] is ___________.Answer: The graph of f (x) = [x] is shown below. It can be seen that, the function f (x) = [x] is discontinuous at all integral values of x i.e. x ∈ Z . Thus, the set of points of discontinuity of f (x) = [x] is the set of integers i.e. Z . The set of points of discontinuity of f (x) = [x] is__the set of integers i.e. Z___.Page No 8.48:Question 20: The set of points of discontinuity of f (x) = \frac{1}{x - [x]} is ________.Answer: The function f (x) = \frac{1}{x - [x]} is discontinuous when x – [x] = 0. x – [x] = 0 ⇒ x = [x] ⇒ x is an integer So, the function f (x) is discontinuous for all x ∈ Z i.e. the set of integers. Thus, the set of points of discontinuity of f (x) = \frac{1}{x - [x]} is the set of integers i.e. Z . The set of points of discontinuity of f (x) = \frac{1}{x - [x]} is____the set of integers i.e. Z_____.Page No 8.49:Question 1: Define continuity of a function at a point.Answer: Continuity at a point: A function f (x) is said to be continuous at a point x = a of its domain, iff \lim_{x \to a} f (x) = f (a) . Thus, f (x) is continuous at x = a . \Leftrightarrow \lim_{x \to a} f (x) = f (a) \Leftrightarrow \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = f (a)Page No 8.49:Question 2: What happens to a function f (x) at x = a, if \lim_{x \to a} f (x) = f (a)?Answer: If f (x) is a function defined in its domain such that \lim_{x \to a} f (x) = f (a), then f (x) becomes continuous at x = a .Page No 8.49:Question 3: Find f (0), so that f (x) = \frac{x}{1 - \sqrt{1 - x}} becomes continuous at x = 0.Answer: If f (x) is continuous at x = 0, then \lim_{x \to 0} f (x) = f (0) ...(1) Given: f (x) = \frac{x}{1 - \sqrt{1 - x}} \Rightarrow f (x) = \frac{x (1 + \sqrt{1 - x})}{(1 - \sqrt{1 - x}) (1 + \sqrt{1 - x})} \Rightarrow f (x) = \frac{x (1 + \sqrt{1 - x})}{1 - (1 - x)} \Rightarrow f (x) = (1 + \sqrt{1 - x}) \lim_{x \to 0} (1 + \sqrt{1 - x}) = f (0) [From eq . (1)] \Rightarrow f (0) = 2 So, for f (0) = 2, the function f(x) becomes continuous at x = 0.Page No 8.49:Question 4: If f (x) = \{\begin{cases} \frac{x}{\sin 3 x}, & x \neq 0 \\ k, & x = 0 \end{cases} is continuous at x = 0, then write the value of k .Answer: If f (x) is continuous at x = 0, then \lim_{x \to 0} f (x) = f (0) \Rightarrow \lim_{x \to 0} \frac{x}{\sin 3 x} = k \Rightarrow \lim_{x \to 0} \frac{1}{\frac{\sin 3 x}{x}} = k \Rightarrow \lim_{x \to 0} \frac{1}{\frac{3 \sin 3 x}{3 x}} = k \Rightarrow \frac{1}{3} (\frac{1}{\lim_{x \to 0} \frac{\sin 3 x}{3 x}}) = k \Rightarrow k = \frac{1}{3}Page No 8.49:Question 5: If the function f (x) = \frac{\sin 10 x}{x}, x \neq 0 is continuous at x = 0, find f (0).Answer: Given: f (x) = \frac{\sin 10 x}{x}, x \neq 0 is continuous at x = 0 . \lim_{x \to 0} f (x) = f (0) \Rightarrow \lim_{x \to 0} \frac{\sin 10 x}{x} = f (0) \Rightarrow \lim_{x \to 0} \frac{10 \sin 10 x}{10 x} = f (0) \Rightarrow 10 \lim_{x \to 0} \frac{\sin 10 x}{10 x} = f (0) \Rightarrow f (0) = 10Page No 8.49:Question 6: If f (x) = \{\begin{cases} \frac{x^{2} - 16}{x - 4}, & if x \neq 4 \\ k, & if x = 4 \end{cases} is continuous at x = 4, find k .Answer: Given: f (x) = \{\begin{cases} \frac{x^{2} - 16}{x - 4}, if x \neq 4 \\ k, if x = 4 \end{cases} If f (x) is continuous at x = 4, then \lim_{x \to 4} f (x) = f (4) \Rightarrow \lim_{x \to 4} (\frac{x^{2} - 16}{x - 4}) = k \Rightarrow \lim_{x \to 4} \frac{(x + 4) (x - 4)}{(x - 4)} = k \Rightarrow \lim_{x \to 4} (x + 4) = k \Rightarrow k = 8Page No 8.49:Question 7: Determine whether f (x) = \{\begin{cases} \frac{\sin x^{2}}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} is continuous at x = 0 or not.Answer: Given: f (x) = \{\begin{cases} \frac{\sin x^{2}}{x}, x \neq 0 \\ 0, x = 0 \end{cases} We have \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{\sin x^{2}}{x} = \lim_{x \to 0} \frac{x \sin x^{2}}{x^{2}} = \lim_{x \to 0} \frac{\sin x^{2}}{x^{2}} \lim_{x \to 0} x = 1 \times 0 = 0 = f (0) \underset{x \to 0}{∴ \lim} f (x) = f (0) Hence, f (x) is continuous at x = 0 .Page No 8.49:Question 8: If f (x) = \{\begin{cases} \frac{1 - \cos x}{x^{2}}, & x \neq 0 \\ k, & x = 0 \end{cases} is continuous at x = 0, find k .Answer: Given: f (x) = \{\begin{cases} \frac{1 - \cos x}{x^{2}}, x \neq 0 \\ k, x = 0 \end{cases} If f (x) is continuous at x = 0, then \lim_{x \to 0} f (x) = f (0) \Rightarrow \lim_{x \to 0} (\frac{1 - \cos x}{x^{2}}) = k \Rightarrow \lim_{x \to 0} (\frac{2 {[\sin (\frac{x}{2})]}^{2}}{4 {(\frac{x}{2})}^{2}}) = k \Rightarrow \frac{1}{2} \lim_{x \to 0} (\frac{{[\sin (\frac{x}{2})]}^{2}}{{(\frac{x}{2})}^{2}}) = k \Rightarrow 1 \times \frac{1}{2} = k \Rightarrow k = \frac{1}{2}Page No 8.49:Question 9: If f (x) = \{\begin{cases} \frac{\sin^{- 1} x}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} is continuous at x = 0, write the value of k .Answer: Given, f (x) = \{\begin{cases} \frac{\sin^{- 1} x}{x}, x \neq 0 \\ k, x = 0 \end{cases} If f (x) is continuous at x = 0, then \lim_{x \to 0} f (x) = f (0) \Rightarrow \lim_{x \to 0} (\frac{\sin^{- 1} x}{x}) = f (0) \Rightarrow \lim_{x \to 0} (\frac{\sin^{- 1} x}{x}) = k \Rightarrow k = 1 [∵ \lim_{x \to 0} (\frac{\sin^{- 1} x}{x}) = 1]Page No 8.49:Question 10: Write the value of b for which f (x) = \{\begin{cases} 5 x - 4 & 0 < x \leq 1 \\ 4 x^{2} + 3 b x & 1 < x < 2 \end{cases} is continuous at x = 1.Answer: Given: f (x) = \{\begin{cases} 5 x - 4, 0 < x \leq 1 \\ 4 x^{2} + 3 b x, 1 < x < 2 \end{cases} If f (x) is continuous at x = 1, then \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) ...(1) Now, \lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} 5 (1 - h) - 4 = 5 - 4 = 1 \lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} 4 {(1 + h)}^{2} + 3 b (1 + h) = 4 + 3 b Also, f (1) = 5 (1) - 4 = 1 \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) [From eq . (1)] \Rightarrow 1 = 4 + 3 b = 1 \Rightarrow 1 = 4 + 3 b \Rightarrow - 3 = 3 b \Rightarrow b = - 1 Thus, for b = - 1, the function f (x) is continuous at x = 1 .Page No 8.49:Question 11: Determine the value of the constant ' k' so that function f (x) = \{\begin{cases} \frac{k x}{|x|}, & if x < 0 \\ 3, & if x \geq 0 \end{cases} is continuous at x = 0.Answer: Given, f (x) = \{\begin{cases} \frac{k x}{|x|} & , i f x < 0 \\ 3 & , i f x ⩾ 0 \end{cases} Since the function is continuous at x = 0, therefore, \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) \Rightarrow \lim_{x \to 0} \frac{- k x}{x} = \lim_{x \to 0} 3 = 3 \Rightarrow - k = 3 \Rightarrow k = - 3Page No 8.49:Question 12: Find the value of k for which the function f (x) = \{\begin{cases} \begin{matrix} \frac{x^{2} + 3 x - 10}{x - 2}, & x \neq 2 \\ k, & x = 2 \end{matrix} \end{cases} is continuous at x = 2Answer: Given, f (x) = \{\begin{cases} \begin{matrix} \frac{x^{2} + 3 x - 10}{x - 2}, & x \neq 2 \\ k, & x = 2 \end{matrix} \end{cases} \lim_{x \to 2^{-}} (\frac{x^{2} + 3 x - 10}{x - 2}) = \lim_{x \to 2^{-}} (x + 5) = 7 f (2) = k \lim_{x \to 2^{+}} (\frac{x^{2} + 3 x - 10}{x - 2}) = \lim_{x \to 2^{+}} (x + 5) = 7 If f (x) is continuous at x = 2, then \lim_{x \to 2^{-}} f (x) = f (2) = \lim_{x \to 2^{+}} f (x) \Rightarrow k = 7View NCERT Solutions for all chapters of Class 12 Login or Create a free accountvar MN_CSS_SUBDOMAIN_URL = "https://css-nm.mnimgs.com/";
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