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# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.8) Exercise 5.8

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.8) Exercise 5.8

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## Access NCERT Solutions for Class 12 Maths Chapter 5- Continuity and Differentiability

### Exercise 5.8

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

Ans: Given function, $f(x) = {x^2} + 2x - 8$, is a polynomial function.

We know that A polynomial function is continuous and differentiable on any interval, the function $f(x) = {x^2} + 2x - 8$ is continuous in the interval $\left[ { - 4,2} \right]$ and differentiable in the interval $\left( { - 4,2} \right)$.

Hence the first and second conditions of Rolle’s Theorem are satisfied.

To check if the third condition is satisfied, the values of the functions at the endpoints of the given interval must coincide.

To derive the value of $f\left( { - 4} \right)$, substitute $- 4$ for $x$, as follows,

$\Rightarrow f\left( { - 4} \right) = {\left( { - 4} \right)^2} + 2\left( { - 4} \right) - 8$

$\Rightarrow f\left( { - 4} \right) = 16 - 8 - 8$

$\Rightarrow f\left( { - 4} \right) = 0$

To derive the value of $f\left( 2 \right)$, substitute 2 for $x$, as follows,

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

$\Rightarrow f\left( 2 \right) = 4 + 4 - 8$

$\Rightarrow f\left( 2 \right) = 0$

Therefore,

$\Rightarrow f\left( { - 4} \right) = 0 = f\left( 2 \right)$.

Hence both the values of $f\left( { - 4} \right)$and $f\left( 2 \right)$coincide and consequently, the third condition of Rolle’s Theorem is satisfied.

Since all the conditions of Rolle’s Theorem are satisfied, it can be stated by the Rolle’s Theorem that there exists a point $c \in \left( { - 4,2} \right)$ such that $f'\left( c \right) = 0$.

To determine $f'\left( c \right)$, derive $f'\left( x \right)$ by differentiating $f\left( x \right)$,

$\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} + 2x - 8} \right)$

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

To obtain the value of $c$, substitute $c$ for $x$ in $f'\left( x \right)$ and equate it to $0$in order to simplify.

$\Rightarrow f'\left( c \right) = 0$

$\Rightarrow 2c + 2 = 0$

$\Rightarrow 2c = - 2$

$\Rightarrow c = - 1$

Since $- 1 \in \left( { - 4,2} \right)$, then $c \in \left( { - 4,2} \right)$ and Rolle’s Theorem is verified for the given polynomial function $f(x) = {x^2} + 2x - 8$, where $x \in \left[ { - 4,2} \right]$.

2. Examine if Rolle’s Theorem is applicable to the functions. Can you say something about the converse of Rolle’s Theorem from this function?

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

Ans: The given function $f\left( x \right) = \left[ x \right]$, is a greatest integer function, which is discontinuous at every integral point.

Hence, in the given interval, which contains the integers $5,6,7,8,9$, $f\left( x \right) = \left[ x \right]$is discontinuous and consequently, the first condition of Rolle’s Theorem is not satisfied.

To check the second condition (differentiability) consider the following limits at an integer $n \in \left( {5,9} \right)$.

The left hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n - 1$, since $h \to {0^ - }$.

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

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

$\Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = - \infty$

The right hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n$, since $h \to {0^ + }$.

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

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

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

Since the left and right hand limit doesn’t coincide, the given function $f\left( x \right) = \left[ x \right]$ is not differentiable in the interval $\left( {5,9} \right)$ and consequently, the second condition of Rolle’s Theorem is not satisfied.

To check if the third condition is satisfied, the values of the functions at the endpoints of the given interval must coincide.

To derive the value of $f\left( 5 \right)$, substitute 5 for $x$, as follows,

$\Rightarrow f\left( 5 \right) = \left[ 5 \right]$

$\Rightarrow f\left( 5 \right) = 5$

To derive the value of $f\left( 9 \right)$, substitute 9 for $x$, as follows,

$\Rightarrow f\left( 9 \right) = \left[ 9 \right]$

$\Rightarrow f\left( 9 \right) = 9$

Therefore,

$\Rightarrow f\left( 5 \right) = 5 \ne 9 = f\left( 9 \right)$.

Hence, the last condition of Rolle’s Theorem is not satisfied.

Therefore, Rolle's Theorem is not applicable on the given function.

For converse of the Rolle’s Theorem, if there exist some $c \in \left( {5,9} \right)$ such that $f'\left( c \right) = 0$, then the function will be continuous on $\left[ {5,9} \right]$, differentiable on $\left( {5,9} \right)$ and $f\left( 5 \right) = f\left( 9 \right)$.

Now, since $f\left( x \right) = \left[ x \right]$is constant in any interval $\left( {n - 1,n} \right)$, where $n$ is an integer, then $f'\left( x \right) = 0$ in the open interval $\left( {n - 1,n} \right)$.

Therefore $f'\left( x \right) = 0$ in the interval $\left( {5,6} \right) \cup \left( {6,7} \right) \cup \left( {7,8} \right) \cup \left( {8,9} \right) \in \left( {5,9} \right)$.

Thus, there exists infinitely many $c \in \left( {5,9} \right)$ such that $f'\left( c \right) = 0$.

But neither $f\left( x \right) = \left[ x \right]$is continuous in $\left[ {5,9} \right]$, nor is it differentiable in $\left( {5,9} \right)$.

Also, $f\left( 5 \right) = 5 \ne 9 = f\left( 9 \right)$.

Hence  $f\left( x \right) = \left[ x \right]$ does not satisfy the converse of Rolle’s Theorem in the interval $\left[ {5,9} \right]$.

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

Ans: The given function $f\left( x \right) = \left[ x \right]$, is a greatest integer function, which is discontinuous at every integral point.

Hence, in the given interval, which contains the integers $- 2, - 1,0,1,2$, $f\left( x \right) = \left[ x \right]$is discontinuous and consequently, the first condition of Rolle’s Theorem is not satisfied.

To check the second condition(differentiability) consider the following limits at an integer $n \in \left( { - 2,2} \right)$.

The left hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n - 1$, since $h \to {0^ - }$.

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

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

$\Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = - \infty$

The right hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n$, since $h \to {0^ + }$.

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

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

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

Since the left and right hand limit doesn’t coincide, the given function $f\left( x \right) = \left[ x \right]$ is not differentiable in the interval $\left( { - 2,2} \right)$ and consequently, the second condition of Rolle’s Theorem is not satisfied.

To check if the third condition is satisfied, the values of the functions at the endpoints of the given interval must coincide.

To derive the value of $f\left( { - 2} \right)$, substitute $- 2$for $x$, as follows,

$\Rightarrow f\left( { - 2} \right) = \left[ { - 2} \right]$

$\Rightarrow f\left( { - 2} \right) = - 2$

To derive the value of $f\left( 2 \right)$, substitute 2 for $x$, as follows,

$\Rightarrow f\left( 2 \right) = \left[ 2 \right]$

$\Rightarrow f\left( 2 \right) = 2$

Therefore,

$\Rightarrow f\left( { - 2} \right) = - 2 \ne 2 = f\left( 2 \right)$.

Hence, the last condition of Rolle’s Theorem is not satisfied.

Therefore, Rolle's Theorem is not applicable on the given function.

For converse of the Rolle’s Theorem, if there exist some $c \in \left( { - 2,2} \right)$ such that $f'\left( c \right) = 0$, then the function will be continuous on $\left[ { - 2,2} \right]$, differentiable on $\left( { - 2,2} \right)$ and $f\left( { - 2} \right) = f\left( 2 \right)$.

Now, since $f\left( x \right) = \left[ x \right]$is constant in any interval $\left( {n - 1,n} \right)$, where $n$ is an integer, then $f'\left( x \right) = 0$ in the open interval $\left( {n - 1,n} \right)$.

Therefore $f'\left( x \right) = 0$ in the interval $\left( { - 2, - 1} \right) \cup \left( { - 1,0} \right) \cup \left( {0,1} \right) \cup \left( {1,2} \right) \in \left( { - 2,2} \right)$.

Thus, there exists infinitely many $c \in \left( { - 2,2} \right)$ such that $f'\left( c \right) = 0$.

But neither $f\left( x \right) = \left[ x \right]$is continuous in $\left[ { - 2,2} \right]$, nor it is differentiable in $\left( { - 2,2} \right)$.

Also, $f\left( { - 2} \right) = - 2 \ne 2 = f\left( 2 \right)$.

Hence  $f\left( x \right) = \left[ x \right]$ does not satisfy the converse of Rolle’s Theorem in the interval $\left[ { - 2,2} \right]$.

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

Ans: Given function, $f(x) = {x^2} - 1$, is a polynomial function.

We know that a polynomial function is continuous and differentiable on any interval, the function $f(x) = {x^2} - 1$ is continuous in the interval $\left[ {1,2} \right]$ and differentiable in the interval $\left( {1,2} \right)$.

Hence the first and second conditions of Rolle’s Theorem are satisfied.

To check if the third condition is satisfied, the values of the functions at the endpoints of the given interval must coincide.

To derive the value of $f\left( 1 \right)$, substitute 1 for $x$, as follows,

$\Rightarrow f\left( 1 \right) = {\left( 1 \right)^2} - 1$

$\Rightarrow f\left( 1 \right) = 1 - 1$

$\Rightarrow f\left( 1 \right) = 0$

To derive the value of $f\left( 2 \right)$, substitute 2 for $x$, as follows,

$\Rightarrow f\left( 2 \right) = {\left( 2 \right)^2} - 1$

$\Rightarrow f\left( 2 \right) = 4 - 1$

$\Rightarrow f\left( 2 \right) = 3$

Therefore,

$\Rightarrow f\left( 1 \right) = 0 \ne 3 = f\left( 2 \right)$.

Hence, the last condition of Rolle’s Theorem is not satisfied.

Therefore, Rolle's Theorem is not applicable on the given function.

For converse of the Rolle’s Theorem, if there exist some $c \in \left( {1,2} \right)$ such that $f'\left( c \right) = 0$, then the function will be continuous on $\left[ {1,2} \right]$, differentiable on $\left( {1,2} \right)$ and $f\left( 1 \right) = f\left( 2 \right)$.

First obtain the derivative of the polynomial function $f(x) = {x^2} - 1$ as follows.

$\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} - 1} \right)$

$\Rightarrow f'\left( x \right) = 2x$

To obtain the value of $c$, substitute $c$ for $x$ in $f'\left( x \right)$ and equate it to $0$ in order to simplify.

$\Rightarrow f'\left( c \right) = 0$

$\Rightarrow 2c = 0$

$\Rightarrow \dfrac{{2c}}{2} = \dfrac{0}{2}$

$\Rightarrow c = 0$

Since $c = 0 \notin \left( {1,2} \right)$, then there does not exist any $c \in \left( {1,2} \right)$ such that $f'\left( c \right) = 0$ and consequently, the converse of Rolle’s Theorem is not applicable on the given function $f(x) = {x^2} - 1$in the interval $\left[ {1,2} \right]$.

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

Ans: Given function $f$ is differentiable in the given interval $\left[ { - 5,5} \right]$. Since any differentiable function is also continuous, $f$ is continuous on $\left[ { - 5,5} \right]$ and differentiable on $\left( { - 5,5} \right)$. Hence, all the conditions of Mean Value Theorem are satisfied.

Therefore, by Mean Value Theorem, it can be stated that there exists a $c \in \left( { - 5,5} \right)$ such that,

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( { - 5} \right)}}{{5 - \left( { - 5} \right)}}$

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( { - 5} \right)}}{{5 + 5}}$

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( { - 5} \right)}}{{10}}$

$\Rightarrow 10 \cdot f'\left( c \right) = f\left( 5 \right) - f\left( { - 5} \right)$

Given that, $f'\left( x \right)$ does not vanish anywhere. Hence $f'\left( x \right) \ne 0$ for every real number $x$.

Therefore,

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

$\Rightarrow 10 \cdot f'\left( c \right) \ne 0$

$\Rightarrow f\left( 5 \right) - f\left( { - 5} \right) \ne 0$

$\Rightarrow f\left( 5 \right) \ne f\left( { - 5} \right)$

Hence $f\left( 5 \right) \ne f\left( { - 5} \right)$ is proved.

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

Ans: Given function, $f\left( x \right) = {x^2} - 4x - 3$, is a polynomial function.

We know that a polynomial function is continuous and differentiable on any interval. The function $f\left( x \right) = {x^2} - 4x - 3$ is continuous in the interval $\left[ {1,4} \right]$ and differentiable in the interval $\left( {1,4} \right)$.

Hence both the conditions of the Mean Value Theorem are satisfied.

Therefore, by Mean Value Theorem, it can be stated that there exists a $c \in \left( {1,4} \right)$ such that,

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}}$

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 4 \right) - f\left( 1 \right)}}{3}$

To derive the value of $f\left( 4 \right)$, substitute 4 for $x$, as follows,

$\Rightarrow f\left( 4 \right) = {\left( 4 \right)^2} - 4\left( 4 \right) - 3$

$\Rightarrow f\left( 4 \right) = 16 - 16 - 3$

$\Rightarrow f\left( 4 \right) = - 3$

To derive the value of $f\left( 1 \right)$, substitute 1 for $x$, as follows,

$\Rightarrow f\left( 1 \right) = {\left( 1 \right)^2} - 4\left( 1 \right) - 3$

$\Rightarrow f\left( 1 \right) = 1 - 4 - 3$

$\Rightarrow f\left( 1 \right) = - 3 - 3$

$\Rightarrow f\left( 1 \right) = - 6$

To determine $f'\left( c \right)$, derive $f'\left( x \right)$ by differentiating $f\left( x \right)$,

$\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} - 4x - 3} \right)$

$\Rightarrow f'\left( x \right) = 2x - 4$

Therefore,

$f'\left( c \right) = \dfrac{{f\left( 4 \right) - f\left( 1 \right)}}{3}$

$\Rightarrow 2c - 4 = \dfrac{{\left( { - 3} \right) - \left( { - 6} \right)}}{3}$

$\Rightarrow 2c - 4 = \dfrac{{ - 3 + 6}}{3}$

$\Rightarrow 2c - 4 = \dfrac{3}{3}$

$\Rightarrow 2c - 4 = 1$

Now in order to derive the value of $c$ solve the equation $2c - 4 = 1$.

$\Rightarrow 2c - 4 = 1$

$\Rightarrow 2c = 1 + 4$

$\Rightarrow 2c = 5$

$\Rightarrow c = \dfrac{5}{2}$

Since $c = \dfrac{5}{2} \in \left( {1,4} \right)$, the Mean Value Theorem is verified for the given function $f\left( x \right) = {x^2} - 4x - 3$in the interval$\left[ {1,4} \right]$.

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 $f'\left( c \right) = 0$.

Ans: Given function, $f\left( x \right) = {x^3} - 5{x^2} - 3x$, is a polynomial function.

We know that a polynomial function is continuous and differentiable on any interval. The function $f\left( x \right) = {x^3} - 5{x^2} - 3x$ is continuous in the interval $\left[ {1,3} \right]$ and differentiable in the interval $\left( {1,3} \right)$.

Hence both the conditions of Mean Value Theorem are satisfied.

Therefore, by Mean Value Theorem, it can be stated that there exists a $c \in \left( {1,3} \right)$ such that,

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}}$

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}$

To derive the value of $f\left( 3 \right)$, substitute 3 for $x$, as follows,

$\Rightarrow f\left( 3 \right) = {\left( 3 \right)^3} - 5{\left( 3 \right)^2} - 3\left( 3 \right)$

$\Rightarrow f\left( 3 \right) = 27 - 5\left( 9 \right) - 9$

$\Rightarrow f\left( 3 \right) = 27 - 45 - 9$

$\Rightarrow f\left( 3 \right) = - 18 - 9$

$\Rightarrow f\left( 3 \right) = - 27$

To derive the value of $f\left( 1 \right)$, substitute 1 for $x$, as follows,

$\Rightarrow f\left( 1 \right) = {\left( 1 \right)^3} - 5{\left( 1 \right)^2} - 3\left( 1 \right)$

$\Rightarrow f\left( 1 \right) = 1 - 5\left( 1 \right) - 3$

$\Rightarrow f\left( 1 \right) = 1 - 5 - 3$

$\Rightarrow f\left( 1 \right) = - 4 - 3$

$\Rightarrow f\left( 1 \right) = - 7$

To determine $f'\left( c \right)$, derive $f'\left( x \right)$ by differentiating $f\left( x \right)$,

$\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^3} - 5{x^2} - 3x} \right)$

$\Rightarrow f'\left( x \right) = 3{x^2} - 5\left( {2x} \right) - 3$

$\Rightarrow f'\left( x \right) = 3{x^2} - 10x - 3$

Therefore,

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}$

$\Rightarrow 3{c^2} - 10c - 3 = \dfrac{{\left( { - 27} \right) - \left( { - 7} \right)}}{2}$

$\Rightarrow 3{c^2} - 10c - 3 = \dfrac{{ - 27 + 7}}{2}$

$\Rightarrow 3{c^2} - 10c - 3 = \dfrac{{ - 20}}{2}$

$\Rightarrow 3{c^2} - 10c - 3 = - 10$

$\Rightarrow 3{c^2} - 10c - 3 + 10 = 0$

$\Rightarrow 3{c^2} - 10c + 7 = 0$

Now in order to derive the value of $c$ solve the equation $3{c^2} - 10c + 7 = 0$.

$\Rightarrow 3{c^2} - 10c + 7 = 0$

$\Rightarrow 3{c^2} - 3c - 7c + 7 = 0$

$\Rightarrow 3c\left( {c - 1} \right) - 7\left( {c - 1} \right) = 0$

$\Rightarrow \left( {c - 1} \right)\left( {3c - 7} \right) = 0$

Therefore either,

$\Rightarrow \left( {c - 1} \right) = 0$

$\Rightarrow c = 1$

Or,

$\Rightarrow \left( {3c - 7} \right) = 0$

$\Rightarrow 3c = 7$

$\Rightarrow c = \dfrac{7}{3}$

Since $1 \notin \left( {1,3} \right)$, the only value of $c$ is $c = \dfrac{7}{3} \in \left( {1,3} \right)$, and consequently the Mean Value Theorem is verified for the given function $f\left( x \right) = {x^3} - 5{x^2} - 3x$in the interval $\left[ {1,3} \right]$.

In order to obtain $c \in \left( {1,3} \right)$for which$f'\left( c \right) = 0$, solve the following equation.

$\Rightarrow f'\left( c \right) = 0$

$\Rightarrow 3{c^2} - 10c - 3 = 0$

To solve this quadratic equation of the form $A{c^2} + Bc + C = 0$, where $A$ is 3, $B$ is $- 10$ and $C$ is $- 3$, apply $c = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$.

$\Rightarrow c = \dfrac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4\left( 3 \right)\left( { - 3} \right)} }}{{2\left( 3 \right)}}$

$\Rightarrow c = \dfrac{{10 \pm \sqrt {100 + 36} }}{6}$

$\Rightarrow c = \dfrac{{10 \pm \sqrt {136} }}{6}$

$\Rightarrow c = \dfrac{{10 \pm \sqrt {2 \times 2 \times 34} }}{6}$

$\Rightarrow c = \dfrac{{10 \pm 2\sqrt {34} }}{6}$

$\Rightarrow c = \dfrac{{10}}{6} \pm \dfrac{{2\sqrt {34} }}{6}$

$\Rightarrow c = \dfrac{5}{3} \pm \dfrac{{\sqrt {34} }}{3}$

Therefore either,

$\Rightarrow c = \dfrac{5}{3} + \dfrac{{\sqrt {34} }}{3}$

$\Rightarrow c \approx \dfrac{5}{3} + \dfrac{{5.83}}{3}$

$\Rightarrow c \approx \dfrac{{10.83}}{3}$

$\Rightarrow c \approx 3.61$

Or,

$\Rightarrow c = \dfrac{5}{3} - \dfrac{{\sqrt {34} }}{3}$

$\Rightarrow c \approx \dfrac{5}{3} - \dfrac{{5.83}}{3}$

$\Rightarrow c \approx \dfrac{{ - 0.83}}{3}$

$\Rightarrow c \approx - 0.28$

Since $3.61 \notin \left( {1,3} \right)$ and $- 0.28 \notin \left( {1,3} \right)$, then there exists no$c \in \left( {1,3} \right)$ such that, $f'\left( c \right) = 0$.

6. Examine the applicability of the Mean Value Theorem for all three functions given in the above exercise 2.

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

Ans: The given function $f\left( x \right) = \left[ x \right]$, is known as a greatest integer function, which is discontinuous at every integral point.

Hence, in the given interval, which contains the integers $5,6,7,8,9$, $f\left( x \right) = \left[ x \right]$is discontinuous and consequently, the first condition of Mean Value Theorem is not satisfied.

To check the second condition(differentiability) consider the following limits at an integer $n \in \left( {5,9} \right)$.

The left hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n - 1$, since $h \to {0^ - }$.

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

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

$\Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = - \infty$

The right hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n$, since $h \to {0^ + }$.

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

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

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

Since the left and right hand limit doesn’t coincide, the given function $f\left( x \right) = \left[ x \right]$ is not differentiable in the interval $\left( {5,9} \right)$ and consequently, the second condition of Mean Value Theorem is not satisfied.

Therefore, on the given function, the Mean Value Theorem is not applicable.

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

Ans: The given function $f\left( x \right) = \left[ x \right]$, is known as a greatest integer function, which is discontinuous at every integral point.

Hence, in the given interval, which contains the integers $- 2, - 1,0,1,2$, $f\left( x \right) = \left[ x \right]$is discontinuous and consequently, the first condition of Mean Value Theorem is not satisfied.

To check the second condition(differentiability) consider the following limits at an integer $n \in \left( { - 2,2} \right)$.

The left hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n - 1$, since $h \to {0^ - }$.

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

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

$\Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {n + h} \right) - f\left( n \right)}}{h} = - \infty$

The right hand limit is as follows,

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

The value of $\left[ n \right]$ is $n$ and $\left[ {n + h} \right]$ is $n$, since $h \to {0^ + }$.

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

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

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

Since the left and right hand limit doesn’t coincide, the given function $f\left( x \right) = \left[ x \right]$ is not differentiable in the interval $\left( { - 2,2} \right)$ and consequently, the second condition of Mean Value Theorem is not satisfied.

Therefore, on the given function, the Mean Value Theorem is not applicable.

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

Ans: Given function, $f(x) = {x^2} - 1$, is a polynomial function.

We know that a polynomial function is continuous and differentiable on any interval. The function $f(x) = {x^2} - 1$ is continuous in the interval $\left[ {1,2} \right]$ and differentiable in the interval $\left( {1,2} \right)$.

Hence both the conditions of the Mean Value Theorem are satisfied, so the Mean Value Theorem is applicable to the given function.

Therefore, by Mean Value Theorem, it can be stated that there exists a $c \in \left( {1,2} \right)$ such that,

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 2 \right) - f\left( 1 \right)}}{{2 - 1}}$

$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 2 \right) - f\left( 1 \right)}}{1}$

$\Rightarrow f'\left( c \right) = f\left( 2 \right) - f\left( 1 \right)$

To derive the value of $f\left( 1 \right)$, substitute 1 for $x$, as follows,

$\Rightarrow f\left( 1 \right) = {\left( 1 \right)^2} - 1$

$\Rightarrow f\left( 1 \right) = 1 - 1$

$\Rightarrow f\left( 1 \right) = 0$

To derive the value of $f\left( 2 \right)$, substitute 2 for $x$, as follows,

$\Rightarrow f\left( 2 \right) = {\left( 2 \right)^2} - 1$

$\Rightarrow f\left( 2 \right) = 4 - 1$

$\Rightarrow f\left( 2 \right) = 3$

To determine $f'\left( c \right)$, derive $f'\left( x \right)$ by differentiating $f\left( x \right)$,

$\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} - 1} \right)$

$\Rightarrow f'\left( x \right) = 2x$

Therefore,

$\Rightarrow f'\left( c \right) = f\left( 2 \right) - f\left( 1 \right)$

$\Rightarrow 2c = 3 - 0$

$\Rightarrow 2c = 3$

$\Rightarrow c = \dfrac{3}{2}$

Since $c = \dfrac{3}{2} \in \left( {1,2} \right)$, the Mean Value Theorem is verified for the given function $f(x) = {x^2} - 1$in the interval $\left[ {1,2} \right]$.

### NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.8

Opting for the NCERT solutions for Ex 5.8 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 5.8 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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NCERT Solutions for Class 12 Math Chapter 5 Exercise 5.8 prepared by expert Mathematics teacher at Vedantu as per the guidelines issued by the CBSE board. The solutions are given in pdf. Download our Class 12 Math Chapter 5 Continuity and Differentiability Ex 5.8 Questions with Solutions  free pdf through the link given below. These solutions will help you to revise the complete syllabus and score more marks in your exams. The solutions of Chapter 5 Exercise 5.5 is based on the topic “ Mean Value Theorem and Rolle’s Theorem”.

### What Is the Mean Value Theorem?

Mean value theorem states that let f = (a,b) R be continuous function on [a,b], and differentiable on (a,b), then there exist some C in (a,b) such that:

f' (c) = f(b) - f(a)b - a

### What Is Rolle’s Theorem?

The Rolle’s theorem states that let f = (a,b) R be continuous function on [a,b], and differentiable on (a,b) such that f(a) = f(b) where a and b are some real numbers. Then there exist some c in (a,b) such thatf'(c) is equal to 0.