RD Sharma Class 12 Exercise 14.2 Mean Value Theorems Solutions Maths

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Class 12 RD Sharma Chapter 14 Exercise 14.2 Solutions material covers all topics and contains clear concepts. It is the best choice for students for exam preparation.

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RD Sharma Class 12th Exercise 14.2 material contains solved questions. It will help students save time and be convenient for revision.

Q: What is the mean value theorem?

If an arc is drawn using two endpoints, at least one point where its tangent is parallel to the secant of the points.

Q: What is Rolle's theorem?

Any differentiable function with equal values at two points will have at least one value where the first derivative is zero. To get more information about this, topic check RD Sharma Class 12 Solutions Chapter 14 Ex 14.2.

Q: What is Lagrange's theorem?

If a subgroup is taken from a group of any order, it will be a divisor of that group. This is Lagrange's theorem. For more info, check RD Sharma Class 12 Solutions Mean Value Theorem Ex 14.2.

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RD Sharma Class 12 Solutions Chapter 14 - Mean Value Theorems - Other Exercise
Mean Value Theorems Excercise:14.2
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RD Sharma Class 12 Solutions Chapter 14 - Mean Value Theorems - Other Exercise

Mean Value Theorems Excercise:14.2

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
$f(x) = x^2 -8x +12$

Being polynomial f(x) is continuous for all x and hence continuous in [2,6]
$f'(x) = 2x -8$ , 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 theorem exercise 14.2 question 2 (ii)

Answer:
$c=2 \in(1,3)$ , Hence, Rolle’s Theorem is verified.
Hint:
$f(x)$ is continuous for all x and hence continuous in [1,3]
Given:
$f(x)=x^{2}-4 x+3$ on [1,3]
Explanation:
We have
$f(x)=x^{2}-4 x+3$

Being polynomial f(x) is continuous for all x and hence continuous in [1,3]
$f^{\prime}(x)=2 x-4$ , which exists in (1,3)

$\therefore f (x )$ is derivable in (1,3)
3.
$\\f(1)=(1)^{2}-4(1)+3\\\\ =1-4+3=-3+3=0\\\\ f(3)=(3)^{2}-4(3)+3\)\\\\\ =9-12+3=-3+3=0\\\\ \therefore f(1)=f(3)$
Thus all the conditions of Rolle’s Theorem are satisfied.
There exists at least one $c \in(1,3)$ such that $f'(c) = 0$
$\\ \Rightarrow 2c - 4 = 0 \\\\ \Rightarrow c = 2 \\\\ \Rightarrow c = 2 \in (1,3)$
Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(iii)

Answer:
$c = \frac{4}{3} \in (1,2)$
Hint:
Using discrimination method, $\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Given:
$f(x)=(x-1)(x-2)^{2}$ on [1,2]
Explanation:
We have,
$f(x)=(x-1)(x-2)^{2}$

Being polynomial f(x) is continuous for all x and hence continuous in [1,2]

$\\\\ f^{\prime}(x)=(x-1) \frac{d}{d x}(x-2)^{2}+(x-2)^{2} \frac{d}{d x}(x-1)\\\\ =(x-1) 2(x-2)^{2-1}(1)+(x-2)^{2}(1) \\\\ \quad=2(x-1)(x-2)+(x-2)^{2}$
Which exists in (1.2)
$\therefore f(x)$ is derivable in (1,2)
3.
$\\ f(1)=(1-1)(1-2)^{2}=0 \\\\ f(2)=(2-1)(2-2)^{2}=0\\\\ \therefore f(1)=f(2)$

Thus all the conditions of Rolle’s Theorem are satisfied.
There exists at least one $c \in(1,2)$ such that $f^{\prime}(c)=0$
Now, $f^{\prime}(c)=0$
$\begin{aligned} &=2(c-1)(c-2)+(c-2)^{2} \\ &=2\left(c^{2}-2 c-c+2\right)+c^{2}+4-4 c \\ &=2\left(c^{2}-3 c+2\right)+c^{2}+4-4 c \\ &=2 c^{2}-6 c+2+c^{2}+4-4 c \\ &=3 c^{2}-10 c+8 \end{aligned}$
Using discrimination method,
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\\c=\frac{-(-10) \pm \sqrt{(-10)^{2}-4 \times 3 \times 8}}{2 \times 3}\\\\ c=\frac{10 \pm \sqrt{100-96}}{6}\\\\ c=\frac{10 \pm \sqrt{4}}{6}=\frac{10 \pm 2}{6}\\\\ =\frac{8}{6}, \frac{12}{6}\\\\ c=\frac{4}{3}, 2\\\\ c=\frac{4}{3} \in(1,2)$
[Neglecting the value 2 ]
Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(iv)

Answer:
$c=\left(1, \frac{1}{3}\right) \in(0,1)$ , hence Rolle’s Theorem is verified.
Hint:
Using discrimination method, $\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Given:
$f(x)=x(x-1)^{2}$ on [0,1]
Explanation:
We have,
$f(x)=x(x-1)^{2}$

Being polynomial f(x) is continuous for all x and hence continuous in [0,1]

2.
$\\f^{\prime}(x)=x \frac{d}{d x}(x-1)^{2}+(x-1)^{2} \frac{d}{d x} x \\\\ =x(2)(x-1)^{1}+(x-1)^{2}(1)$
$\\=2 x(x-1)+(x-1)^{2} \\ =2 x^{2}-2 x+x^{2}+1-2 x \\ =3 x^{2}-4 x+1 \\ =3 x^{2}-3 x-x+1 \\$
3.
$\\f(0)=0(0-1)^{2}=0 \\\\ \quad f(1)=1(1-1)^{2}=0 \\\\ \therefore f(0) =f(1)$
Thus all the conditions of Rolle’s Theorem are satisfied.
There exists at least one $c \in(0,1)$ such that $f^{\prime}(c)=0$
Now, $f^{\prime}(c)=0$
$3 c^{2}-4 c+1$
Using discrimination method,
$\\ \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\\\\\ \begin{aligned} c &=\frac{-(-4) \pm \sqrt{16-12}}{6} \\\\ c &=\frac{4 \pm \sqrt{4}}{6} \\\\ c &=\frac{4 \pm 2}{6} \\\\ &=\frac{6}{6}, \frac{2}{6} \\\\ c &=1, \frac{1}{3} \\\\ c &=\left(1, \frac{1}{3}\right) \in(0,1) \end{aligned}$
Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(v)

Answer:
$c=\left(\frac{2-\sqrt{7}}{3}, \frac{2+\sqrt{7}}{3}\right) \in[-1,2]$ , Hence Rolle’s Theorem is verified.
Hint:
Using discrimination method, $\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Given:
$f(x)=\left(x^{2}-1\right)(x-2)$ on [1,-2]
Explanation:
We have,
$f(x)=\left(x^{2}-1\right)(x-2)$

Being polynomial f(x) is continuous for all x and hence continuous in [-1,2]

$\\f^{\prime}(x) =\left(x^{2}-1\right) \frac{d}{d x}(x-2)+(x-2) \frac{d}{d x}\left(x^{2}-1\right) \\\\ =\left(x^{2}-1\right)(1)+(x-2)(2 x) \\\\ =x^{2}-1+2 x^{2}-4 x \\\\ =3 x^{2}-4 x-1 \\\\$

3.
$\\ f(-1)=\left((-1)^{2}-1\right)(-1-2)=(1-1)(-1-2)=0\\\\ f(2)=\left(2^{2}-1\right)(2-2)=(4-1)(0)=0 \\\\ \therefore f(-1)=f(2)$
Thus all the conditions of Rolle’s Theorem are satisfied.
There exists at least one $c \in(-1,2)$ such that $f^{\prime}(c)=0$
Now, $f^{\prime}(c)=0$
$3 c^{2}-4 c-1$

Using discrimination method,
$\begin{array}{l} \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\\\ c=\frac{4 \pm \sqrt{16+12}}{6}=\frac{4 \pm \sqrt{28}}{6} \\\\ c=\frac{4 \pm 2 \sqrt{7}}{6} \\\\ c=\frac{2(2 \pm \sqrt{7})}{6} \\\\ c=\left(\frac{2-\sqrt{7}}{3}, \frac{2+\sqrt{7}}{3}\right) \in[-1,2] \end{array}$
Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(vii)

Answer:
$c=\frac{-5}{2} \in(-3,-2)$ , hence Rolle’s Theorem is verified.
Hint:
f(x) is continuous for all x and hence continuous in [-3,-2]
Given:
$f(x)=x^{2}+5 x+6$ on [-3,-2]
Explanation:
We have,
$f(x)=x^{2}+5 x+6$

Being polynomial f(x) is continuous for all x and hence continuous in [-3,-2]
$f^{\prime}(x)=2 x+5$ , which exists in [-3,-2]

$\therefore f(x)$ is derivable in (-3,-2)
3. $f(-3)=(-3)^{2}+5(-3)+6$
$=9-15+6=-6+6=0$
$f(-2)=(-2)^{2}+5(-2)+6$
$=4-10+6=-6+6=0$
$\therefore f(-3) =f(-2)$
Thus all the conditions of Rolle’s Theorem are satisfied.
There exists at least one $c \in(-3,-2)$ such that $f^{\prime}(c)=0$ Now,
$f^{\prime}(c)=0\\\\ 2 c+5=0\\\\ 2 c=-5\\\\ c=\frac{-5}{2}\\\\ c = \frac{-5}{2}\in (-3,-2)$
Hence, Rolle’s Theorem is verified.

Mean Value Theoram exercise 14.2 question 1(i)

Answer: $\frac{5}{2}$
Hint:You must know the formula of Lagrange’s Mean Value Theorem.
Given: $f(x)=x^{2}-1 \text { on }[2,3]$
Solution:
$f(x)=x^{2}-1$
$f(x)$ is a polynomial function.
It is continuous in [2, 3]
$f^{\prime}(x)=2 x$
(Which is defined in [2, 3])
$f(x)$ is differentiable in [2, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.
Now, there exists at least one value,
$\begin{aligned} &c \in[2,3] \\ &f^{\prime}(c)=\frac{f(3)-f(2)}{3-2} \end{aligned}$
$\begin{aligned} &2 c=\frac{\left(3^{2}-1\right)-\left(2^{2}-1\right)}{1} \\ &2 c=(9-1)-(4-1) \end{aligned}$
$\begin{aligned} &2 c=8-3 \\ &2 c=5 \\ &c=\frac{5}{2} \end{aligned}$

Mean Value Theoram exercise 14.2 question 1(ii)

Answer: $\frac{1}{3}$
Hint:You must know the formula of Lagrange’s Mean Value Theorem.
Given: $f(x)=x^{3}-2 x^{2}-x+3 \text { on }[0,1]$
Solution: $f(x)=x^{3}-2 x^{2}-x+3$

$f(x)$ is a polynomial function.
It is continuous in [0, 1]
$f^{\prime}(x)=3 x^{2}-4 x-1$
(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,
$\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \end{aligned}$
$3 c^{2}-4 c-1=\frac{\left((1)^{3}-2(1)^{2}-1+3\right)-\left((0)^{3}-2(0)^{2}-0+3\right)}{1}$
$\begin{aligned} &3 c^{2}-4 c-1=(1-2-1+3)-(3) \\ &3 c^{2}-4 c-1=1-3 \\ &3 c^{2}-4 c-1=-2 \\ &3 c^{2}-4 c-1+2=0 \\ &3 c^{2}-4 c+1=0 \end{aligned}$
$\begin{aligned} &3 c^{2}-3 c-c+1=0 \\ &3 c(c-1)-(c-1)=0 \\ &(3 c-1)(c-1)=0 \end{aligned}$
Therefore, $c=\frac{1}{3} \text { and } \mathrm{c}=1$

Mean Value Theoram exercise 14.2 question 1(iii)

Answer: $\frac{3}{2}$
Hint:You must know the formula of Lagrange’s Mean Value Theorem.
Given: $f(x)=x(x-1) \text { on }[1,2]$
Solution: $f(x)=x(x-1)=x^{2}-x$

$f(x)$ is a polynomial function.
It is continuous in [1, 2]
$f^{\prime}(x)=2 x-1$
(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,
$\begin{aligned} &c \in[1,2] \\ &f^{\prime}(c)=\frac{f(2)-f(1)}{2-1} \\ &2 c-1=\frac{\left(2^{2}-2\right)-\left(1^{2}-1\right)}{1} \end{aligned}$
$\begin{aligned} &2 c-1=(4-2)-(1-1) \\ &2 c-1=2-0 \\ &2 c-1=2 \\ &2 c=2+1 \\ &c=\frac{3}{2} \end{aligned}$

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 $\left(\frac{9}{2}, \frac{1}{4}\right) .$

Mean Value Theoram exercise 14.2 question 6

Answer: $\left(\frac{1}{2}, \frac{3}{4}\right)$
Hint: You must know the slope of tangent’

Given: $y=x^{2}+x$ parallel to chord: $\left [ 0,0 \right ],\left [ 1,2 \right ]$
Solution:
$y=x^{2}+x,[0,0],[1,2]$
$\frac{d y}{d x}=2 x+1$
Slope of tangent $= 2x + 1$
Slope of line joining [0, 0] and [1, 2] = = 2
The tangent is parallel to this line:
$\therefore$ Slope is equal
$\begin{aligned} &2 x+1=2 \\ &2 x=2-1 \\ &2 x=1 \\ &x=\frac{1}{2} \end{aligned}$
$x=\frac{1}{2}$
$\therefore \quad y=\left(\frac{1}{2}\right)^{2}+\frac{1}{2}$
$\begin{aligned} &=\left(\frac{1}{4}\right)+\frac{1}{2} \\ &=\frac{3}{4} \end{aligned}$
Hence, the points are $\left(\frac{1}{2}, \frac{3}{4}\right)$

Mean Value Theoram exercise 14.2 question 7

Answer: $\left(\frac{7}{2}, \frac{1}{4}\right)$
Hint: You must know the value of tangent parabola.

Given: $y=(x-3)^{2}$ parallel to chord: $(3,0)(4,1)$
Solution:
$y=(x-3)^{2}$
Since tangent is parallel to chord joining (3, 0) and (4, 1), so, we get:
$2(x-3)=\frac{1-0}{4-3}$
$\begin{aligned} &2 x-6=\frac{1}{1} \\ &2 x=7 \\ &x=\frac{7}{2} \end{aligned}$
$\begin{aligned} \text {When} \; \; \; \; \; &x=\frac{7}{2}, \text { then } &y=\left(\frac{7}{2}-3\right)^{2} \end{aligned}$
$\begin{aligned} &=\left(\frac{7-6}{2}\right)^{2} \\ &=\left(\frac{1}{2}\right)^{2} \\ &=\frac{1}{4} \end{aligned}$
Hence, the point is $\left(\frac{7}{2}, \frac{1}{4}\right)$

Mean Value Theoram exercise 14.2 question 8

Answer: $\left(\pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3} \sqrt{\frac{7}{3}}\right)$
Hint: You must know the slope of the tangent.

Given: $y=x^{3}-3 x ;(1,-2),(2,2)$
Solution:
$\begin{aligned} &y=x^{3}-3 x \\ &\frac{d y}{d x}=3 x^{2}-3 \end{aligned}$
Slope of tangent $=3 x^{2}-3$
Slope of line joining (1, -2) and (2, 2) $=\frac{2-(-2)}{2-1}$
$\begin{aligned} &=\frac{2+2}{1} \\ &=4 \end{aligned}$
The tangent is parallel to the line: $y=x^{3}-x$
Slope is equal to $3 x^{2}-3=4$
$\begin{gathered} 3 x^{2}=7 \\ x^{2}=\frac{7}{3} \\ x=\pm \sqrt{\frac{7}{3}} \\ y=x^{3}-x \end{gathered}$
$\begin{aligned} &=\left(\sqrt{\frac{7}{3}}\right)^{3}+3 \sqrt{\frac{7}{3}} \\ &=\sqrt{\frac{7}{3}}\left(\frac{7}{3}-3\right) \\ &=\sqrt{\frac{7}{3}}\left(\frac{7-9}{3}\right) \\ &=\frac{2}{3} \sqrt{\frac{7}{3}} \end{aligned}$
Points are $\left(\pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3} \sqrt{\frac{7}{3}}\right)$

Mean Value Theoram exercise 14.2 question 9 maths

Answer: $\left(\pm \sqrt{\frac{13}{3}} \cdot\left(\frac{13}{3}\right)^{\frac{1}{2}}+1\right)$
Hint: You must know the slope of the tangent.

Given: $y=x^{3}+1 ;(1,2),(3,28)$
Solution:
$\begin{aligned} &y=x^{3}+1 \\ &\frac{d y}{d x}=3 x^{2} \end{aligned}$
Slope of tangent = 3x2
Slope of line joining (1, 2) and (3, 28) $=\frac{28-2}{3-1}$
$\begin{aligned} &=\frac{26}{2} \\ &=13 \end{aligned}$
The tangent is parallel to the line:
Slope is equal to $3x^{2} = 13$
$\begin{aligned} &x^{2}=\frac{13}{3} \\ &x=\pm \sqrt{\frac{13}{3}} \end{aligned}$
$\Rightarrow \quad y=x^{3}+1$
$\begin{aligned} &=\left(\sqrt{\frac{13}{3}}\right)^{3}+1 \\ &=\left(\frac{13}{3}\right)^{\frac{3}{2}}+1 \end{aligned}$
Points are $\left(\pm \sqrt{\frac{13}{3}} ,\left (\frac{13}{3}\right)^{\frac{1}{2}}+1\right)$

Mean Value Theoram exercise 14.2 question 10

Answer: Point P $\left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)$
Hint: Find $\frac{dy}{dx}$ and then use this formula for slope of tangent.

Given: $\mathrm{x}=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \quad 0<\theta<\frac{\pi}{2} .$
The tangent to C is parallel to the chord joining the points (a, 0) and (0, a).
Solution:
$\begin{aligned} &x=a \cos ^{3} \theta \\ &\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta \\ &y=a \sin ^{3} \theta \\ &\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta \end{aligned}$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{3 a \sin ^{2} \theta \cdot \cos \theta}{-3 a \cos ^{2} \theta \cdot \sin \theta} \\ &\Rightarrow \frac{d y}{d x}=-\tan \theta \end{aligned}$
Now, slope of chord $=\frac{a-0}{0-a}$
Slope of chord $=\frac{a}{-a}=-1$
Slope of chord $=\frac{d y}{d x}$
$\begin{aligned} &\Rightarrow-1=-\tan \theta \\ &\Rightarrow \tan \theta=1 \\ &\Rightarrow \tan \theta=\tan \frac{\pi}{4} \\ &\Rightarrow \theta=\frac{\pi}{4} \end{aligned}$
Now, $x=a \cos ^{3} \theta$
$\begin{aligned} &=a \cos ^{3} \frac{\pi}{4} \\ &=a\left(\frac{1}{\sqrt{2}}\right)^{2} \\ &=\frac{a}{2 \sqrt{2}} \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{aligned}$
Now, $y=a \sin ^{3} \theta$
$\begin{aligned} &=a \sin ^{3} \frac{\pi}{4} \\ &=a\left(\frac{1}{\sqrt{2}}\right)^{2} \\ &=\frac{a}{2 \sqrt{2}} \quad\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{aligned}$
Hence Point P is $\left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)$

Mean Value Theoram exercise 14.2 question 11

Answer:
Hint: Use Lagrange’s mean value theorem formula.

Given:Use Lagrange’s mean value theorem on $(b-a) \sec ^{2} a<\tan b-\tan a<(b-a) \sec ^{2} b$ , where $0<a<b<\frac{\pi}{2}$ .
Solution:
$\text { Let } f(x)=\tan x \text { in }(a, b) \text { where } 0<a<b<\frac{\pi}{2}$
$\therefore$ f (x) is continuous in [a, b] and differentiable in (a, b)
Now, $f^{\prime}(x)=\sec ^{2} x$
$\Rightarrow f^{\prime}(c)=\sec ^{2} c$
By Mean Value Theorem,
$\begin{aligned} f^{\prime}(c) &=\frac{f(b)-f(a)}{b-a} \\ f^{\prime}(c) &=\frac{\tan b-\tan a}{b-a} \end{aligned}$ $c \in(a, b)$
$\begin{aligned} &\Rightarrow a<c<b \\ &\Rightarrow \sec ^{2} a<\sec ^{2} c<\sec ^{2} b \\ &\Rightarrow \sec ^{2} a<\frac{\tan b-\tan a}{b-a}<\sec ^{2} b \end{aligned}$

Hence, proved.

RD Sharma Class 12th Exercise 14.2 has 26 questions, 24 of which are Level 1 and two are Level 2. The Level 1 sums are simple and can be completed in a short time. This exercise is based on Lagrange's Mean Value theorem, calculating the tangents to curve, parabola, and other essential functional problems.

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