RD Sharma Class 12 Exercise 14.2 Mean Value Theorems Solutions Maths - Download PDF Free Online

Mean value theorem exercise 14.1 question 2(i)

Answer:
$c = 4 \in (2,6)$ , Hence, Rolle’s Theorem is verified.
Hint:
f(x) is continuous for all x and hence continuous in [2,6]
Given:
$f(x) = x^2 -8x +12$ on [2,6]
Explanation:
We have
$f(x) = x^2 -8x +12$

1. Being polynomial f(x) is continuous for all x and hence continuous in [2,6]
2. $f'(x) = 2x -8$ , which exists in [2,6]

$\therefore f(x)$is derivable in (2,6)
3.
$\\f(2) = (2) ^2 - 8(2) + 12 \\\\= 4- 16 +12 = 12 + 12 = 0 \\\\ f (6) = (6) ^ 2 - 8 (6) + 12 \\\\ 36 - 48 +12 \\\\ = -12 + 12 = 0 \\\\ \therefore f(2) = f (6)$
Thus all the conditions of Rolle’s Theorem are satisfied.
There exists at least one$c \in (2,6)$such that$f' (c) = 0$
$\\ \Rightarrow 2c- 8 = 0 \\\\ \Rightarrow 2c = 8 \\\\ \Rightarrow c = 4 \\\\ \therefore c = 4 \in (2,6 )$
Hence, Rolle’s Theorem is verified.

Mean Value Theoram exercise 14.2 question 1(iv) maths

Answer: $\frac{1}{2}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x^{2}-3 x+2 \text { on }[-1,2]$
Solution:
$f(x)=x^{2}-3 x+2$

$f(x)$ is a polynomial function.
It is continuous in [-1, 2]
$f^{\prime}(x)=2 x-3$
(Which is defined in [-1, 2])
$\therefore f(x)$ is differentiable in [-1, 2]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[-1,2] \\ &f^{\prime}(c)=\frac{f(2)-f(-1)}{2-(-1)} \end{aligned}$
$\begin{aligned} &2 c-3=\frac{\left(2^{2}-3(2)+2\right)-\left((-1)^{2}-3(-1)+2\right)}{2+1} \\ &2 c-3=\frac{(4-6+2)-(1+3+2)}{3} \end{aligned}$
$\begin{aligned} &2 c-3=\frac{0-6}{3} \\ &2 c-3=\frac{-6}{3} \\ &2 c-3=-2 \\ &2 c=-2+3 \\ &2 c=1 \\ &c=\frac{1}{2} \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(v)

Answer: 2
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=2 x^{2}-3 x+1 \text { on }[1,3]$
Solution:
$f(x)=2 x^{2}-3 x+1$

$f(x)$ is a polynomial function.
It is continuous in [1, 3]
$f^{\prime}(x)=4 x-3$
(Which is defined in [1, 3])
$\therefore f(x)$ is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[1,3] \\ &f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\ &4 c-3=\frac{\left(2(3)^{2}-3(3)+1\right)-\left(2(1)^{2}-3(1)+1\right)}{2} \end{aligned}$
$\begin{aligned} &4 c-3=\frac{(2(9)-9+1)-(2-3+1)}{2} \\ &4 c-3=\frac{(18-9+1)-(2-3+1)}{2} \end{aligned}$
$\begin{aligned} &4 c-3=\frac{10-0}{2} \\ &4 c-3=5 \\ &4 c=5+3 \\ &c=\frac{8}{4} \\ &c=2 \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(vi)

Answer: 3
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x^{2}-2 x+4 \text { on }[1,5]$
Solution:
$f(x)=x^{2}-2 x+4$

$f(x)$ is a polynomial function.
It is continuous in [1, 5]
$f^{\prime}(x)=2 x-2$
(Which is defined in [1, 5])
$\therefore f(x)$ is differentiable in [1, 5]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[1,5] \\ &f^{\prime}(c)=\frac{f(5)-f(1)}{5-1} \\ &2 c-2=\frac{\left((5)^{2}-2(5)+4\right)-\left(2(1)^{2}-2(1)+4\right)}{4} \end{aligned}$
$\begin{aligned} &2 c-2=\frac{(25-10+4)-(1-2+4)}{4} \\ &2 c-2=\frac{19-3}{4} \end{aligned}$
$\begin{aligned} &2 c-2=\frac{16}{4} \\ &2 c-2=4 \end{aligned}$
$\begin{aligned} &2 c=4+2 \\ &2 c=6 \\ &c=\frac{6}{2} \\ &c=3 \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(vii)

Answer: $\frac{1}{2}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=2 x-x^{2} \text { on }[0,1]$
Solution:
$f(x)=2 x-x^{2}$

$f(x)$ is a polynomial function.
It is continuous in [0, 1]
$f^{\prime}(x)=2-2 x$
(Which is defined in [0, 1])
$\therefore f(x)$ is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \\ &2-2 c=\frac{\left(2(1)-(1)^{2}\right)-\left(2(0)-(0)^{2}\right)}{1} \end{aligned}$
$\begin{aligned} &2-2 c=(2-1)-0 \\ &2-2 c=1 \\ &2 c=2-1 \end{aligned}$
$\begin{aligned} &2 c=1 \\ &c=\frac{1}{2} \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(viii) maths

Answer: $4 \pm \frac{2}{\sqrt{3}}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=(x-1)(x-2)(x-3) \text { on }[0,4]$
Solution:$\begin{aligned} &f(x)=(x-1)(x-2)(x-3) \\ &f(x)=x^{3}-6 x^{2}+11 x-6 \end{aligned}$

$f(x)$ is a polynomial function.
It is continuous in [0, 4]
$f^{\prime}(x)=3 x^{2}-12 x+11$
(Which is defined in [0, 4])
$\therefore f(x)$ is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[0,4] \\ &f^{\prime}(c)=\frac{f(4)-f(0)}{4-0} \end{aligned}$
$\begin{aligned} &3 c^{2}-12 c+11=\frac{(4-1)(4-2)(4-3)-(0-1)(0-2)(0-3)}{4} \\ &3 c^{2}-12 c+11=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4} \end{aligned}$
$\begin{aligned} &3 c^{2}-12 c+11=\frac{6+6}{4} \\ &3 c^{2}-12 c+11=\frac{12}{4} \end{aligned}$
$\begin{aligned} &3 c^{2}-12 c+11=3 \\ &3 c^{2}-12 c+11-3=0 \\ &3 c^{2}-12 c+8=0 \end{aligned}$
Quadratic Equation,
$\begin{aligned} &c=\frac{12 \pm \sqrt{144-96}}{6} \\ &c=\frac{12 \pm \sqrt{48}}{6} \end{aligned}$
$\begin{aligned} &c=\frac{12 \pm 4 \sqrt{3}}{6} \\ &c=4 \pm \frac{2 \sqrt{3}}{3} \\ &c=4 \pm \frac{2}{\sqrt{3}} \end{aligned}$
Both values lie between [0, 4]

Mean Value Theoram exercise 14.2 question 1(ix)

Answer: $\pm \frac{1}{\sqrt{2}}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given:$f(x)=\sqrt{25-x^{2}} \text { on }[-3,4]$
Solution:
$f(x)=\sqrt{25-x^{2}}$

$f(x)$ is a polynomial function.
It is continuous in [-3, 4]
$\begin{aligned} &f^{\prime}(x)=\frac{1}{2}\left(25-x^{2}\right)^{\frac{-1}{2}}(-2 x) \\ &f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}} \end{aligned}$
(Which is defined in [-3, 4])
$\therefore f(x)$ is differentiable in [-3, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[-3,4] \\ &f^{\prime}(c)=\frac{f(4)-f(-3)}{4-(-3)} \end{aligned}$
$\begin{aligned} &\frac{-c}{\sqrt{25-c^{2}}}=\frac{\sqrt{25-(4)^{2}}-\sqrt{25-(-3)^{2}}}{4+3} \\ &\frac{-c}{\sqrt{25-c^{2}}}=\frac{\sqrt{25-16}-\sqrt{25-9}}{7} \end{aligned}$
$\begin{aligned} &\frac{-c}{\sqrt{25-c^{2}}}=\frac{\sqrt{9}-\sqrt{16}}{7} \\ &\frac{-c}{\sqrt{25-c^{2}}}=\frac{3-4}{7} \\ &\frac{-c}{\sqrt{25-c^{2}}}=\frac{-1}{7} \end{aligned}$
$\begin{aligned} &-7 c=-\sqrt{25-c^{2}} \\ &7 c=\sqrt{25-c^{2}} \end{aligned}$
Squaring on both sides,
$\begin{aligned} &49 c^{2}=25-c^{2} \\ &49 c^{2}+c^{2}-25=0 \\ &50 c^{2}-25=0 \end{aligned}$
$\begin{aligned} &2 c^{2}-1=0 \\ &2 c^{2}=1 \\ &c^{2}=\frac{1}{2} \\ &c^{2}=\pm \frac{1}{\sqrt{2}} \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(x)

Answer: $\sqrt{\frac{4}{\pi}-1}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=\tan ^{-1} x \text { on }[0,1]$
Solution:
$f(x)=\tan ^{-1} x$

$f(x)$ is a polynomial function.
It is continuous in [0, 1]
$f^{\prime}(x)=\frac{1}{\left(1+x^{2}\right)}$
(Which is defined in [0, 1])
$\therefore f(x)$ is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \\ &\frac{1}{1+c^{2}}=\frac{\tan ^{-1}(1)-\tan ^{-1}(0)}{1} \end{aligned}$
$\begin{aligned} &\frac{1}{1+c^{2}}=\frac{\frac{\pi}{4}-0}{1} \\ &\frac{1}{1+c^{2}}=\frac{\pi}{4} \end{aligned}$
$\begin{aligned} &\pi\left(1+c^{2}\right)=4 \\ &\pi+\pi c^{2}=4 \\ &\pi c^{2}=4-\pi \end{aligned}$
$\begin{aligned} &c^{2}=\frac{4-\pi}{\pi} \\ &c=\sqrt{\frac{4-\pi}{\pi}} \end{aligned}$
$c=\sqrt{\frac{4}{\pi}-1}$

Mean Value Theoram exercise 14.2 question 1(xi)

Answer: $\sqrt{3}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x+\frac{1}{x} \text { on }[1,3]$
Solution:
$f(x)=x+\frac{1}{x}$

$f(x)$ is a polynomial function.
It is continuous in [1, 3]
$f^{\prime}(x)=1-\frac{1}{x^{2}}$
(Which is defined in [1, 3])
$\therefore f(x)$ is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[1,3] \\ &f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\ &1-\frac{1}{c^{2}}=\frac{3+\frac{1}{3}-1-\frac{1}{1}}{2} \end{aligned}$
$1-\frac{1}{c^{2}}=\frac{3+\frac{1}{3}-1-1}{2}$
$\begin{aligned} &1-\frac{1}{c^{2}}=\frac{9+1-3-3}{2 \times 3} \\ &1-\frac{1}{c^{2}}=\frac{10-6}{6} \end{aligned}$
$\begin{aligned} &1-\frac{1}{c^{2}}=\frac{4}{6} \\ &1-\frac{1}{c^{2}}=\frac{2}{3} \end{aligned}$
$\begin{aligned} &\frac{1}{c^{2}}=1-\frac{2}{3} \\ &\frac{1}{c^{2}}=\frac{3-2}{3} \end{aligned}$
$\begin{aligned} &\frac{1}{c^{2}}=\frac{1}{3} \\ &c^{2}=3 \\ &c=\sqrt{3} \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(xii)

Answer: $\frac{-8+4 \sqrt{13}}{3}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x(x+4)^{2} \text { on }[0,4]$
Solution:
$\begin{aligned} &f(x)=x(x+4)^{2} \\ &f(x)=x\left(x^{2}+16+8 x\right) \\ &f(x)=x^{3}+8 x^{2}+16 x \end{aligned}$

$f(x)$ is a polynomial function.
It is continuous in [0, 4]
$f^{\prime}(x)=3 x^{2}+16 x+16$
(Which is defined in [0, 4])
$\therefore f(x)$ is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[0,4] \\ &f^{\prime}(c)=\frac{f(4)-f(0)}{4-0} \end{aligned}$
$\begin{aligned} &3 c^{2}+16 c+16=\frac{(4)^{3}+8(4)^{2}+16(4)-(0)^{3}+8(0)^{2}+16(0)}{4} \\ &3 c^{2}+16 c+16=\frac{64+128+64-0}{4} \end{aligned}$
$\begin{aligned} &3 c^{2}+16 c+16=\frac{256}{4} \\ &3 c^{2}+16 c+16=64 \\ &3 c^{2}+16 c+16-64=0 \\ &3 c^{2}+16 c-48=0 \end{aligned}$
$\begin{aligned} &c=\frac{-16 \pm \sqrt{(16)^{2}-4(3)(-48)}}{6} \\ &c=\frac{-16 \pm \sqrt{256+576}}{6} \end{aligned}$
$c=\frac{-16 \pm \sqrt{832}}{6}$
$c=\frac{-8 \pm 4 \sqrt{13}}{3}$
But in [0, 4], only $\frac{-8 \pm 4 \sqrt{13}}{3}$ exists.

Mean Value Theoram exercise 14.2 question 1(xiii)

Answer: $\sqrt{6}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=\sqrt{x^{2}-4} \text { on }[2,4]$
Solution:
$f(x)=\sqrt{x^{2}-4}$

$f(x)$ is a polynomial function.
It is continuous in [2, 4]
$\begin{aligned} &f^{\prime}(x)=\frac{1}{2}\left(\frac{2 x}{\sqrt{x^{2}-4}}\right) \\ &f^{\prime}(x)=\frac{x}{\sqrt{x^{2}-4}} \end{aligned}$
(Which is defined in [2, 4])
$\therefore f(x)$ is differentiable in [2, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[2,4] \\ &f^{\prime}(c)=\frac{f(4)-f(2)}{4-2} \end{aligned}$
$\begin{aligned} &\frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{(4)^{2}-4}-\sqrt{(2)^{2}-4}}{2} \\ &\frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{16-4}-\sqrt{4-4}}{2} \end{aligned}$
$\begin{aligned} &\frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{12}-0}{2} \\ &\frac{c}{\sqrt{c^{2}-4}}=\frac{2 \sqrt{3}}{2} \end{aligned}$
$\begin{aligned} &\frac{c}{\sqrt{c^{2}-4}}=\sqrt{3} \\ &c=\sqrt{3\left(c^{2}-4\right)} \end{aligned}$
Squaring on both sides,
$\begin{aligned} &c^{2}=3\left(c^{2}-4\right) \\ &c^{2}=3 c^{2}-12 \\ &3 c^{2}-c^{2}=12 \\ &2 c^{2}=12 \\ &c^{2}=6 \\ &c=\sqrt{6} \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(xiv) maths

Answer: 2
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x^{2}+x-1 \text { on }[0,4]$
Solution:
$f(x)=x^{2}+x-1$

$f(x)$ is a polynomial function.
It is continuous in [0, 4]
$f^{\prime}(x)=2 x+1$
(Which is defined in [0, 4])
$\therefore f(x)$ is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[0,4] \\ &f^{\prime}(c)=\frac{f(4)-f(2)}{4-2} \end{aligned}$
$\begin{aligned} &2 c+1=\frac{\left((4)^{2}+4-1\right)-\left((0)^{2}+0-1\right)}{4} \\ &2 c+1=\frac{(16+4-1)-(0+0-1)}{4} \end{aligned}$
$\begin{aligned} &2 c+1=\frac{19+1}{4} \\ &2 c+1=\frac{20}{4} \end{aligned}$
$\begin{aligned} &2 c+1=5 \\ &2 c=5-1 \\ &2 c=4 \\ &c=\frac{4}{2} \\ &c=2 \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(xv)

Answer: $\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given:$f(x)=\sin x-\sin 2 x-x \text { on }[0, \pi]$
Solution:
$f(x)=\sin x-\sin 2 x-x$

$f(x)$ is a polynomial function.
It is continuous in $\left [ 0,\pi \right ]$
$f^{\prime}(x)=\cos x-2 \cos 2 x-1$
(Which is defined in $\left [ 0,\pi \right ]$)
$\therefore f(x)$ is differentiable in $\left [ 0,\pi \right ]$. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[0, \pi] \\ &f^{\prime}(c)=\frac{f(\pi)-f(0)}{\pi-0} \end{aligned}$
$\begin{aligned} &\cos c-2 \cos 2 c-1=\frac{(\sin \pi-\sin 2 \pi-\pi)-(\sin 0-\sin 0-0)}{\pi} \\ &\cos c-2 \cos 2 c-1=\frac{(0-0-\pi)-(0-0-0)}{\pi} \end{aligned}$
$\begin{aligned} &\cos c-2 \cos 2 c-1=\frac{-\pi}{\pi} \\ &\cos c-2 \cos 2 c-1=-1 \\ &\cos c-2 \cos 2 c=-1+1 \end{aligned}$
$\begin{aligned} &\cos c-2 \cos 2 c=0 \\ &\cos c-2\left(2 \cos ^{2} c-1\right)=0 \\ &\cos c-4 \cos ^{2} c+2=0 \\ &4 \cos ^{2} c-\cos c-2=0 \end{aligned}$
$\begin{aligned} &\cos c=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 4 \times(-2)}}{2 \times 4} \\ &\cos c=\frac{1 \pm \sqrt{1+32}}{8} \end{aligned}$
$\begin{aligned} &\cos c=\frac{1 \pm \sqrt{33}}{8} \\ &c=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(xvi)

Answer: $\frac{7}{3}$
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x^{3}-5 x^{2}-3 x \text { on }[1,3]$
Solution:
$f(x)=x^{3}-5 x^{2}-3 x$

$f(x)$ is a polynomial function.
It is continuous in [1, 3]
$f^{\prime}(x)=3 x^{2}-10 x-3$
(Which is defined in [1, 3])
$\therefore f(x)$ is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[1,3] \\ &f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \end{aligned}$
$\begin{aligned} &3 c^{2}-10 c-3=\frac{\left((3)^{3}-5(3)^{2}-3(3)\right)-\left((1)^{3}-5(1)^{2}-3(1)\right)}{2} \\ &3 c^{2}-10 c-3=\frac{(27-45-9)-(1-5-3)}{2} \end{aligned}$
$\begin{aligned} &3 c^{2}-10 c-3=\frac{(27-54)-(1-8)}{2} \\ &3 c^{2}-10 c-3=\frac{(-27)-(-7)}{2} \end{aligned}$
$\begin{aligned} &3 c^{2}-10 c-3=\frac{-27+7}{2} \\ &3 c^{2}-10 c-3=\frac{-20}{2} \end{aligned}$
$\begin{aligned} &3 c^{2}-10 c-3=-10 \\ &3 c^{2}-10 c-3+10=0 \\ &3 c^{2}-10 c+7=0 \\ &3 c^{2}-3 c-7 c+7=0 \end{aligned}$
$\begin{aligned} &3 c(c-1)-7(c-1)=0 \\ &(3 c-7)(c-1)=0 \end{aligned}$
$c=\frac{7}{3} \text { And } c=1$
$c=\frac{7}{3}$ Value exists.

Mean Value Theoram exercise 14.2 question 2

Answer: Not applicable
Hint: Using Lagrange’s mean theorem and limits.

Given: $f(x)=|x| \text { on }[-1,1]$
Solution:
Lagrange’s mean value theorem states if a function$f(x)$ is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that
Differentiability at x = 0:
LHD =
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h} \\ &\lim _{x \rightarrow 0^{-}} \frac{-(0-h)-0}{-h} \end{aligned}$
$=\lim _{x \rightarrow 0^{-}} \frac{h-0}{-h}$$\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{h}{-h} \\ =&-1 \end{aligned}$

RHD = $\lim _{x \rightarrow 0^{+}} \frac{f(0-h)-f(0)}{-h}$

$\begin{aligned} &=\lim _{x \rightarrow 0^{-}} \frac{(0-h)-0}{-h} \\ &=\lim _{x \rightarrow 0^{-}} \frac{-h-0}{-h} \\ &=\lim _{x \rightarrow 0^{-}} \frac{-h}{-h} \\ &=1 \end{aligned}$

$\mathrm{LHS} \neq \mathrm{RHS}$

$\therefore f(x)$ is not differentiable at x = 0

$\therefore$Lagrange’s mean value theorem is not applicable for function f (x) = |x| on [-1, 1]

Mean Value Theoram exercise 14.2 question 3

Answer:
Hint: Check the continuity of function

Given: $f(x)=\frac{1}{x} \text { on }[-1,1]$
Solution:
$f(x)=\frac{1}{x} \text { on }[-1,1]$
It is clear $x\neq 0$
$\Rightarrow$ f (x) exists for all the values of x except 0
$\Rightarrow$ f (x) is the discontinuous at x = 0
So, f (x) is not continuous in [-1, 1]
Thus, the Lagrange’s mean value theorem is not applicable for the function:
$f(x)=\frac{1}{x} \text { on }[-1,1]$

Mean Value Theoram exercise 14.2 question 4

Answer: Not applicable
Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=\frac{1}{4 x-1}, 1 \leq x \leq 4$
Solution:
$f(x)=\frac{1}{4 x-1} \text { on }[1,4]$
Here, $4 x-1>0$

f^'(x) has the unique values for all x except $\frac{1}{4}$

So, f (x) is continuous in [1, 4]

$f(x)=\frac{1}{4 x-1}$
Differentiate:
$\begin{aligned} &f(x)=(-1)(4 x-1)^{-2}(4) \\ &f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}} \end{aligned}$

f (x) is differentiable in (1, 4)

So, there exists a point c $\in$ (1, 4)

$\begin{aligned} &f^{\prime}(c)=\frac{f(4)-f(1)}{4-1} \\ &f^{\prime}(c)=\frac{f(4)-f(1)}{3} \end{aligned}$

$\frac{-4}{(4 c-1)^{2}}=\frac{\frac{1}{4(4)-1}-\frac{1}{4(1)-1}}{3}$

$\frac{-4}{(4 c-1)^{2}}=\frac{\frac{1}{15}-\frac{1}{3}}{3}$

$-3(4)=(4 c-1)^{2}\left(\frac{1}{15}-\frac{1}{3}\right)$

$\begin{aligned} &-12=(4 c-1)^{2}\left(\frac{3-15}{45}\right) \\ &-12=(4 c-1)^{2}\left(\frac{-12}{45}\right) \end{aligned}$

$-12 \times \frac{45}{-12}=(4 c-1)^{2}$

$\begin{aligned} &(4 c-1)^{2}=45 \\ &4 c-1=\sqrt{45} \\ &4 c-1=\pm 3 \sqrt{5} \end{aligned}$

$\begin{aligned} &c=\frac{\pm 3 \sqrt{5}+1}{4} \\ &c=\frac{3 \sqrt{5}+1}{4} \approx 1.92 \in(1,4) \end{aligned}$
Thus, Lagrange’s Theorem is verified.

Mean Value Theoram exercise 14.2 question 5 maths

Answer: $\left(\frac{9}{2}, \frac{1}{4}\right)$
Hint: You must know the value of tangent parabola.

Given:$y=(x-4)^{2}$ parallel to chord: $(4,0)(5,1)$
Solution:
$y=(x-4)^{2}$
Since tangent is parallel to chord joining (3, 0) and (4, 1), so, we get:
$2(x-4)=\frac{1-0}{5-4}$
$\begin{aligned} &2 x-8=\frac{1}{1} \\ &2 x=9 \end{aligned}$
$x=\frac{9}{2}$
$\begin{aligned} &\text { When } &x=\frac{9}{2}, \text { then } y=\left(\frac{9}{2}-4\right)^{2} \end{aligned}$
$\begin{aligned} &=\left(\frac{9-8}{2}\right)^{2} \\ &=\left(\frac{1}{2}\right)^{2} \\ &=\frac{1}{4} \end{aligned}$
Hence, the point is $\left(\frac{9}{2}, \frac{1}{4}\right) .$