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Ncert Solutions Maths class 12th Maths Class 12th Continuity and Differentiability

113. 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?

$f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

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

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113. Solution:

By Rolle’s Theorem, for a function $f [a, b] \to R,$ if

f is continuous on $[a, b]$

f is differentiable on $(a, b)$

f(a)= f(b)

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

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

$f (x) = [x]$ for $x \in [5, 9]$

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

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

$\Rightarrow$ f(x) is not continuous in $[5, 9]$

Also, $\begin{array}{l} f (5) = [5] = 5, a n d, f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \end{array}$

The differentiability of f in $(5, 9)$ is checked as follows.

Let n be an integer such that $n \in (5, 9)$ .

The left hand lim

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113. Solution:

By Rolle’s Theorem, for a function $f [a, b] \to R,$ if

f is continuous on $[a, b]$

f is differentiable on $(a, b)$

f(a)= f(b)

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

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

$f (x) = [x]$ for $x \in [5, 9]$

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

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

$\Rightarrow$ f(x) is not continuous in $[5, 9]$

Also, $\begin{array}{l} f (5) = [5] = 5, a n d, f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \end{array}$

The differentiability of f in $(5, 9)$ is checked as follows.

Let n be an integer such that $n \in (5, 9)$ .

The left hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - 1 - n}{h} = \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

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

$∴$ f is not differentiable in (5,9).

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

Hence, Rolle’s Theorem is not applicable for $f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

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

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

$\Rightarrow$ f(x) is not continuous in $[- 2, 2]$ .

Also, $\begin{array}{l} f (- 2) = [- 2] = - 2, a n d, f (2) = [2] = 2 \\ ∴ f (- 2) \neq f (2) \end{array}$

The differentiability of f in $(- 2, 2)$ is checked as follows.

Let n be an integer such that $n \in (- 2, 2)$ .

The left hand limit of f at x=n is,

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

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

$∴$ f is not differentiable in (-2,2).

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

Hence, Rolle’s Theorem is not applicable for $f (x) = [x]$ for $x \in [- 2, 2]$

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

It is evident that f, being a polynomial function, is continuous in $[1, 2]$ and is differentiable on (1,2).

$\begin{array}{l} f (1) = 1^{2} = 0 \\ f (2) = 2^{2} - 1 = 3 \\ ∴ f (1) \neq f (2) \end{array}$

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

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

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113. 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?

$f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

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

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113. 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? f(x)=[x] for x∈[5,9] f(x)=[x] for x∈[−2,2] f(x)=x2−1 for x∈[1,2]

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113. 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?

$f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

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