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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.
$\text{f(x)}=[\text{x}]\text{ for x}\in[-2,\ 2]$
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Answer

By Rolle's Theorem, for a function  $\text{f}:[\text{a},\ \text{b}]\rightarrow\text{R},\text{if}$
  1. f is continuous on [a, b]
  2. f is differentiabie on (a, b)
  3. f(a) = f(b)
then, there exists some $\text{c}\in(\text{a},\ \text{b})$ such that f'(c) = 0
Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three two conditions of the hypothesis.
$\text{f(x)}=[\text{x}]\text{ for x}\in[-2,\ 2]$ If 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, f(x) is not continuous in [-2, 2]. $\therefore\ \text{f}(-2)\neq\text{f}(2)$ 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, $\lim\limits_{\text{h} \rightarrow0^-}\frac{\text{f(n}+\text{h)}-\text{f(n)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$ The right hand limit of f at x = n is, $\lim\limits_{\text{h} \rightarrow0^+}\frac{\text{f(n}+\text{h)}-\text{f(n)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}0=0$ Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n. $\therefore\ $ f is not differentiable in (-2, 2). It is observed that f dose not satisfy all the conditions of the hypothesis of Rolles's Theorem. Hence, Rolle's Theorem is not applicable for $\text{f(x)}=[\text{x}]\text{ for x}\in[-2,\ 2].$

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