Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 14 - Mean Value Theorems

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Page No 14.17:

Question 1:

Verify Lagrange's mean value theorem for the following functions on the indicated intervals. In each case find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem
(i) f(x) = x² − 1 on [2, 3]
(ii) f(x) = x³ − 2x² − x + 3 on [0, 1]
(iii) f(x) = x(x −1) on [1, 2]
(iv) f(x) = x² − 3x + 2 on [−1, 2]
(v) f(x) = 2x² − 3x + 1 on [1, 3]
(vi) f(x) = x² − 2x + 4 on [1, 5]
(vii) f(x) = 2x − x² on [0, 1]
(viii) f(x) = (x − 1)(x − 2)(x − 3) on [0, 4]
(ix) $f (x) = \sqrt{25 - x^{2}}$ on [−3, 4]
(x) f(x) = tan⁻¹ x on [0, 1]
(xi) $f (x) = x + \frac{1}{x} on [1, 3]$
(xii) f(x) = x(x + 4)² on [0, 4]
(xiii) $f (x) = \sqrt{x^{2} - 4} on [2, 4]$
(xiv) f(x) = x²+ x − 1 on [0, 4]
(xv) f(x) = sin x − sin 2x − x on [0, π]
(xvi) f(x) = x³ − 5x² − 3x on [1, 3]

Answer:

(i) We have

$f (x) = x^{2} - 1$

Since a polynomial function is everywhere continuous and differentiable, $f (x)$ is continuous on $[2, 3]$ and differentiable on $(2, 3)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (2, 3)$ such that
$f' (c) = \frac{f (3) - f (2)}{3 - 2}$

Now, $f (x) = x^{2} - 1$
$\Rightarrow f' (x) = 2 x$ , $f (3) = {(3)}^{2} - 1 = 8$ , $f (2) = {(2)}^{2} - 1 = 3$

∴ $f' (x) = \frac{f (3) - f (2)}{3 - 2}$

$\Rightarrow 2 x = \frac{8 - 3}{1} \Rightarrow x = \frac{5}{2}$
Thus, $c = \frac{5}{2} \in (2, 3)$ such that $f' (c) = \frac{f (3) - f (2)}{3 - 2}$ .

Hence, Lagrange's theorem is verified.

(ii) We have,

$f (x) = x^{3} - 2 x^{2} - x + 3 = 0$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (0, 1)$ such that
$f' (c) = \frac{f (1) - f (0)}{1 - 0}$

Now, $f (x) = x^{3} - 2 x^{2} - x + 3 = 0$
$\Rightarrow f' (x) = 3 x^{2} - 4 x - 1$ , $f (1) = 1$ , $f (0) = 3$

∴ $f' (x) = \frac{f (1) - f (0)}{1 - 0}$

$\Rightarrow 3 x^{2} - 4 x - 1 = \frac{1 - 3}{1} \Rightarrow 3 x^{2} - 4 x - 1 + 2 = 0 \Rightarrow 3 x^{2} - 4 x + 1 = 0 \Rightarrow 3 x^{2} - 3 x - x + 1 = 0 \Rightarrow (3 x - 1) (x - 1) = 0 \Rightarrow x = \frac{1}{3}, 1$
Thus, $c = \frac{1}{3} \in (0, 1)$ such that $f' (c) = \frac{f (1) - f (0)}{1 - 0}$ .

Hence, Lagrange's theorem is verified.

(iii) We have,

$f (x) = x (x - 1)$ which can be rewritten as $f (x) = x^{2} - x$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[1, 2]$ and differentiable on $(1, 2)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (1, 2)$ such that
$f' (c) = \frac{f (2) - f (1)}{2 - 1}$

Now, $f (x) = x^{2} - x$

$\Rightarrow f' (x) = 2 x - 1$ , $f (2) = 2$ , $f (1) = 0$

∴ $f' (x) = \frac{f (2) - f (1)}{2 - 1}$

$\Rightarrow 2 x - 1 = \frac{2 - 0}{2 - 1} \Rightarrow 2 x - 1 - 2 = 0 \Rightarrow 2 x = 3 \Rightarrow x = \frac{3}{2}$
Thus, $c = \frac{3}{2} \in (1, 2)$ such that $f' (c) = \frac{f (2) - f (1)}{2 - 1}$ .

Hence, Lagrange's theorem is verified.

(iv) We have,

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

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[- 1, 2]$ and differentiable on $(- 1, 2)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (- 1, 2)$ such that
$f' (c) = \frac{f (2) - f (- 1)}{2 + 1} = \frac{f (2) - f (- 1)}{3}$

Now, $f (x) = x^{2} - 3 x + 2$
$\Rightarrow f' (x) = 2 x - 3$ , $f (2) = 0$ , $f (- 1) = {(- 1)}^{2} - 3 (- 1) + 2 = 6$

∴ $f' (x) = \frac{f (2) - f (- 1)}{3}$

$\Rightarrow 2 x - 3 = - 2 \Rightarrow 2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$
Thus, $c = \frac{1}{2} \in (- 1, 2)$ such that $f' (c) = \frac{f (2) - f (- 1)}{2 - (- 1)}$ .

Hence, Lagrange's theorem is verified.

(v) We have,

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

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (1, 3)$ such that
$f' (c) = \frac{f (3) - f (1)}{3 - 1} = \frac{f (3) - f (1)}{2}$

Now, $f (x) = 2 x^{2} - 3 x + 1$
$\Rightarrow f' (x) = 4 x - 3$ , $f (3) = 10$ , $f (1) = 2 {(1)}^{2} - 3 (1) + 1 = 0$

∴ $f' (x) = \frac{f (3) - f (1)}{2}$

$\Rightarrow 4 x - 3 = \frac{10 - 0}{2} \Rightarrow 4 x - 3 - 5 = 0 \Rightarrow x = 2$
Thus, $c = 2 \in (1, 3)$ such that $f' (c) = \frac{f (3) - f (1)}{3 - 1}$ .

Hence, Lagrange's theorem is verified.

(vi) We have,

$f (x) = x^{2} - 2 x + 4$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[1, 5]$ and differentiable on $(1, 5)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (1, 5)$ such that
$f' (c) = \frac{f (5) - f (1)}{5 - 1} = \frac{f (5) - f (1)}{4}$

Now, $f (x) = x^{2} - 2 x + 4$
$\Rightarrow f' (x) = 2 x - 2$ , $f (5) = 25 - 10 + 4 = 19$ , $f (1) = 1 - 2 + 4 = 3$

∴ $f' (x) = \frac{f (5) - f (1)}{4}$

$\Rightarrow 2 x - 2 = \frac{19 - 3}{4} \Rightarrow 2 x - 2 - 4 = 0 \Rightarrow x = \frac{6}{2} = 3$
Thus, $c = 3 \in (1, 5)$ such that $f' (c) = \frac{f (5) - f (1)}{5 - 1}$ .

Hence, Lagrange's theorem is verified.

(vii) We have,

$f (x) = 2 x - x^{2}$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (0, 1)$ such that
$f' (c) = \frac{f (1) - f (0)}{1 - 0} = \frac{f (1) - f (0)}{1}$

Now, $f (x) = 2 x - x^{2}$
$\Rightarrow f' (x) = 2 - 2 x$ , $f (1) = 2 - 1 = 1$ , $f (0) = 0$

∴ $f' (x) = \frac{f (1) - f (0)}{1}$

$\Rightarrow 2 - 2 x = \frac{1 - 0}{1} \Rightarrow - 2 x = 1 - 2 \Rightarrow x = \frac{1}{2}$
Thus, $c = \frac{1}{2} \in (0, 1)$ such that $f' (c) = \frac{f (1) - f (0)}{1 - 0}$ .

Hence, Lagrange's theorem is verified.

(viii) We have,

$f (x) = (x - 1) (x - 2) (x - 3)$ which can be rewritten as $f (x) = x^{3} - 6 x^{2} + 11 x - 6$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[0, 4]$ and differentiable on $(0, 4)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $c \in (0, 4)$ such that
$f' (c) = \frac{f (4) - f (0)}{4 - 0} = \frac{f (4) - f (0)}{4}$

Now, $f (x) = x^{3} - 6 x^{2} + 11 x - 6$
$\Rightarrow f' (x) = 3 x^{2} - 12 x + 11$ , $f (0) = - 6$ , $f (4) = 64 - 96 + 44 - 6 = 6$

∴ $f' (x) = \frac{f (4) - f (0)}{4 - 0}$

$\Rightarrow 3 x^{2} - 12 x + 11 = \frac{6 + 6}{4} \Rightarrow 3 x^{2} - 12 x + 8 = 0 \Rightarrow x = 2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}$
Thus, $c = 2 \pm \frac{2}{\sqrt{3}} \in (0, 4)$ such that $f' (c) = \frac{f (4) - f (0)}{4 - 0}$ .

Hence, Lagrange's theorem is verified.

(ix) We have,

$f (x) = \sqrt{25 - x^{2}}$

Here, $f (x)$ will exist,
if
$25 - x^{2} \geq 0 \Rightarrow x^{2} \leq 25 \Rightarrow - 5 \leq x \leq 5$

Since for each $x \in [- 3, 4]$ , the function $f (x)$ attains a unique definite value.
So, $f (x)$ is continuous on $[- 3, 4]$

Also, $f' (x) = \frac{1}{2 \sqrt{25 - x^{2}}} (- 2 x) = \frac{- x}{\sqrt{25 - x^{2}}}$ exists for all $x \in (- 3, 4)$

So, $f (x)$ is differentiable on $(- 3, 4)$ .

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (- 3, 4)$ such that

$f' (c) = \frac{f (4) - f (- 3)}{4 + 3} = \frac{f (4) - f (- 3)}{7}$

Now, $f (x) = \sqrt{25 - x^{2}}$

$f' (x) = \frac{- x}{\sqrt{25 - x^{2}}}$ , $f (- 3) = 4$ , $f (4) = 3$

∴ $f' (x) = \frac{f (4) - f (- 3)}{4 + 3}$

$\Rightarrow \frac{- x}{\sqrt{25 - x^{2}}} = \frac{3 - 4}{7} \Rightarrow 49 x^{2} = 25 - x^{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}$
Thus, $c = \pm \frac{1}{\sqrt{2}} \in (- 3, 4)$ such that $f' (c) = \frac{f (4) - f (- 3)}{4 - (- 3)}$ .

Hence, Lagrange's theorem is verified.

(x) We have,

$f (x) = \tan^{- 1} x$

Clearly, $f (x)$ is continuous on $[0, 1]$ and derivable on $(0, 1)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (- 3, 4)$ such that

$f' (c) = \frac{f (1) - f (0)}{1 - 0} = \frac{f (1) - f (0)}{1}$

Now, $f (x) = \tan^{- 1} x$

$f' (x) = \frac{1}{1 + x^{2}}$ , $f (1) = \frac{π}{4}$ , $f (0) = 0$

∴ $f' (x) = \frac{f (1) - f (0)}{1 - 0}$

$\Rightarrow \frac{1}{1 + x^{2}} = \frac{π}{4} - 0 \Rightarrow (\frac{4}{π} - 1) = x^{2} \Rightarrow x = \pm \sqrt{\frac{4 - π}{π}}$
Thus, $c = \sqrt{\frac{4 - π}{π}} \in (0, 1)$ such that $f' (c) = \frac{f (1) - f (0)}{1 - 0}$ .

Hence, Lagrange's theorem is verified.

(xi) We have,

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

Clearly, $f (x)$ is continuous on $[1, 3]$ and derivable on $(1, 3)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (1, 3)$ such that

$f' (c) = \frac{f (3) - f (1)}{3 - 1} = \frac{f (3) - f (1)}{2}$

Now, $f (x) = \frac{x^{2} + 1}{x}$

$f' (x) = \frac{x^{2} - 1}{x^{2}}$ , $f (1) = 2$ , $f (3) = \frac{10}{3}$

∴ $f' (x) = \frac{f (3) - f (1)}{2}$

$\Rightarrow \frac{x^{2} - 1}{x^{2}} = \frac{4}{6} \Rightarrow \frac{x^{2} - 1}{x^{2}} = \frac{2}{3} \Rightarrow 3 x^{2} - 3 = 2 x^{2} \Rightarrow x = \pm \sqrt{3}$
Thus, $c = \sqrt{3} \in (1, 3)$ such that $f' (c) = \frac{f (3) - f (1)}{3 - 1}$ .

Hence, Lagrange's theorem is verified.

(xii) We have,

$f (x) = x {(x + 4)}^{2} = x (x^{2} + 16 + 8 x) = x^{3} + 8 x^{2} + 16 x$

Since $f (x)$ is a polynomial function which is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[0, 4]$ and derivable on $(0, 4)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (0, 4)$ such that

$f' (c) = \frac{f (4) - f (0)}{4 - 0} = \frac{f (4) - f (0)}{4}$

Now, $f (x) = x^{3} + 8 x^{2} + 16 x$

$f' (x) = 3 x^{2} + 16 x + 16$ , $f (4) = 64 + 128 + 64 = 256$ , $f (0) = 0$

∴ $f' (x) = \frac{f (4) - f (0)}{4 - 0}$

$\Rightarrow 3 x^{2} + 16 x + 16 = \frac{256}{4} \Rightarrow 3 x^{2} + 16 x - 48 = 0 \Rightarrow x = - \frac{4}{3} (2 + \sqrt{13}), \frac{4}{3} (\sqrt{13} - 2)$
Thus, $c = \frac{- 8 + 4 \sqrt{13}}{3} \in (0, 4)$ such that $f' (c) = \frac{f (4) - f (0)}{4 - 0}$ .

Hence, Lagrange's theorem is verified.

(xiii) We have,

$f (x) = \sqrt{x^{2} - 4}$

Here, $f (x)$ will exist,
if
$x^{2} - 4 \geq 0 \Rightarrow x \leq - 2 o r x \geq 2$

Since for each $x \in [2, 4]$ , the function $f (x)$ attains a unique definite value.
So, $f (x)$ is continuous on $[2, 4]$

Also, $f' (x) = \frac{1}{2 \sqrt{x^{2} - 4}} (2 x) = \frac{x}{\sqrt{x^{2} - 4}}$ exists for all $x \in (2, 4)$

So, $f (x)$ is differentiable on $(2, 4)$ .

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (2, 4)$ such that

$f' (c) = \frac{f (4) - f (2)}{4 - 2} = \frac{f (4) - f (2)}{2}$

Now, $f (x) = \sqrt{x^{2} - 4}$

$f' (x) = \frac{x}{\sqrt{x^{2} - 4}}$ , $f (4) = 2 \sqrt{3}$ , $f (2) = 0$

∴ $f' (x) = \frac{f (4) - f (2)}{4 - 2}$

$\Rightarrow \frac{x}{\sqrt{x^{2} - 4}} = \frac{2 \sqrt{3}}{2} \Rightarrow \frac{x}{\sqrt{x^{2} - 4}} = \sqrt{3} \Rightarrow \frac{x^{2}}{x^{2} - 4} = 3 \Rightarrow x^{2} = 3 x^{2} - 12 \Rightarrow x^{2} = 6 \Rightarrow x = \pm \sqrt{6}$
Thus, $c = \sqrt{6} \in (2, 4)$ such that $f' (c) = \frac{f (4) - f (2)}{4 - 2}$ .

Hence, Lagrange's theorem is verified.

(xiv) We have,

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

Since polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[0, 4]$ and differentiable on $(0, 4)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (0, 4)$ such that

$f' (c) = \frac{f (4) - f (0)}{4 - 0} = \frac{f (4) - f (0)}{4}$

Now, $f (x) = x^{2} + x - 1$

$f' (x) = 2 x + 1$ , $f (4) = 19$ , $f (0) = - 1$

∴ $f' (x) = \frac{f (4) - f (0)}{4 - 0}$

$\Rightarrow 2 x + 1 = \frac{20}{4} \Rightarrow 2 x + 1 = 5 \Rightarrow 2 x = 4 \Rightarrow x = 2$
Thus, $c = 2 \in (0, 4)$ such that $f' (c) = \frac{f (4) - f (0)}{4 - 0}$ .

Hence, Lagrange's theorem is verified.

(xv) We have,

$f (x) = \sin x - \sin 2 x - x$

Since $\sin x, \sin 2 x & x$ are everywhere continuous and differentiable

Therefore, $f (x)$ is continuous on $[0, π]$ and differentiable on $(0, π)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (0, π)$ such that

$f' (c) = \frac{f (π) - f (0)}{π - 0} = \frac{f (π) - f (0)}{π}$

Now, $f (x) = \sin x - \sin 2 x - x$

$f' (x) = \cos x - 2 \cos 2 x - 1$ , $f (π) = - π$ , $f (0) = 0$

∴ $f' (x) = \frac{f (π) - f (0)}{π - 0}$

$\Rightarrow \cos x - 2 \cos 2 x - 1 = - 1 \Rightarrow \cos x - 2 \cos 2 x = 0 \Rightarrow \cos x - 4 \cos^{2} x = - 2 \Rightarrow 4 \cos^{2} x - \cos x - 2 = 0 \Rightarrow \cos x = \frac{1}{8} (1 \pm \sqrt{33}) \Rightarrow x = \cos^{- 1} [\frac{1}{8} (1 \pm \sqrt{33})]$
Thus, $c = \cos^{- 1} (\frac{1 \pm \sqrt{33}}{8}) \in (0, π)$ such that $f' (c) = \frac{f (π) - f (0)}{π - 0}$ .

Hence, Lagrange's theorem is verified.

(xvi) We have,

$f (x) = x^{3} - 5 x^{2} - 3 x$

Since polynomial function is everywhere continuous and differentiable

Therefore, $f (x)$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (1, 3)$ such that

$f' (c) = \frac{f (3) - f (1)}{3 - 1} = \frac{f (3) - f (1)}{2}$

Now, $f (x) = x^{3} - 5 x^{2} - 3 x$

$f' (x) = 3 x^{2} - 10 x - 3$ , $f (3) = - 27$ , $f (1) = - 7$

∴ $f' (x) = \frac{f (3) - f (1)}{2}$
$\Rightarrow 3 x^{2} - 10 x - 3 = \frac{- 20}{2} \Rightarrow 3 x^{2} - 10 x + 7 = 0 \Rightarrow x = 1, \frac{7}{3}$

Thus, $c = \frac{7}{3} \in (1, 3)$ such that $f' (c) = \frac{f (3) - f (1)}{3 - 1}$ .

Hence, Lagrange's theorem is verified.

Page No 14.18:

Question 2:

Discuss the applicability of Lagrange's mean value theorem for the function
f(x) = | x | on [−1, 1]

Answer:

Given:
$f (x) = |x|$

If Lagrange's theorem is applicable for the given function, then $f (x)$ is continuous on $[- 1, 1]$ and differentiable on $(- 1, 1)$ .

But it is known that $f (x) = |x|$ is not differentiable at $x = 0 \in (- 1, 1)$ .

Thus, our supposition is wrong.
Therefore, Lagrange's theorem is not applicable for the given function.

Page No 14.18:

Question 3:

Show that the lagrange's mean value theorem is not applicable to the function
f(x) = $\frac{1}{x}$ on [−1, 1]

Answer:

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

Clearly, $f (x)$ does not exist for x = 0

Thus, the given function is discontinuous on $[- 1, 1]$ .

Hence, Lagrange's mean value theorem is not applicable for the given function on $[- 1, 1]$ .1x

Page No 14.18:

Question 4:

Verify the hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = $\frac{1}{4 x - 1},$ 1≤ x ≤ 4.

Answer:

The given function is $f (x) = \frac{1}{4 x - 1}$ .

Since for each $x \in [1, 4]$ , the function attains a unique definite value, $f (x)$ is continuous on $[1, 4]$ .

Also, $f' (x) = \frac{- 4}{{(4 x - 1)}^{2}}$ exists for all $x \in [1, 4]$

Thus, both the conditions of Lagrange's mean value theorem are satisfied.

Consequently, there exists some $c \in (1, 4)$ such that
$f' (c) = \frac{f (4) - f (1)}{4 - 1} = \frac{f (4) - f (1)}{3}$

Now,
$f (x) = \frac{1}{4 x - 1}$ $\Rightarrow$ $f' (x) = \frac{- 4}{{(4 x - 1)}^{2}}$ , $f (4) = \frac{1}{15}, f (1) = \frac{1}{3}$

$∴$ $f' (x) = \frac{f (4) - f (1)}{4 - 1}$
$\Rightarrow f' (x) = \frac{\frac{1}{15} - \frac{1}{3}}{4 - 1} = \frac{- 4}{45} \Rightarrow \frac{- 4}{{(4 x - 1)}^{2}} = \frac{- 4}{45} \Rightarrow {(4 x - 1)}^{2} = 45 \Rightarrow 16 x^{2} - 8 x - 44 = 0 \Rightarrow 4 x^{2} - 2 x - 11 = 0 \Rightarrow x = \frac{1}{4} (1 \pm 3 \sqrt{5})$

Thus, $c = \frac{1}{4} (1 + 3 \sqrt{5}) \in (1, 4)$ such that $f' (c) = \frac{f (4) - f (1)}{4 - 1}$ .

Hence, Lagrange's theorem is verified.

Page No 14.18:

Question 5:

Find a point on the parabola y = (x − 4)², where the tangent is parallel to the chord joining (4, 0) and (5, 1).

Answer:

Let:
$f (x) = {(x - 4)}^{2} = x^{2} - 8 x + 16$

The tangent to the curve is parallel to the chord joining the points $(4, 0)$ and $(5, 1)$ .

Assume that the chord joins the points $(a, f (a))$ and $(b, f (b))$ .

$∴$ $a = 4, b = 5$

The polynomial function is everywhere continuous and differentiable.

So, $x^{2} - 8 x + 16$ is continuous on $[4, 5]$ and differentiable on $(4, 5)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in (4, 5)$ such that $f' (c) = \frac{f (5) - f (4)}{5 - 4}$ .

Now,
$f (x) = x^{2} - 8 x + 16 \Rightarrow$ $f' (x) = 2 x - 8$ , $f (5) = 1, f (4) = 0$

$∴$ $f' (x) = \frac{f (5) - f (4)}{5 - 4}$ $\Rightarrow$ $2 x - 8 = \frac{1}{1} \Rightarrow 2 x = 9 \Rightarrow x = \frac{9}{2}$

Thus, $c = \frac{9}{2} \in (4, 5)$ such that $f' (c) = \frac{f (5) - f (4)}{5 - 4}$ .

Clearly,
$f (c) = {(\frac{9}{2} - 4)}^{2} = \frac{1}{4}$

Thus, $(c, f (c))$ , i.e. $(\frac{9}{2}, \frac{1}{4})$ , is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

Page No 14.18:

Question 6:

Find a point on the curve y = x² + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2).

Answer:

Let:
$f (x) = x^{2} + x$

The tangent to the curve is parallel to the chord joining the points $(0, 0)$ and $(1, 2)$ .

Assume that the chord joins the points $(a, f (a))$ and $(b, f (b))$ .

$∴$ $a = 0, b = 1$

The polynomial function is everywhere continuous and differentiable.

So, $f (x) = x^{2} + x$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in (0, 1)$ such that $f' (c) = \frac{f (1) - f (0)}{1 - 0}$ .

Now,
$f (x) = x^{2} + x$ $\Rightarrow$ $f' (x) = 2 x + 1$ , $f (1) = 2, f (0) = 0$

$∴$ $f' (x) = \frac{f (1) - f (0)}{1 - 0}$ $\Rightarrow$ $2 x + 1 = \frac{2 - 0}{1 - 0} \Rightarrow 2 x = 1 \Rightarrow x = \frac{1}{2}$

Thus, $c = \frac{1}{2} \in (0, 1)$ such that $f' (c) = \frac{f (1) - f (0)}{1 - 0}$ .

Clearly,
$f (c) = {(\frac{1}{2})}^{2} + \frac{1}{2} = \frac{3}{4}$ .

Thus, $(c, f (c))$ , i.e. $(\frac{1}{2}, \frac{3}{4})$ , is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

Page No 14.18:

Question 7:

Find a point on the parabola y = (x − 3)², where the tangent is parallel to the chord joining (3, 0) and (4, 1).

Answer:

Let:
$f (x) = {(x - 3)}^{2} = x^{2} - 6 x + 9$

The tangent to the curve is parallel to the chord joining the points $(3, 0)$ and $(4, 1)$ .

Assume that the chord joins the points $(a, f (a))$ and $(b, f (b))$ .

$∴$ $a = 3, b = 4$

The polynomial function is everywhere continuous and differentiable.

So, $f (x) = x^{2} - 6 x + 9$ is continuous on $[3, 4]$ and differentiable on $(3, 4)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in (3, 4)$ such that $f' (c) = \frac{f (4) - f (3)}{4 - 3}$ .

Now,
$f (x) = x^{2} - 6 x + 9$ $\Rightarrow$ $f' (x) = 2 x - 6$ , $f (3) = 0, f (4) = 1$

$∴$ $f' (x) = \frac{f (4) - f (3)}{4 - 3}$ $\Rightarrow$ $2 x - 6 = \frac{1 - 0}{4 - 3} \Rightarrow 2 x = 7 \Rightarrow x = \frac{7}{2}$

Thus, $c = \frac{7}{2} \in (3, 4)$ such that $f' (c) = \frac{f (4) - f (3)}{4 - 3}$ .

Clearly,
$f (c) = {(\frac{7}{2} - 3)}^{2} = \frac{1}{4}$

Thus, $(c, f (c))$ , i.e. $(\frac{7}{2}, \frac{1}{4})$ , is a point on the given curve where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1)$ .

Page No 14.18:

Question 8:

Find the points on the curve y = x³ − 3x, where the tangent to the curve is parallel to the chord joining (1, −2) and (2, 2).

Answer:

Let:
$f (x) = x^{3} - 3 x$

The tangent to the curve is parallel to the chord joining the points $(1, - 2)$ and $(2, 2)$ .

Assume that the chord joins the points $(a, f (a))$ and $(b, f (b))$ .

$∴$ $a = 1, b = 2$

The polynomial function is everywhere continuous and differentiable.

So, $f (x) = x^{3} - 3 x$ is continuous on $[1, 2]$ and differentiable on $(1, 2)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in (1, 2)$ such that $f' (c) = \frac{f (2) - f (1)}{2 - 1}$ .

Now,
$f (x) = x^{3} - 3 x$ $\Rightarrow$ $f' (x) = 3 x^{2} - 3$ , $f (1) = - 2, f (2) = 2$

$∴$ $f' (x) = \frac{f (2) - f (1)}{2 - 1}$ $\Rightarrow$ $3 x^{2} - 3 = \frac{2 + 2}{2 - 1} \Rightarrow 3 x^{2} = 7 \Rightarrow x = \pm \sqrt{\frac{7}{3}}$

Thus, $c = \pm \sqrt{\frac{7}{3}}$ such that $f' (c) = \frac{f (2) - f (1)}{2 - 1}$ .

Clearly,
$f (\sqrt{\frac{7}{3}}) = [{(\frac{7}{3})}^{\frac{3}{2}} - 3 \sqrt{\frac{7}{3}}] = \sqrt{\frac{7}{3}} [\frac{7}{3} - 3] = \sqrt{\frac{7}{3}} [\frac{- 2}{3}] = \frac{- 2}{3} \sqrt{\frac{7}{3}}$ and $f (- \sqrt{\frac{7}{3}}) = \frac{2}{3} \sqrt{\frac{7}{3}}$

∴ $f (c) = \mp \frac{2}{3} \sqrt{\frac{7}{3}}$

Thus, $(c, f (c))$ , i.e. $(\pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3} \sqrt{\frac{7}{3}})$ , are points on the given curve where the tangent is parallel to the chord joining the points $(1, - 2)$ and $(2, 2)$ .

Page No 14.18:

Question 9:

Find a point on the curve y = x³ + 1 where the tangent is parallel to the chord joining (1, 2) and (3, 28).

Answer:

Let:
$f (x) = x^{3} + 1$

The tangent to the curve is parallel to the chord joining the points $(1, 2)$ and $(3, 28)$ .

Assume that the chord joins the points $(a, f (a))$ and $(b, f (b))$ .

$∴$ $a = 1, b = 3$

The polynomial function is everywhere continuous and differentiable.

So, $f (x) = x^{3} + 1$ is continuous on $[1, 3]$ and differentiable on $(1, 3)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in (1, 3)$ such that $f' (c) = \frac{f (3) - f (1)}{3 - 1}$ .

Now,
$f (x) = x^{3} + 1$ $\Rightarrow$ $f' (x) = 3 x^{2}$ , $f (1) = 2, f (3) = 28$

$∴$ $f' (x) = \frac{f (3) - f (1)}{3 - 1}$ $\Rightarrow$ $3 x^{2} = \frac{26}{2} \Rightarrow 3 x^{2} = 13 \Rightarrow x = \pm \sqrt{\frac{13}{3}}$

Thus, $c = \sqrt{\frac{13}{3}}$ such that $f' (c) = \frac{f (3) - f (1)}{3 - 1}$ .

Clearly,
$f (c) = [{(\frac{13}{3})}^{\frac{3}{2}} + 1]$

Thus, $(c, f (c))$ , i.e. $(\sqrt{\frac{13}{3}}, 1 + {(\frac{13}{3})}^{\frac{3}{2}})$ , is a point on the given curve where the tangent is parallel to the chord joining the points $(1, 2)$ and $(3, 28)$ .

Page No 14.18:

Question 10:

Let C be a curve defined parametrically as $x = a \cos^{3} θ, y = a \sin^{3} θ, 0 \leq θ \leq \frac{π}{2}$ . Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a). [CBSE 2014]

Answer:

$As, x = a \cos^{3} θ \Rightarrow \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ And, y = a \sin^{3} θ \Rightarrow \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ So, \frac{d y}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ} = - \tan θ For the tangent to be parallel to the chord joining the points (a, 0) and (0, a), \frac{d y}{d x} = \frac{0 - a}{a - 0} \Rightarrow - \tan θ = - 1 \Rightarrow \tan θ = 1 \Rightarrow θ = \frac{π}{4} Now, x = a \cos^{3} \frac{π}{4} = a {(\frac{1}{\sqrt{2}})}^{3} = \frac{a}{2 \sqrt{2}} and y = a \sin^{3} \frac{π}{4} = a {(\frac{1}{\sqrt{2}})}^{3} = \frac{a}{2 \sqrt{2}} So, the point P on the curve C is (\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}) .$

Page No 14.18:

Question 11:

Using Lagrange's mean value theorem, prove that

(b − a) sec² a < tan b − tan a < (b − a) sec² b

where 0 < a < b < $\frac{π}{2}$ .

Answer:

Consider, the function
$f (x) = \tan x, x \in [a, b], 0 < a < b < \frac{π}{2}$

Clearly, $f (x)$ is continuous on $[a, b]$ and derivable on $(a, b)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, $c \in (a, b)$ such that $f' (c) = \frac{f (b) - f (a)}{b - a}$ .

Now,
$f (x) = \tan x$ $\Rightarrow$ $f' (x) = s e c^{2} x$ , $f (a) = \tan a, f (b) = \tan b$

$∴$ $f' (c) = \frac{f (b) - f (a)}{b - a}$ $\Rightarrow$ $s e c^{2} c = \frac{\tan b - \tan a}{b - a} . . . (1)$

Now,
$c \in (a, b) \Rightarrow a < c < b \Rightarrow s e c^{2} a < s e c^{2} c < s e c^{2} b [∵ s e c^{2} x is increasing in (0, \frac{π}{2})] \Rightarrow s e c^{2} a < \frac{\tan b - \tan a}{b - a} < s e c^{2} b [from (1)] \Rightarrow (b - a) s e c^{2} a < \tan b - \tan a < (b - a) s e c^{2} b$

Hence proved.

Page No 14.19:

Question 1:

If the polynomial equation
$a_{0} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_{2} x^{2} + a_{1} x + a_{0} = 0$
n positive integer, has two different real roots α and β, then between α and β, the equation
$n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + . . . + a_{1} = 0 has$
(a) exactly one root
(b) almost one root
(c) at least one root
(d) no root

Answer:

(c) at least one root

We observe that, $n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + . . . + a_{1} = 0$ is the derivative of the
polynomial $a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_{2} x^{2} + a_{1} x + a_{0} = 0$

Polynomial function is continuous every where in R and consequently derivative in R
Therefore, $a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_{2} x^{2} + a_{1} x + a_{0}$ is continuous on $[α, β]$ and derivative on $(α, β)$ .
Hence, it satisfies the both the conditions of Rolle's theorem.

By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function $f (x)$ , there exists at least one root of its derivative.

Hence, the equation $n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + . . . + a_{1} = 0$ will have at least one root between $α and β$ .

Page No 14.19:

Question 2:

If 4a + 2b + c = 0, then the equation 3ax² + 2bx + c = 0 has at least one real root lying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these

Answer:

(c) (0, 2)

$Let f (x) = a x^{3} + b x^{2} + c x + d . . . . . (1) f (0) = d f (2) = 8 a + 4 b + 2 c + d = 2 (4 a + 2 b + c) + d = d (∵ (4 a + 2 b + c) = 0)$

f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

By Rolle's Theorem,
$f' (α) = 0 for 0 < α < 2 Now, f' (x) = 3 a x^{2} + 2 b x + c \Rightarrow f' (α) = 3 a α^{2} + 2 b α + c = 0 Equation (1) has atleast one root in the interval (0, 2) . Thus, f' (x) must have root in the interval (0, 2) .$

Page No 14.19:

Question 3:

For the function f (x) = x + $\frac{1}{x}$ , x ∈ [1, 3], the value of c for the Lagrange's mean value theorem
is
(a) 1
(b) $\sqrt{3}$
(c) 2
(d) none of these

Answer:

(b) $\sqrt{3}$

We have
$f (x) = x + \frac{1}{x} = \frac{x^{2} + 1}{x}$

Clearly, $f (x)$ is continuous on $[1, 3]$ and derivable on $(1, 3)$ .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in (1, 3)$ such that

$f' (c) = \frac{f (3) - f (1)}{3 - 1} = \frac{f (3) - f (1)}{2}$

Page No 14.19:

Question 4:

If from Lagrange's mean value theorem, we have

$f' (x_{1}) = \frac{f' (b) - f (a)}{b - a}, then$
(a) a < x₁ ≤ b
(b) a ≤ x₁ < b
(c) a < x₁ < b
(d) a ≤ x₁ ≤ b

Answer:

(c) a < x₁ < b

In the Lagrange's mean value theorem, $c \in (a, b)$ such that $f' (c) = \frac{f (b) - f (a)}{b - a}$ .

So, if there is $x_{1}$ such that $f' (x_{1}) = \frac{f (b) - f (a)}{b - a}$ , then $x_{1} \in (a, b)$ .
$\Rightarrow a < x_{1} < b$

Page No 14.19:

Question 5:

Rolle's theorem is applicable in case of ϕ (x) = a^sin^x, a > a in
(a) any interval
(b) the interval [0, π]
(c) the interval (0, π/2)
(d) none of these

Answer:

(b) the interval [0, π]1−5√2

The given function is $ϕ (x) = a^{\sin x}$ , where a > 0.

Differentiating the given function with respect to x, we get

$f' (x) = \log a (\cos x a^{\sin x})$

$\Rightarrow f' (c) = \log a (\cos c a^{\sin c})$

$Let f' (c) = 0 \Rightarrow \log a (\cos c a^{\sin c}) = 0 \Rightarrow \cos c a^{\sin c} = 0 \Rightarrow \cos c = 0 \Rightarrow c = \frac{π}{2}$
∴ $c \in (0, π)$

Also, the given function is derivable and hence continuous on the interval $[0, π]$ .

Hence, the Rolle's theorem is applicable on the given function in the interval $[0, π]$ .

Page No 14.19:

Question 6:

The value of c in Rolle's theorem when
f (x) = 2x³ − 5x² − 4x + 3, x ∈ [1/3, 3] is

(a) 2

(b) $- \frac{1}{3}$

(c) −2

(d) $\frac{2}{3}$

Answer:

(a) 2

Given:
$f (x) = 2 x^{3} - 5 x^{2} - 4 x + 3$

Differentiating the given function with respect to x, we get

$f' (x) = 6 x^{2} - 10 x - 4 \Rightarrow f' (c) = 6 c^{2} - 10 c - 4 ∴ f' (c) = 0 \Rightarrow 3 c^{2} - 5 c - 2 = 0 \Rightarrow 3 c^{2} - 6 c + c - 2 = 0 \Rightarrow 3 c (c - 2) + c - 2 = 0 \Rightarrow (3 c + 1) (c - 2) = 0 \Rightarrow c = 2, \frac{- 1}{3} ∴ c = 2 \in (\frac{1}{3}, 3)$

Thus, $c = 2 \in (\frac{1}{3}, 3)$ for which Rolle's theorem holds.

Hence, the required value of c is 2.

Page No 14.19:

Question 7:

When the tangent to the curve y = x log x is parallel to the chord joining the points (1, 0) and (e, e), the value of x is

(a) e¹^/1−e

(b) e^{(e−1)(2e−1)}

(c) $e^{\frac{2 e - 1}{e - 1}}$

(d) $\frac{e - 1}{e}$

Answer:

(a) e¹^/1−e

Given:
$y = f (x) = x \log x$

Differentiating the given function with respect to x, we get

$f' (x) = 1 + \log x$

$\Rightarrow$ Slope of the tangent to the curve = $1 + \log x$

Also,
Slope of the chord joining the points $(1, 0) and (e, e)$ , (m) = $\frac{e}{e - 1}$

The tangent to the curve is parallel to the chord joining the points $(1, 0) and (e, e)$ .

$∴ m = 1 + \log x$

$\Rightarrow \frac{e}{e - 1} = 1 + \log x$

$\Rightarrow \frac{e}{e - 1} - 1 = \log x \Rightarrow \frac{e - e + 1}{e - 1} = \log x \Rightarrow \frac{1}{e - 1} = \log x \Rightarrow x = e^{\frac{1}{e - 1}}$

Page No 14.19:

Question 8:

The value of c in Rolle's theorem for the function $f (x) = \frac{x (x + 1)}{e^{x}}$ defined on [−1, 0] is
(a) 0.5

(b) $\frac{1 + \sqrt{5}}{2}$

(c) $\frac{1 - \sqrt{5}}{2}$

(d) −0.5

Answer:

(c) $\frac{1 - \sqrt{5}}{2}$ 1−5√2

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

Differentiating the given function with respect to x, we get

$f' (x) = \frac{e^{x} (2 x + 1) - x (x + 1) e^{x}}{{(e^{x})}^{2}} \Rightarrow f' (x) = \frac{(2 x + 1) - x (x + 1)}{e^{x}} \Rightarrow f' (x) = \frac{2 x + 1 - x^{2} - x}{e^{x}} \Rightarrow f' (x) = \frac{- x^{2} + x + 1}{e^{x}} \Rightarrow f' (c) = \frac{- c^{2} + c + 1}{e^{c}} ∴ f' (c) = 0 \Rightarrow \frac{- c^{2} + c + 1}{e^{c}} = 0 \Rightarrow c^{2} - c - 1 = 0 \Rightarrow c = \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} ∴ c = \frac{1 - \sqrt{5}}{2} \in (- 1, 0)$

Hence, the required value of c is $\frac{1 - \sqrt{5}}{2}$ .

Page No 14.20:

Question 9:

The value of c in Lagrange's mean value theorem for the function f (x) = x (x − 2) when x ∈ [1, 2] is
(a) 1
(b) 1/2
(c) 2/3
(d) 3/2

Answer:

(d) $\frac{3}{2}$

We have
f (x) = x (x − 2)

It can be rewritten as $f (x) = x^{2} - 2 x$ .

We know that a polynomial function is everywhere continuous and differentiable.

Since $f (x)$ is a polynomial , it is continuous on $[1, 2]$ and differentiable on $(1, 2)$ .

Thus, $f (x)$ satisfies both the conditions of Lagrange's theorem on $[1, 2]$ .

So, there must exist at least one real number c $\in (1, 2)$ such that

$f' (c) = \frac{f (2) - f (1)}{2 - 1} = \frac{f (2) - f (1)}{1}$

Now, $f (x) = x^{2} - 2 x$
$\Rightarrow f' (x) = 2 x - 2$ ,
and $f (1) = - 1, f (2) = 0$

$∴ f' (x) = \frac{f (2) - f (1)}{2 - 1}$
$\Rightarrow f' (x) = \frac{0 + 1}{1} \Rightarrow 2 x - 2 = 1 \Rightarrow x = \frac{3}{2}$

∴ $c = \frac{3}{2} \in (1, 2)$

Page No 14.20:

Question 10:

The value of c in Rolle's theorem for the function f (x) = x³ − 3x in the interval [0, $\sqrt{3}$ ] is
(a) 1
(b) −1
(c) 3/2
(d) 1/3

Answer:

(a) 1

The given function is $f (x) = x^{3} - 3 x$ .
This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0, $\sqrt{3}$ ] and derivable on (0, $\sqrt{3}$ ).

Differentiating the given function with respect to x, we get

$f' (x) = 3 x^{2} - 3 \Rightarrow f' (c) = 3 c^{2} - 3 ∴ f' (c) = 0 \Rightarrow 3 c^{2} - 3 = 0 \Rightarrow c^{2} = 1 \Rightarrow c = \pm 1$

Thus, $c = 1 \in [0, \sqrt{3}]$ for which Rolle's theorem holds.

Hence, the required value of c is 1.

Page No 14.20:

Question 11:

If f (x) = e^x sin x in [0, π], then c in Rolle's theorem is

(a) π/6

(b) π/4

(c) π/2

(d) 3π/4

Answer:

(d) 3π/4

The given function is $f (x) = e^{x} \sin x$ .

Differentiating the given function with respect to x, we get

$f' (x) = e^{x} \cos x + \sin x e^{x} \Rightarrow f' (c) = e^{c} \cos c + \sin c e^{c} Now, e^{x} cosx is continuous and derivable in R . Therefore, it is continuous on [0, π] and derivable on (0, π) . ∴ f' (c) = 0 \Rightarrow e^{c} (\cos c + \sin c) = 0 \Rightarrow (\cos c + \sin c) = 0 (∵ e^{c} \neq 0) \Rightarrow \tan c = - 1 \Rightarrow c = \frac{3 π}{4}, \frac{7 π}{4}, . . . ∴ c = \frac{3 π}{4} \in (0, π)$

Thus, $c = \frac{3 π}{4} \in (0, π)$ for which Rolle's theorem holds.

Hence, the required value of c is 3π/4.

Page No 14.20:

Question 1:

A function f (x) = 1 + $\frac{1}{x}$ is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is ______________.

Answer:

The function $f (x) = 1 + \frac{1}{x}$ is defined on the interval [1, 3].

f(x) is continuous on [1, 3] and differentiable on (1, 3).

So, by mean value theorem there must exist at least one real number c ∈ (1, 3) such that

$f' (c) = \frac{f (3) - f (1)}{3 - 1}$

$\Rightarrow - \frac{1}{c^{2}} = \frac{\frac{4}{3} - 2}{3 - 1} [f (x) = 1 + \frac{1}{x} \Rightarrow f' (x) = - \frac{1}{x^{2}}]$

$\Rightarrow - \frac{1}{c^{2}} = \frac{- \frac{2}{3}}{2}$

$\Rightarrow \frac{1}{c^{2}} = \frac{1}{3}$

$\Rightarrow c^{2} = 3$

$\Rightarrow c = \pm \sqrt{3}$

Thus, $c = \sqrt{3} \in (1, 3)$ such that $f' (c) = \frac{f (3) - f (1)}{3 - 1}$ .

Hence, a point in the interval where the given function satisfies the mean value theorem is $\sqrt{3}$ .

A function f (x) = 1 + $\frac{1}{x}$ is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is $\sqrt{3}$ .

Page No 14.20:

Question 2:

For the function f(x) = 8x² - 7x + 5, x ∈ [-6, 6], the value of c for the lagrange's mean value theorem is __________________.

Answer:

The given function is $f (x) = 8 x^{2} - 7 x + 5$ .

f(x) is a polynomial function.

We know that a polynomial function is everywhere continuous and differentiable. So, f(x) is continuous on [−6, 6] and differentiable on (−6, 6). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (−6, 6) such that

$f' (c) = \frac{f (6) - f (- 6)}{6 - (- 6)}$

Now, $f (x) = 8 x^{2} - 7 x + 5$

$\Rightarrow f' (x) = 16 x - 7$

$∴ f' (c) = \frac{f (6) - f (- 6)}{6 - (- 6)}$

$\Rightarrow 16 c - 7 = \frac{[8 \times {(6)}^{2} - 7 \times 6 + 5] - [8 \times {(- 6)}^{2} - 7 \times (- 6) + 5]}{12}$

$\Rightarrow 16 c - 7 = \frac{- 84}{12} = - 7$

$\Rightarrow 16 c = 0$

$\Rightarrow c = 0$

Thus, c = 0 ∈ (−6, 6) such that $f' (c) = \frac{f (6) - f (- 6)}{6 - (- 6)}$ .

Hence, the value of c is 0.

For the function f(x) = 8x² − 7x + 5, x ∈ [−6, 6], the value of c for the Lagrange's mean value theorem is ___0___.

Page No 14.20:

Question 3:

If the function f(x) = x³ – 6x² + ax + b defined on [1, 3] satisfies Roll's theorem for c = $2 + \frac{1}{\sqrt{3}},$ then a = ___________, b = __________.

Answer:

The given function is f(x) = x³ − 6x² + ax + b.

It is given that f(x) defined on [1, 3] satisfies Rolle's theorem for $c = 2 + \frac{1}{\sqrt{3}}$ .

$∴ f (1) = f (3)$ and $f' (c) = 0$

Now,

$f (1) = f (3)$

$\Rightarrow 1 - 6 + a + b = 27 - 54 + 3 a + b$

$\Rightarrow - 5 + a = - 27 + 3 a$

$\Rightarrow 2 a = 22$

$\Rightarrow a = 11$

Also,

f(x) = x³ − 6x² + ax + b

$\Rightarrow f' (x) = 3 x^{2} - 12 x + a$

$∴ f' (c) = 0$

$\Rightarrow 3 c^{2} - 12 c + a = 0$

$\Rightarrow 3 c^{2} - 12 c + 11 = 0$       (a = 11)

Since both equations f(1) = f(3) and $f' (c) = 3 c^{2} - 12 c + 11 = 0$ are independent of b, so b can taken any real value.

∴ a = 11 and b ∈ R

Thus, if the function  f(x) = x³ – 6x² + ax + b defined on [1, 3] satisfies Roll's theorem for $c = 2 + \frac{1}{\sqrt{3}}$ , then a = 11 and b ∈ R.

If the function  f(x) = x³ – 6x² + ax + b defined on [1, 3] satisfies Roll's theorem for c = $2 + \frac{1}{\sqrt{3}},$ then a = ___11___, b = ___R___.

Page No 14.20:

Question 4:

It is given that for the function f(x) = x³ - 6x² + ax + b on [1, 3], Rolle's theorem holds with c = $2 + \frac{1}{\sqrt{3}} .$ If f(1) = f(3) = 0, then a =_______, b =________.

Answer:

The given function is f(x) = x³ − 6x² + ax + b.

It is given that Rolle's theorem holds for f(x) defined on [1, 3] with $c = 2 + \frac{1}{\sqrt{3}}$ .

$f (1) = f (3) = 0$      (Given)

$∴ f (1) = 0$

$\Rightarrow 1 - 6 + a + b = 0$

$\Rightarrow a + b = 5$         .....(1)

Also,

$f (3) = 0$

$\Rightarrow 27 - 54 + 3 a + b = 0$

$\Rightarrow 3 a + b = 27$      .....(2)

Solving (1) and (2), we get

a = 11 and b = −6

It can be verified that for a = 11 and b = −6, $f' (2 + \frac{1}{\sqrt{3}}) = 0$ .

Thus, the values of a and b are 11 and −6, respectively.

It is given that for the function  f(x) = x³ − 6x² + ax + b on [1, 3], Rolle's theorem holds with c = $2 + \frac{1}{\sqrt{3}} .$ If f(1) = f(3) = 0, then a = ___11___, b =___−6___.

Page No 14.20:

Question 5:

For the function f(x) = log_ex, x ∈ [1, 2], the value of c for the lagrange's mean value theorem is _______________.

Answer:

The given function is f(x) = log_ex.

Now, f(x) = log_ex is differentiable and so continuous for all x > 0. So, f(x) is continuous on [1, 2] and differentiable on (1, 2). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (1, 2) such that

$f' (c) = \frac{f (2) - f (1)}{2 - 1}$

f(x) = log_ex

$\Rightarrow f' (x) = \frac{1}{x}$

$∴ f' (c) = \frac{f (2) - f (1)}{2 - 1}$

$\Rightarrow \frac{1}{c} = \frac{\log_{e} 2 - \log_{e} 1}{2 - 1}$

$\Rightarrow \frac{1}{c} = \log_{e} 2 - 0 (\log_{a} 1 = 0, a > 0)$

$\Rightarrow c = \frac{1}{\log_{e} 2} = \log_{2} e (\log_{b} a = \frac{1}{\log_{a} b})$

$\Rightarrow c = \log_{2} e \in (1, 2) (2 < e < 4 \Rightarrow \log_{2} 2 < \log_{2} e < \log_{2} 4 \Rightarrow 1 < \log_{2} e < 2)$

Thus, $c = \log_{2} e \in (1, 2)$ such that $f' (c) = \frac{f (2) - f (1)}{2 - 1}$ .

Hence, the value of c is $\log_{2} e$ .

For the function f(x) = log_ex, x ∈ [1, 2], the value of c for the Lagrange's mean value theorem is $\log_{2} e$ .

Page No 14.20:

Question 6:

The value of c in Rolle's theorem for the function f(x) = x³ - 3x in the interval [0, $\sqrt{3}$ ] is _______________.

Answer:

The given function is f(x) = x³ − 3x.

f(x) is a polynomial function. We know that a polynomial function is everywhere continuous and differentiable.

So, f(x) is continuous on $[0, \sqrt{3}]$ and differentiable on $(0, \sqrt{3})$ .

Also, f(0) = 0 and $f (\sqrt{3}) = {(\sqrt{3})}^{3} - 3 \sqrt{3} = 3 \sqrt{3} - 3 \sqrt{3} = 0$

$∴ f (0) = f (\sqrt{3})$

Thus, all the conditions of Rolle's theorem are satisfied.

So, there exist a real number c ∈ $(0, \sqrt{3})$ such that $f' (c) = 0$ .

f(x) = x³ − 3x

$\Rightarrow f' (x) = 3 x^{2} - 3$

$∴ f' (c) = 0$

$\Rightarrow 3 c^{2} - 3 = 0$

$\Rightarrow c^{2} = 1$

$\Rightarrow c = \pm 1$

Thus, c = 1 ∈ $(0, \sqrt{3})$ such that $f' (c) = 0$ .

Hence, the value of c is 1.

The value of c in Rolle's theorem for the function f(x) = x³ − 3x in the interval $[0, \sqrt{3}]$ is ___1___.

Page No 14.20:

Question 1:

If f (x) = Ax² + Bx + C is such that f (a) = f (b), then write the value of c in Rolle's theorem.

Answer:

We have
$f (x) = A x^{2} + B x + C$

Differentiating the given function with respect to x, we get
$f' (x) = 2 A x + B$
$\Rightarrow f' (c) = 2 A c + B$
$∴ f' (c) = 0 \Rightarrow 2 A c + B = 0 \Rightarrow c = \frac{- B}{2 A} . . . (1)$

$∵ f (a) = f (b) ∴ A a^{2} + B a + C = A b^{2} + b B + C \Rightarrow A a^{2} + B a = A b^{2} + b B \Rightarrow A (a^{2} - b^{2}) + B (a - b) = 0 \Rightarrow A (a - b) (a + b) + B (a - b) = 0 \Rightarrow (a - b) [A (a + b) + B] = 0 \Rightarrow a = b, A = \frac{- B}{(a + b)} \Rightarrow (a + b) = \frac{- B}{A} (∵ a \neq b)$

From (1), we have

$c = \frac{a + b}{2}$

Page No 14.20:

Question 2:

State Rolle's theorem.

Answer:

Rolle's Theorem:

Let f be a real valued function defined on the closed interval $[a, b]$ such that
(i) it is continuous on the closed interval $[a, b]$ ,
(ii) it is differentiable on the open interval $(a, b),$ and
(iii) $f (a) = f (b)$
Then, there exists a real number $c \in (a, b)$ such that $f' (c) = 0$ .

Page No 14.20:

Question 3:

State Lagrange's mean value theorem.

Answer:

Lagrange's Mean Value Theorem:

Let $f (x)$ be a function defined on $[a, b]$ such that
(i) it is continuous on $[a, b]$ and
(ii) it is differentiable on $(a, b)$ .

Then, there exists a real number $c \in (a, b)$ such that $f' (c) = \frac{f (b) - f (a)}{b - a}$ .

Page No 14.20:

Question 4:

If the value of c prescribed in Rolle's theorem for the function
f (x) = 2x (x − 3)ⁿ on the interval $[0, 2 \sqrt{3}] is \frac{3}{4},$ write the value of n (a positive integer).

Answer:

We have
$f (x) = 2 x {(x - 3)}^{n}$

Differentiating the given function with respect to x, we get

$f' (x) = 2 [x n {(x - 3)}^{n - 1} + {(x - 3)}^{n}] \Rightarrow f' (x) = 2 {(x - 3)}^{n} [\frac{x n}{(x - 3)} + 1] \Rightarrow f' (c) = 2 {(c - 3)}^{n} [\frac{c n}{(c - 3)} + 1]$

Given:
$f' (\frac{3}{4}) = 0$

$∴ 2 {(\frac{- 9}{4})}^{n} [\frac{\frac{3}{4} n}{(\frac{- 9}{4})} + 1] = 0 \Rightarrow 2 {(\frac{- 9}{4})}^{n} [\frac{- n}{3} + 1] = 0 \Rightarrow [\frac{- n}{3} + 1] = 0 \Rightarrow - n + 3 = 0 \Rightarrow n = 3$

Page No 14.20:

Question 5:

Find the value of c prescribed by Lagrange's mean value theorem for the function
$f (x) = \sqrt{x^{2} - 4}$ defined on [2, 3].

Answer:

We have

$f (x) = \sqrt{x^{2} - 4}$

Here, $f (x)$ will exist, if
$x^{2} - 4 \geq 0 \Rightarrow x \leq - 2 or x \geq 2$

Since for each $x \in [2, 3]$ , the function $f (x)$ attains a unique definite value, $f (x)$ is continuous on $[2, 3]$ .

Also, $f' (x) = \frac{1}{2 \sqrt{x^{2} - 4}} (2 x) = \frac{x}{\sqrt{x^{2} - 4}}$ exists for all $x \in (2, 3)$ .

So, $f (x)$ is differentiable on $(2, 3)$ .

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists $c \in (2, 3)$ such that

$f' (c) = \frac{f (3) - f (2)}{3 - 2} = \frac{f (3) - f (2)}{1}$

Now,
$f (x) = \sqrt{x^{2} - 4}$

$f' (x) = \frac{x}{\sqrt{x^{2} - 4}}$ , $f (3) = \sqrt{5}$ , $f (2) = 0$

∴ $f' (x) = \frac{f (3) - f (2)}{3 - 2}$

$\Rightarrow \frac{x}{\sqrt{x^{2} - 4}} = \sqrt{5} \Rightarrow \frac{x^{2}}{x^{2} - 4} = 5 \Rightarrow x^{2} = 5 x^{2} - 20 \Rightarrow 4 x^{2} = 20 \Rightarrow x = \pm \sqrt{5}$
Thus, $c = \sqrt{5} \in (2, 3)$ such that $f' (c) = \frac{f (3) - f (2)}{3 - 2}$ .

Hence, Lagrange's theorem is verified.

Page No 14.8:

Question 1:

Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
(i) f(x) = 3 + (x − 2)^2/3 on [1, 3]

(ii) f(x) = [x] for −1 ≤ x ≤ 1, where [x] denotes the greatest integer not exceeding x

(iii) f(x) = sin $\frac{1}{x}$ for −1 ≤ x ≤ 1

(iv) f(x) = 2x² − 5x + 3 on [1, 3]

(v) f(x) = x²^/3 on [−1, 1]

(vi) $f (x) = \{\begin{cases} - 4 x + 5, & 0 \leq x \leq 1 \\ 2 x - 3, & 1 < x \leq 2 \end{cases}$

Answer:

(i) The given function is $f (x) = 3 + {(x - 2)}^{\frac{2}{3}}$ .

Differentiating with respect to x, we get

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

Clearly, we observe that for x=2 $\in [1, 3]$ , $f' (x)$ does not exist.

Therefore, $f (x)$ is not derivable on $[1, 3]$ .

Hence, Rolle's theorem is not applicable for the given function.

(ii) The given function is $f (x) = [x]$ .
The domain of f is given to be $[- 1, 1]$ .

Let $c \in [- 1, 1]$ such that c is not an integer.
Then,
$\lim_{x \to c} f (x) = f (c)$

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

Now, let $c = 0$ .

Then,

$\lim_{x \to 0^{-}} f (x) = - 1 \neq 0 = f (0)$

Thus, f is discontinuous at x = 0.

Therefore, $f (x)$ is not continuous in $[- 1, 1]$ .

Rolle's theorem is not applicable for the given function.

(iii) The given function is $f (x) = \sin \frac{1}{x}$ .
The domain of f is given to be $[- 1, 1]$ .

It is known that $\lim_{x \to 0} \sin \frac{1}{x}$ does not exist.

Thus, $f (x)$ is discontinuous at x = 0 on $[- 1, 1]$ .

Hence, Rolle's theorem is not applicable for the given function.

(iv) The given function is $f (x) = 2 x^{2} - 5 x + 3$ on $[1, 3]$ .
The domain of f is given to be $[1, 3]$ .
It is a polynomial function.
Thus, it is everywhere derivable and hence continuous.

But
$f (1) = 0 and f (3) = 6 \Rightarrow f (3) \neq f (1)$

Hence, Rolle's theorem is not applicable for the given function.

(v) The given function is $f (x) = x^{\frac{2}{3}}$ on $[- 1, 1]$ .
The domain of f is given to be $[- 1, 1]$ .

Differentiating $f (x)$ with respect to x, we get

$f' (x) = \frac{2}{3} x^{- \frac{1}{3}}$
We observe that at $x = 0$ , $f' (x)$ is not defined.
Hence, Rolle's theorem is not applicable for the given function.

(vi) The given function is
$f (x) = \{\begin{cases} - 4 x + 5, 0 \leq x \leq 1 \\ 2 x - 3, 1 < x \leq 2 \end{cases}$

At x = 0, we have

$\lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} f (1 - h) = \lim_{h \to 0} [- 4 (1 - h) + 5] = 1$
And
$\lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} f (1 + h) = \lim_{h \to 0} [2 (1 + h) - 3] = - 1$

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

Thus, $f (x)$ is discontinuous at $x = 1$ .
Hence, Rolle's theorem is not applicable for the given function.

Page No 14.9:

Question 2:

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = x² − 8x + 12 on [2, 6]

(ii) f(x) = x² − 4x + 3 on [1, 3]

(iii) f(x) = (x − 1) (x − 2)² on [1, 2]

(iv) f(x) = x(x − 1)² on [0, 1]

(v) f(x) = (x² − 1) (x − 2) on [−1, 2]

(vi) f(x) = x(x − 4)² on the interval [0, 4]

(vii) f(x) = x(x −2)² on the interval [0, 2]

(viii) f(x) = x² + 5x + 6 on the interval [−3, −2]

Answer:

(i) Given:
$f (x) = x^{2} - 8 x + 12$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[2, 6]$ .

Also,
$f (2) = {(2)}^{2} - 8 (2) + 12 = 4 - 16 + 12 = 0 f (6) = {(6)}^{2} - 8 (6) + 12 = 36 - 48 + 12 = 0 ∴ f (2) = f (6) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in (2, 6)$ such that $f' (c) = 0$ .

We have
$f (x) = x^{2} - 8 x + 12 \Rightarrow f' (x) = 2 x - 8 ∴ f' (x) = 0 \Rightarrow 2 x - 8 = 0 \Rightarrow x = 4$

Thus, $c = 4 \in (2, 6) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(ii) Given:
$f (x) = x^{2} - 4 x + 3$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[1, 3]$ .

Also,
$f (1) = {(1)}^{2} - 4 (1) + 3 = 1 - 4 + 3 = 0 f (3) = {(3)}^{2} - 4 (3) + 3 = 9 - 12 + 3 = 0 ∴ f (1) = f (3) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in (1, 3)$ such that $f' (c) = 0$ .

We have
$f (x) = x^{2} - 4 x + 3 \Rightarrow f' (x) = 2 x - 4 ∴ f' (x) = 0 \Rightarrow 2 x - 4 = 0 \Rightarrow x = 2$

Thus, $c = 2 \in (1, 3) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(iii) Given:
$f (x) = (x - 1) {(x - 2)}^{2}$
i.e. $f (x) = x^{3} + 4 x - 4 x^{2} - x^{2} - 4 + 4 x$
i.e. $f (x) = x^{3} - 5 x^{2} + 8 x - 4$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[1, 2]$ .

Also,
$f (1) = {(1)}^{3} - 5 {(1)}^{2} + 8 (1) - 4 = 0 f (2) = {(2)}^{3} - 5 {(2)}^{2} + 8 (2) - 4 = 0 ∴ f (1) = f (2) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in (1, 2)$ such that $f' (c) = 0$ .

We have
$f (x) = x^{3} + 8 x - 5 x^{2} - 4 \Rightarrow f' (x) = 3 x^{2} + 8 - 10 x ∴ f' (x) = 0 \Rightarrow 3 x^{2} - 10 x + 8 = 0 \Rightarrow 3 x^{2} - 6 x - 4 x + 8 = 0 \Rightarrow 3 x (x - 2) - 4 (x - 2) = 0 \Rightarrow (x - 2) (3 x - 4) \Rightarrow x = 2, \frac{4}{3}$

Thus, $c = \frac{4}{3} \in (1, 2) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(iv) Given:
$f (x) = x {(x - 1)}^{2}$
$\Rightarrow f (x) = x (x^{2} - 2 x + 1)$
$∴ f (x) = (x^{3} - 2 x^{2} + x)$

We know that a polynomial function is everywhere derivable and hence continuous.

So, $f (x)$ being a polynomial function is continuous and derivable on $[0, 1]$ .

Also,
$f (0) = f (1) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in (0, 1)$ such that $f' (c) = 0$ .

We have
$f (x) = x^{3} - 2 x^{2} + x \Rightarrow f' (x) = 3 x^{2} - 4 x + 1 ∴ f' (x) = 0 \Rightarrow 3 x^{2} - 4 x + 1 = 0 \Rightarrow 3 x^{2} - 3 x - x + 1 = 0 \Rightarrow 3 x (x - 1) - 1 (x - 1) = 0 \Rightarrow (x - 1) (3 x - 1) = 0 \Rightarrow x = 1, \frac{1}{3}$

Thus, $c = \frac{1}{3} \in (0, 1) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(v) Given:
$f (x) = (x^{2} - 1) (x - 2)$
i.e. $f (x) = x^{3} - 2 x^{2} - x + 2$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[- 1, 2]$ .

Also,
$f (- 1) = f (2) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in (- 1, 2)$ such that $f' (c) = 0$ .

We have
$f (x) = x^{3} - x - 2 x^{2} + 2 \Rightarrow f' (x) = 3 x^{2} - 4 x - 1 ∴ f' (x) = 0 \Rightarrow 3 x^{2} - 4 x - 1 = 0 \Rightarrow x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 \times 3 \times (- 1)}}{2 \times 3} \Rightarrow x = \frac{4 \pm \sqrt{16 + 12}}{6} \Rightarrow x = \frac{4 \pm \sqrt{28}}{6} \Rightarrow x = \frac{4 \pm 2 \sqrt{7}}{6} \Rightarrow x = \frac{2 \pm \sqrt{7}}{3} \Rightarrow x = \frac{1}{3} (2 - \sqrt{7}), \frac{1}{3} (2 + \sqrt{7})$

Thus, $c = \frac{1}{3} (2 - \sqrt{7}), \frac{1}{3} (2 + \sqrt{7}) \in (- 1, 2) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(vi) Given function is $f (x) = x {(x - 4)}^{2}$ , which can be rewritten as $f (x) = x^{3} - 8 x^{2} + 16 x$ .

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[0, 4]$ .

Also,
$f (0) = f (4) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in [0, 4]$ such that $f' (c) = 0$ .

We have
$f (x) = x^{3} - 8 x^{2} + 16 x \Rightarrow f' (x) = 3 x^{2} - 16 x + 16 ∴ f' (x) = 0 \Rightarrow 3 x^{2} - 16 x + 16 = 0 \Rightarrow 3 x^{2} - 12 x - 4 x + 16 = 0 \Rightarrow 3 x (x - 4) - 4 (x - 4) = 0 \Rightarrow (x - 4) (3 x - 4) \Rightarrow x = 4, \frac{4}{3}$

Thus, $c = \frac{4}{3} \in (0, 4) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(vii) The given function is $f (x) = x {(x - 2)}^{2}$ , which can be rewritten as $f (x) = x^{3} - 4 x^{2} + 4 x$ .

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[0, 2]$ .

Also,
$f (0) = f (2) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in [0, 2]$ such that $f' (c) = 0$ .

We have
$f (x) = x^{3} - 4 x^{2} + 4 x \Rightarrow f' (x) = 3 x^{2} - 8 x + 4 When f' (x) = 0 3 x^{2} - 8 x + 4 = 0 \Rightarrow 3 x^{2} - 6 x - 2 x + 4 = 0 \Rightarrow 3 x (x - 2) - 2 (x - 2) = 0 \Rightarrow (x - 2) (3 x - 2) \Rightarrow x = 2, \frac{2}{3}$

Thus, $c = \frac{2}{3} \in (0, 2) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(viii) Given function is $f (x) = x^{2} + 5 x + 6$ .

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f (x)$ is continuous and derivable on $[- 3, - 2]$ .
Also,
$f (- 3) = {(- 3)}^{2} + 5 (- 3) + 6 = 9 - 15 + 6 = 0 f (- 2) = {(- 2)}^{2} + 5 (- 2) + 6 = 4 - 10 + 6 = 0 ∴ f (- 3) = f (- 2) = 0$

Thus, all the conditions of the Rolle's theorem are satisfied.

Now, we have to show that there exists $c \in [- 3, - 2]$ such that $f' (c) = 0$ .

We have
$f (x) = x^{2} + 5 x + 6 \Rightarrow f' (x) = 2 x + 5 ∴ f' (x) = 0 \Rightarrow 2 x + 5 = 0 \Rightarrow x = \frac{- 5}{2}$

Thus, $c = \frac{- 5}{2} \in (- 3, - 2) such that f' (c) = 0$ .

Hence, Rolle's theorem is verified.

Page No 14.9:

Question 3:

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]

(ii) f(x) = sin 2x on [0, π/2]

(iii) f(x) = cos 2x on [−π/4, π/4]

(iv) f(x) = e^x sin x on [0, π]

(v) f(x) = e^xcos x on [−π/2, π/2]

(vi) f(x) = cos 2x on [0, π]

(vii) f(x) = $\frac{\sin x}{e^{x}}$ on 0 ≤ x ≤ π

(viii) f(x) = sin 3x on [0, π]

(ix) f(x) = ${e^{1 - x}}^{2}$ on [−1, 1]

(x) f(x) = log (x² + 2) − log 3 on [−1, 1]

(xi) f(x) = sin x + cos x on [0, π/2]

(xii) f(x) = 2 sin x + sin 2x on [0, π]

(xiii) $f (x) = \frac{x}{2} - \sin \frac{π x}{6} on [- 1, 0]$

(xiv) $f (x) = \frac{6 x}{π} - 4 \sin^{2} x on [0, π / 6]$

(xv) f(x) = 4^sin^x on [0, π]

(xvi) f(x) = x² − 5x + 4 on [1, 4]

(xvii) f(x) = sin⁴ x + cos⁴ x on $[0, \frac{π}{2}]$

(xviii) f(x) = sin x − sin 2x on [0, π]

Answer:

(i) The given function is $f (x) = \cos 2 (x - \frac{π}{4}) = \cos (2 x - \frac{π}{2}) = \sin 2 x$ .

Since $\sin 2 x$ is everywhere continuous and differentiable.

Therefore, $\sin 2 x$ is continuous on $[0, \frac{π}{2}]$ and differentiable on $(0, \frac{π}{2})$ .

Also,
$f (\frac{π}{2}) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

We have
$f (x) = \sin 2 x \Rightarrow f' (x) = 2 \cos 2 x$

$∴ f' (x) = 0 \Rightarrow 2 \cos 2 x = 0 \Rightarrow \cos 2 x = 0 \Rightarrow x = \frac{π}{4}$

Thus, $c = \frac{π}{4} \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(ii) The given function is $f (x) = \sin 2 x$ .

Since $\sin 2 x$ is everywhere continuous and differentiable.

Therefore, $\sin 2 x$ is continuous on $[0, \frac{π}{2}]$ and differentiable on $(0, \frac{π}{2})$ .

Also,
$f (\frac{π}{2}) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

We have
$f (x) = \sin 2 x \Rightarrow f' (x) = 2 \cos 2 x$

$∴ f' (x) = 0 \Rightarrow 2 \cos 2 x = 0 \Rightarrow \cos 2 x = 0 \Rightarrow x = \frac{π}{4}$

Thus, $c = \frac{π}{4} \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(iii)

The given function is $f (x) = \cos 2 x$ .

Since $\cos 2 x$ is everywhere continuous and differentiable, $\cos 2 x$ is continuous on $[\frac{- π}{4}, \frac{π}{4}]$ and differentiable on $(\frac{- π}{4}, \frac{π}{4})$ .

Also,
$f (\frac{π}{4}) = f (\frac{- π}{4}) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (\frac{- π}{4}, \frac{π}{4})$ such that $f' (c) = 0$ .

We have
$f (x) = \cos 2 x \Rightarrow f' (x) = - 2 \sin 2 x$

$∴ f' (x) = 0 \Rightarrow - 2 \sin 2 x = 0 \Rightarrow \sin 2 x = 0 \Rightarrow \sin 2 x = 0 \Rightarrow x = 0$

Since $c = 0 \in (\frac{- π}{4}, \frac{π}{4})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(iv)

The given function is $f (x) = e^{x} \sin x$ .

Since $\sin x & e^{x}$ are everywhere continuous and differentiable.

Therefore, being a product of these two, $f (x)$ is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = e^{x} \sin x \Rightarrow f' (x) = e^{x} (\sin x + \cos x)$

$∴ f' (x) = 0 \Rightarrow e^{x} (\sin x + \cos x) = 0 \Rightarrow \sin x + \cos x = 0 \Rightarrow \tan x = - 1 \Rightarrow x = π - \frac{π}{4} = \frac{3 π}{4}$

Since $c = \frac{3 π}{4} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(v)

The given function is $f (x) = e^{x} \cos x$ .

Since $\cos x & e^{x}$ are everywhere continuous and differentiable, $f (x)$ being a product of these two is continuous on $[\frac{- π}{2}, \frac{π}{2}]$ and differentiable on $(\frac{- π}{2}, \frac{π}{2})$ .

Also,
$f (\frac{- π}{2}) = f (\frac{π}{2}) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (\frac{- π}{2}, \frac{π}{2})$ such that $f' (c) = 0$ .

We have
$f (x) = e^{x} \cos x \Rightarrow f' (x) = e^{x} (\cos x - \sin x)$

$∴ f' (x) = 0 \Rightarrow e^{x} (\cos x - \sin x) = 0 \Rightarrow \sin x - \cos x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{π}{4}$

Since $c = \frac{π}{4} \in (\frac{- π}{2}, \frac{π}{2})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(vi)

The given function is $f (x) = \cos 2 x$ .

Since $\cos 2 x$ is everywhere continuous and differentiable.

Therefore, $f (x)$   is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 1$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = \cos 2 x \Rightarrow f' (x) = - 2 \sin 2 x$

$∴ f' (x) = 0 \Rightarrow - 2 \sin 2 x = 0 \Rightarrow \sin 2 x = 0 \Rightarrow 2 x = π \Rightarrow x = \frac{π}{2}$

Thus, $c = \frac{π}{2} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(vii)

The given function is $f (x) = \frac{\sin x}{e^{x}}$ .

Since $\cos x and e^{x}$ are everywhere continuous and differentiable, being the quotient of these two, $f (x)$ is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = \frac{\sin x}{e^{x}} \Rightarrow f' (x) = \frac{\cos x - \sin x}{e^{x}}$

$∴ f' (x) = 0 \Rightarrow \frac{\cos x - \sin x}{e^{x}} = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{π}{4}$

Thus, $c = \frac{π}{4} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(viii)

The given function is $f (x) = \sin 3 x$ .

Since $\sin 3 x$ is everywhere continuous and differentiable, $\sin 3 x$ is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = \sin 3 x \Rightarrow f' (x) = 3 \cos 3 x$

$∴ f' (x) = 0 \Rightarrow 3 \cos 3 x = 0 \Rightarrow \cos 3 x = 0 \Rightarrow 3 x = \frac{π}{2}, \frac{3 π}{2}, . . . . \Rightarrow x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}$

Thus, $c = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(ix)

The given function is $f (x) = e^{1 - x^{2}}$ .

Since exponential function is everywhere continuous and differentiable, $e^{1 - x^{2}}$ is continuous on $[- 1, 1]$ and differentiable on $(- 1, 1)$ .

Also,
$f (1) = f (- 1) = 1$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (- 1, 1)$ such that $f' (c) = 0$ .

We have
$f (x) = e^{1 - x^{2}} \Rightarrow f' (x) = - 2 x e^{1 - x^{2}}$

$∴ f' (x) = 0 \Rightarrow - 2 x e^{1 - x^{2}} = 0 \Rightarrow x = 0$

Thus, $c = 0 \in (- 1, 1)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(x)

The given function is $f (x) = \log (x^{2} + 2) - \log 3$ , which can be rewritten as $f (x) = \log (\frac{x^{2} + 2}{3})$ .

Since logarithmic function is differentiable and so continuous in its domain, $f (x) = \log (\frac{x^{2} + 2}{3})$ is continuous on $[- 1, 1]$ and differentiable on $(- 1, 1)$ .

Also,
$f (1) = f (- 1) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (- 1, 1)$ such that $f' (c) = 0$ .

We have
$f (x) = \log (\frac{x^{2} + 2}{3}) \Rightarrow f' (x) = \frac{3 (2 x)}{x^{2} + 2} = \frac{6 x}{x^{2} + 2}$

$∴ f' (x) = 0 \Rightarrow \frac{6 x}{x^{2} + 2} = 0 \Rightarrow x = 0$

Thus, $c = 0 \in (- 1, 1)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xi)

The given function is $f (x) = \sin x + \cos x$ .
Since $\sin x and \cos x$ are everywhere continuous and differentiable, $f (x) = \sin x + \cos x$ is continuous on $[0, \frac{π}{2}]$ and differentiable on $(0, \frac{π}{2})$ .

Also,
$f (\frac{π}{2}) = f (0) = 1$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

We have
$f (x) = \sin x + \cos x \Rightarrow f' (x) = \cos x - \sin x$

$∴ f' (x) = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{π}{4}$

Thus, $c = \frac{π}{4} \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xii)
The given function is $f (x) = 2 \sin x + \sin 2 x$ .

Since $\sin x & \sin 2 x$ are everywhere continuous and differentiable, $f (x)$   is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = 2 \sin x + \sin 2 x \Rightarrow f' (x) = 2 \cos x + 2 \cos 2 x$

$∴ f' (x) = 0 \Rightarrow 2 \cos x + 2 \cos 2 x = 0 \Rightarrow \cos x + \cos 2 x = 0 \Rightarrow \cos x + 2 \cos^{2} x - 1 = 0 \Rightarrow 2 \cos^{2} x + \cos x - 1 = 0 \Rightarrow (\cos x + 1) (2 \cos x - 1) = 0 \Rightarrow \cos x = - 1, \cos x = \frac{1}{2} \Rightarrow \cos x = cosπ, \cos x = \frac{π}{3} \Rightarrow x = π, \frac{π}{3}$

Thus, $c = \frac{π}{3} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xiii)

The given function is $f (x) = \frac{x}{2} - \sin \frac{πx}{6}$ .

Since $\sin x & \frac{x}{2}$ are everywhere continuous and differentiable, $f (x)$   is continuous on $[- 1, 0]$ and differentiable on $(- 1, 0)$ .

Also,
$f (- 1) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (- 1, 0)$ such that $f' (c) = 0$ .

We have
$f (x) = \frac{x}{2} - \sin \frac{πx}{6} \Rightarrow f' (x) = \frac{1}{2} - \frac{π}{6} \cos \frac{πx}{6}$

$∴ f' (x) = 0 \Rightarrow \frac{1}{2} - \frac{π}{6} \cos \frac{π x}{6} = 0 \Rightarrow \cos \frac{π x}{6} = \frac{3}{π} \Rightarrow x = \frac{- 6}{π} \cos^{- 1} (\frac{3}{π})$

Thus, $c = \frac{- 6}{π} \cos^{- 1} (\frac{3}{π}) \in (- 1, 0)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xiv)

The given function is $f (x) = \frac{6 x}{π} - 4 \sin^{2} x$ .

Since $\sin^{2} x & x$ are everywhere continuous and differentiable, $f (x)$   is continuous on $[0, \frac{π}{6}]$ and differentiable on $(0, \frac{π}{6})$ .

Also,
$f (\frac{π}{6}) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, \frac{π}{6})$ such that $f' (c) = 0$ .

We have
$f (x) = \frac{6 x}{π} - 4 \sin^{2} x \Rightarrow f' (x) = \frac{6}{π} - 8 \sin x \cos x$

$∴ f' (x) = 0 \Rightarrow \frac{6}{π} - 8 \sin x \cos x = 0 \Rightarrow \sin 2 x = \frac{3}{2 π} \Rightarrow x = \frac{1}{2} \sin^{- 1} (\frac{3}{2 π})$

Thus, $c = \frac{1}{2} \sin^{- 1} (\frac{3}{2 π}) \in (0, \frac{π}{6})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xv)

The given function is $f (x) = 4^{\sin x}$ .

Since sine function and exponential function are everywhere continuous and differentiable, $f (x)$   is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 1$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = 4^{\sin x} \Rightarrow f' (x) = 4^{\sin x} (\cos x) \log 4$

$∴ f' (x) = 0 \Rightarrow 4^{\sin x} (\cos x) \log 4 = 0 \Rightarrow 4^{\sin x} \cos x = 0 \Rightarrow \cos x = 0 \Rightarrow x = \frac{π}{2}$

Thus, $c = \frac{π}{2} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xvi)

According to Rolle’s theorem, if f(x) is a real valued function defined on [a, b] such that it is continuous on [a, b], it is differentiable on (a, b) and f(a) = f(b), then there exists a real number c ∈(a, b) such that f(c) = 0.

Now, f(x) is defined for all x ∈[1, 4].
At each point of [1, 4], the limit of f(x) is equal to the value of the function. Therefore, f(x) is continuous on [1, 4].

Also, exists for all x ∈(1, 4).

So, f(x) is differentiable on (1, 4).

Also,
f(1) = f(4) = 0

Thus, all the three conditions of Rolle’s theorem are satisfied.

Now, we have to show that there exists c ∈(1, 4) such that.

We have

$∴ f' (x) = 0 \Rightarrow 2 x - 5 = 0 \Rightarrow x = \frac{5}{2}$

[Since ∈(1, 4) such that]

Hence, Rolle’s theorem is verified.

(xvii)

The given function is $f (x) = \sin^{4} x + \cos^{4} x$ .
Since $\sin x and \cos x$ are everywhere continuous and differentiable, $f (x) = \sin^{4} x + \cos^{4} x$ is continuous on $[0, \frac{π}{2}]$ and differentiable on $(0, \frac{π}{2})$ .

Also,
$f (\frac{π}{2}) = f (0) = 1$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

We have
$f (x) = \sin^{4} x + \cos^{4} x \Rightarrow f' (x) = 4 \sin^{3} x \cos x - 4 \cos^{3} x \sin x$

$∴ f' (x) = 0 \Rightarrow 4 \sin^{3} x \cos x - 4 \cos^{3} x \sin x = 0 \Rightarrow \sin^{3} x \cos x - \cos^{3} x \sin x = 0 \Rightarrow \tan^{3} x - \tan x = 0 \Rightarrow \tan x (\tan^{2} x - 1) = 0 \Rightarrow \tan x = 0, \tan^{2} x = 1 \Rightarrow \tan x = 0, \tan x = \pm 1 \Rightarrow x = 0, x = \frac{π}{4}, \frac{3 π}{4}$

Thus, $c = \frac{π}{4} \in (0, \frac{π}{2})$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

(xviii)

The given function is $f (x) = \sin x - \sin 2 x$ .

Since $\sin x and \sin 2 x$ are everywhere continuous and differentiable, $f (x)$ is continuous on $[0, π]$ and differentiable on $(0, π)$ .

Also,
$f (π) = f (0) = 0$

Thus, $f (x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in (0, π)$ such that $f' (c) = 0$ .

We have
$f (x) = \sin x - \sin 2 x \Rightarrow f' (x) = \cos x - 2 \cos 2 x$

$∴ f' (x) = 0 \Rightarrow \cos x - 2 \cos 2 x = 0 \Rightarrow \cos x - 2 \cos^{2} x + 1 = 0 \Rightarrow 2 \cos^{2} x - \cos x - 1 = 0 \Rightarrow (\cos x - 1) (2 \cos x + 1) = 0 \Rightarrow \cos x = 1, \cos x = \frac{- 1}{2} \Rightarrow \cos x = \cos \frac{π}{2}, \cos x = \frac{2 π}{3} \Rightarrow x = \frac{π}{2}, \frac{2 π}{3}$

Thus, $c = \frac{π}{2}, \frac{2 π}{3} \in (0, π)$ such that $f' (c) = 0$ .

Hence, Rolle's theorem is verified.

Page No 14.9:

Question 4:

Using Rolle's theorem, find points on the curve y = 16 − x², x ∈ [−1, 1], where tangent is parallel to x-axis.

Answer:

The equation of the curve is
$y = 16 - x^{2}$ . ...(1)

Let P $(x_{1}, y_{1})$ be a point on it where the tangent is parallel to the x-axis.

Then,
${(\frac{d y}{d x})}_{P} = 0$ ...(2)

Differentiating (1) with respect to x, we get

$\frac{d y}{d x} = - 2 x \Rightarrow {(\frac{d y}{d x})}_{P} = - 2 x_{1} \Rightarrow - 2 x_{1} = 0 (from (2)) \Rightarrow x_{1} = 0$

$P (x_{1}, y_{1})$ lies on the curve $y = 16 - x^{2}$ .

$∴$ $y_{1} = 16 - {x_{1}}^{2}$

When $x_{1} = 0$ ,
$y_{1} = 16$

Hence, $(0, 16)$ is the required point.

Page No 14.9:

Question 5:

At what points on the following curves, is the tangent parallel to x-axis?
(i) y = x² on [−2, 2]
(ii) y = $e^{1 - x^{2}}$ on [−1, 1]
(iii) y = 12 (x + 1) (x − 2) on [−1, 2].

Answer:

(i) Let $f (x) = x^{2}$
Since $f (x)$ is a polynomial function, it is continuous on $[- 2, 2]$ and differentiable on $(- 2, 2)$ .

Also, $f (2) = f (- 2) = 4$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c $\in (- 2, 2)$ for which $f' (c) = 0$ .

But $f' (c) = 0 \Rightarrow 2 c = 0 \Rightarrow c = 0$

$∴ f (c) = f (0) = 0$

By the geometrical interpretation of Rolle's theorem, $(0, 0)$ is the point on $y = x^{2}$ , where the tangent is parallel to the x-axis.

(ii) Let $f (x) = e^{1 - x^{2}}$
Since $f (x)$ is an exponential function, which is continuous and derivable on its domain, $f (x)$ is continuous on $[- 1, 1]$ and differentiable on $(- 1, 1)$ .

Also, $f (1) = f (- 1) = 1$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c $\in (- 1, 1)$ for which $f' (c) = 0$ .

But $f' (c) = 0 \Rightarrow - 2 c e^{1 - c^{2}} = 0 \Rightarrow c = 0 (∵ e^{1 - c^{2}} \neq 0)$

$∴ f (c) = f (0) = e$

By the geometrical interpretation of Rolle's theorem, $(0, e)$ is the point on $y = e^{1 - x^{2}}$ where the tangent is parallel to the x-axis.

(iii) Let $f (x) = 12 (x + 1) (x - 2)$ ...(1)

$\Rightarrow$ $f (x) = 12 (x^{2} - x - 2)$
$\Rightarrow$ $f (x) = 12 x^{2} - 12 x - 24$

Since $f (x)$ is a polynomial function, $f (x)$ is continuous on $[- 1, 2]$ and differentiable on $(- 1, 2)$ .

Also, $f (2) = f (- 1) = 0$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c $\in (- 1, 2)$ for which $f' (c) = 0$ .

But $f' (c) = 0 \Rightarrow 24 c - 12 = 0 \Rightarrow c = \frac{1}{2}$

$∴ f (c) = f (\frac{1}{2}) = - 12 (\frac{3}{2}) (\frac{3}{2}) = - 27$ (using (1))

By the geometrical interpretation of Rolle's theorem, $(\frac{1}{2}, - 27)$ is the point on $y = 12 (x + 1) (x - 2)$ where the tangent is parallel to the x-axis.

Page No 14.9:

Question 6:

If f : [−5, 5] → R is differentiable and if f' (x) doesnot vanish anywhere, then prove that f (−5) ± f (5).

Answer:

It is given thatis a differentiable function.
Every differentiable function is a continuous function. Thus,
(a) f is continuous in [−5, 5].
(b) f is differentiable in (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that

It is also given thatdoes not vanish anywhere.

Hence proved.

Page No 14.9:

Question 7:

Examine if Rolle's theorem is applicable to any one of the following functions.
(i) f (x) = [x] for x ∈ [5, 9]
(ii) f (x) = [x] for x ∈ [−2, 2]
Can you say something about the converse of Rolle's Theorem from these functions?

Answer:

By Rolle’s theorem, for a function, if

(a) f is continuous on [a, b],

(b) f is differentiable on (a, b) and

then there exists some c ∈ (a, b) such that .

Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

(i)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9.

Thus, f (x) is not continuous on [5, 9].

The differentiability of f on (5, 9) is checked in the following way.

Let n be an integer such that n ∈ (5, 9).

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, f is not differentiable on (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on.

(ii)

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2.

Thus, f (x) is not continuous on [−2, 2].

The differentiability of f on (−2, 2) is checked in the following way.

Let n be an integer such that n ∈ (−2, 2).

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, f is not differentiable on (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on.

Page No 14.9:

Question 8:

It is given that the Rolle's theorem holds for the function f(x) = x³ + bx² + cx, x $\in$ [1, 2] at the point x = $\frac{4}{3}$ . Find the values of b and c.

Answer:

As, the Rolle's theorem holds for the function f(x) = x³ + bx² + cx, x $\in$ [1, 2] at the point x = $\frac{4}{3}$

$So, f (1) = f (2) \Rightarrow {(1)}^{3} + b {(1)}^{2} + c (1) = {(2)}^{3} + b {(2)}^{2} + c (2) \Rightarrow 1 + b + c = 8 + 4 b + 2 c \Rightarrow 3 b + c + 7 = 0 . . . . . (i) And f' (\frac{4}{3}) = 0 \Rightarrow 3 {(\frac{4}{3})}^{2} + 2 b (\frac{4}{3}) + c = 0 [As, f' (x) = 3 x^{2} + 2 b x + c] \Rightarrow \frac{16}{3} + \frac{8 b}{3} + c = 0 \Rightarrow 8 b + 3 c + 16 = 0 . . . . . (ii) (ii) - (i) \times 3, we ge 8 b - 9 b + 16 - 21 = 0 \Rightarrow - b - 5 = 0 \Rightarrow b = - 5 Substituting b = - 5 in (i), we get 3 (- 5) + c + 7 = 0 \Rightarrow - 15 + c + 7 = 0 \Rightarrow c = 8$

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