# NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.8

## Chapter 5 Ex.5.8 Question 1

Verify Rolle’s Theorem for the function $$f\left( x \right) = {x^2} + 2x - 8,x \in \left[ { - 4,2} \right]$$

### Solution

Given, $$f\left( x \right) = {x^2} + 2x - 8$$, being polynomial function is continuous in $$\left[ { - 4,2} \right]$$ and also differentiable in $$\left( { - 4,2} \right)$$.

\begin{align}f\left( { - 4} \right) &= {\left( { - 4} \right)^2} + 2.\left( { - 4} \right) - 8\\ &= 16 - 8 - 8\\ &= 0\end{align}

\begin{align}f\left( 2 \right) &= {\left( 2 \right)^2} + 2 \times 2 - 8\\ &= 4 + 4 - 8\\ &= 0\end{align}

Therefore, $$f\left( { - 4} \right) = f\left( 2 \right) = 0$$

The value of $$f\left( x \right)$$at $$- 4$$ and 2 coincides.

Rolle’s Theorem states that there is a point $$c \in \left( { - 4,2} \right)$$ such that $$f'\left( c \right) = 0$$

$$f\left( x \right) = {x^2} + 2x - 8$$

Therefore, $$f'\left( x \right) = 2x + 2$$

Hence,

\begin{align}f'\left( c \right) &= 0\\2c + 2 &= 0\\c &= - 1\end{align}

Thus, $$c = - 1 \in \left( { - 4,2} \right)$$

Hence, Rolle’s Theorem is verified.

## Chapter 5 Ex.5.8 Question 2

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s Theorem from these examples?

(i) $$f\left( x \right) = \left[ x \right]{\;\rm{ for\; }} x \in \left[ {5,9} \right]$$

(ii) $$f\left( x \right) = \left[ x \right]{\;\rm{ for\; }}x \in \left[ { - 2,2} \right]$$

(iii) $$f\left( x \right) = {x^2} - 1{\;\rm{ for\;}}x \in \left[ {1,2} \right]$$

### Solution

By Rolle’s Theorem, $$f:\left[ {a,b} \right] \to {\bf{R}}$$,

If

(a) $$f$$ is continuous on $$\left[ {a,b} \right]$$

(b) $$f$$ is continuous on $$\left( {a,b} \right)$$

(c) $$f\left( a \right) = f\left( b \right)$$

Then, there exists some $$c \in \left( {a,b} \right)$$ such that $$f'\left( c \right) = 0$$

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

(i) $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]$$

Since, the given function $$f\left( x \right)$$ is not continuous at every integral point.

In general, $$f\left( x \right)$$ is not continuous at $$x = 5$$ and $$x = 9$$

Therefore, $$f\left( x \right)$$ is not continuous in $$\left[ {5,9} \right]$$

Also, $$f\left( 5 \right) = \left[ 5 \right] = 5$$ and $$f\left( 9 \right) = \left[ 9 \right] = 9$$

Thus, $$f\left( 5 \right) \ne f\left( 9 \right)$$

The differentiability of $$f$$ in $$\left( {5,9} \right)$$ is checked as follows.

Let $$n$$ be an integer such that $$n \in \left( {5,9} \right)$$

The LHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty$$

The RHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$$

Since LHD and RHD of $$f$$ at $$x = n$$ are not equal, f is not differentiable at $$x = n$$

Therefore, $$f$$ is not differentiable in $$\left( {5,9} \right)$$.

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

Thus, Rolle’s Theorem is not applicable for $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]$$.

(ii) $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]$$

Since, the given function $$f\left( x \right)$$ is not continuous at every integral point.

In general, $$f\left( x \right)$$ is not continuous at $$x = - 2$$ and $$x = 2$$

Therefore, $$f\left( x \right)$$ is not continuous in $$\left[ { - 2,2} \right]$$

Also, $$f\left( { - 2} \right) = \left[ { - 2} \right] = - 2$$ and $$f\left( 2 \right) = \left[ 2 \right] = 2$$

Thus, $$f\left( { - 2} \right) \ne f\left( 2 \right)$$

The differentiability of $$f$$ in $$\left( { - 2,2} \right)$$ is checked as follows.

Let $$n$$ be an integer such that $$n \in \left( { - 2,2} \right)$$

The LHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty$$

The RHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$$

Since LHD and RHD of $$f$$ at $$x = n$$ are not equal, f is not differentiable at $$x = n$$

Therefore, $$f$$ is not differentiable in $$\left( { - 2,2} \right)$$.

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

Thus, Rolle’s Theorem is not applicable for $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]$$

(iii) $$f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]$$

Since, $$f$$ being a polynomial function is continuous in $$\left[ {1,2} \right]$$ and is differentiable in $$\left( {1,2} \right)$$.

Thus,

\begin{align}f\left( 1 \right)& = {\left( 1 \right)^2} - 1 = 0\\f\left( 2 \right)& = {\left( 2 \right)^2} - 1 = 3\end{align}

Therefore, $$f\left( 1 \right) \ne f\left( 2 \right)$$

Since, $$f$$ does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $$f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]$$.

## Chapter 5 Ex.5.8 Question 3

If $$f:\left[ { - 5,5} \right] \to {\bf{R}}$$ is a differentiable function and if $$f'\left( x \right)$$ does not vanish anywhere, then prove that $$f\left( { - 5} \right) \ne f\left( 5 \right)$$.

### Solution

Given, $$f:\left[ { - 5,5} \right] \to {\bf{R}}$$ is a differentiable function.

Since every differentiable function is a continuous function, we obtain

$$f$$ is continuous on $$\left[ { - 5,5} \right]$$

$$f$$ is continuous on $$\left( { - 5,5} \right)$$

Thus, by the Mean Value Theorem, there exists $$c \in \left( { - 5,5} \right)$$ such that

\begin{align} &\Rightarrow \; f'\left( c \right) = \frac{{f\left( 5 \right) - f\left( { - 5} \right)}}{{5 - \left( { - 5} \right)}}\\& \Rightarrow \; 10f'\left( c \right) = f\left( 5 \right) - f\left( { - 5} \right)\end{align}

It is also given that $$f'\left( x \right)$$ does not vanish anywhere.

Therefore, $$f'\left( c \right) \ne 0$$

Thus,

\begin{align}& \Rightarrow \; 10f'\left( c \right) \ne 0\\ &\Rightarrow \; f\left( 5 \right) - f\left( { - 5} \right) \ne 0\\ &\Rightarrow \; f\left( 5 \right) \ne f\left( { - 5} \right)\end{align}

Hence proved.

## Chapter 5 Ex.5.8 Question 4

Verify Mean Value Theorem, if $$f\left( x \right) = {x^2} - 4x - 3$$ in the integral $$\left[ {a,b} \right]$$, where $$a = 1$$ and $$b = 4$$.

### Solution

Given, $$f\left( x \right) = {x^2} - 4x - 3$$

$$f$$, being a polynomial function, is continuous in $$\left[ {1,4} \right]$$ and is differentiable in $$\left( {1,4} \right)$$, whose derivative is $$2x - 4$$.

Thus,

\begin{align}f\left( 1 \right) &= {1^2} - 4 \times 1 - 3 = - 6\\f\left( 4 \right) &= {4^2} - 4 \times 4 - 3 = - 3\end{align}

Therefore,

\begin{align}\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}& = \frac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}}\\ &= \frac{{ - 3 - \left( { - 6} \right)}}{3}\\& = \frac{3}{3}\\& = 1\end{align}

Mean Value Theorem states that there is a point $$c \in \left( {1,4} \right)$$ such that $$f'\left( c \right) = 1$$

Hence,

\begin{align}& \Rightarrow \; f'\left( c \right) = 1\\& \Rightarrow \; 2c - 4 = 1\\& \Rightarrow \; c = \frac{5}{2}\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{where }}c = \frac{5}{2} \in \left( {1,4} \right)} \right]\end{align}

Thus, mean value theorem is verified for the given function.

## Chapter 5 Ex.5.8 Question 5

Verify Mean Value Theorem, if $$f\left( x \right) = {x^3} - 5{x^2} - 3x$$ in the interval $$\left[ {a,b} \right]$$ where $$a = 1$$ and$$b = 3$$. Find all $$c \in \left( {1,3} \right)$$ for which $$f'\left( c \right) = 0$$.

### Solution

Given, f is $$f\left( x \right) = {x^3} - 5{x^2} - 3x$$

$$f$$, being a polynomial function, is continuous in $$\left[ {1,3} \right]$$ and is differentiable in $$\left( {1,3} \right)$$, whose derivative is $$3{x^2} - 10x - 3$$

Thus,

\begin{align}f\left( 1 \right) &= {1^3} - 5 \times {1^2} - 3 \times 1 = - 7\\f\left( 3 \right) &= {3^3} - 5 \times {3^2} - 3 \times 3 = - 27\end{align}

Therefore,

\begin{align}\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} &= \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}}\\ &= \frac{{ - 27 - \left( { - 7} \right)}}{{3 - 1}}\\& = - 10\end{align}

Mean Value Theorem states that there exists a point $$c \in \left( {1,3} \right)$$ such that $$f'\left( c \right) = - 10$$

Hence,

\begin{align}& \Rightarrow \; f'\left( c \right) = - 10\\& \Rightarrow \; 3{c^2} - 10c - 3 = - 10\\ &\Rightarrow \; 3{c^2} - 10c + 7 = 0\\ &\Rightarrow \; 3{c^2} - 3c - 7c + 7 = 0\\& \Rightarrow \; 3c\left( {c - 1} \right) - 7\left( {c - 1} \right) = 0\\& \Rightarrow \; \left( {c - 1} \right)\left( {3c - 7} \right) = 0\\ &\Rightarrow \; c = 1,\frac{7}{3} \qquad \left[ {{\text{where }}c = \frac{7}{3} \in \left( {1,3} \right)} \right]\end{align}

Thus, Mean Value Theorem is verified for the given function and $$c = \frac{7}{3} \in \left( {1,3} \right)$$ is the only point for which $$f'\left( c \right) = 0$$.

## Chapter 5 Ex.5.8 Question 6

Examine the applicability of Mean Value Theorem for all three functions given

(i) $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]$$

(ii) $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]$$

(iii) $$f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]$$

### Solution

Mean Value Theorem states that for a function $$f:\left[ {a,b} \right] \to {\bf{R}}$$, if

$$f$$ is continuous on $$\left[ {a,b} \right]$$

$$f$$ is continuous on $$\left( {a,b} \right)$$

Then there exists some $$c \in \left( {a,b} \right)$$ such that $$f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$$

Thus, Mean Value Theorem is not applicable to those functions that do not satisfy any of three conditions of the hypothesis.

(i) $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]$$

Since, the given function $$f\left( x \right)$$ is not continuous at every integral point.

In general, $$f\left( x \right)$$ is not continuous at $$x = 5$$ and $$x = 9$$

Therefore, $$f\left( x \right)$$ is not continuous in $$\left[ {5,9} \right]$$

The differentiability of $$f$$ in $$\left( {5,9} \right)$$ is checked as follows.

Let $$n$$ be an integer such that $$n \in \left( {5,9} \right)$$

The LHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty$$

The RHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$$

Since LHD and RHD of $$f$$ at $$x = n$$ are not equal, $$f$$ is not differentiable at $$x = n$$

Therefore, $$f$$ is not differentiable in $$\left( {5,9} \right)$$.

It is observed that $$f$$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ {5,9} \right]$$

(ii) $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]$$

Since, the given function $$f\left( x \right)$$ is not continuous at every integral point.

In general, $$f\left( x \right)$$ is not continuous at $$x = - 2$$ and $$x = 2$$

Therefore, $$f\left( x \right)$$ is not continuous in $$\left[ { - 2,2} \right]$$

The differentiability of $$f$$ in $$\left( { - 2,2} \right)$$ is checked as follows.

Let n be an integer such that $$n \in \left( { - 2,2} \right)$$

The LHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - 1}}{h} = \infty$$

The RHD of $$f$$ at $$x = n$$ is

$$\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left[ {n + h} \right] - \left[ n \right]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$$

Since LHD and RHD of $$f$$ at $$x = n$$ are not equal, $$f$$ is not differentiable at $$x = n$$

Therefore, $$f$$ is not differentiable in $$\left( { - 2,2} \right)$$ .

It is observed that $$f$$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for $$f\left( x \right) = \left[ x \right]{\text{ for }}x \in \left[ { - 2,2} \right]$$.

(iii) $$f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]$$

Since, $$f$$ being a polynomial function is continuous in $$\left[ {1,2} \right]$$ and is differentiable in $$\left( {1,2} \right)$$

It is observed that $$f$$ satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $$f\left( x \right) = {x^2} - 1{\text{ for }}x \in \left[ {1,2} \right]$$.

It can be proved as follows.

We have, $$f\left( x \right) = {x^2} - 1$$

Then,

\begin{align}f\left( 1 \right) &= {\left( 1 \right)^2} - 1 = 0,\\f\left( 2 \right) &= {\left( 2 \right)^2} - 1 = 3\end{align}

Therefore,

\begin{align}\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} &= \frac{{f\left( 2 \right) - f\left( 1 \right)}}{{2 - 1}} = \frac{{3 - 0}}{1}\\& = 3\end{align}

Hence, $$f'\left( x \right) = 2x$$

Thus,

\begin{align} &\Rightarrow \; f'\left( c \right) = 3\\& \Rightarrow \; 2c = 3\\ &\Rightarrow \; c = \frac{3}{2}\\ &\Rightarrow \; c = 1.5 \qquad \left[ {{\text{where }}1.5 \in \left[ {1,2} \right]} \right]\end{align}

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