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Mean Value Theorems — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems5 Marks
Question
Explain if Rolle's theorem is applicable to any one of the following functions.
  1. $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
  2. $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$
Can you say something about the converse of Rolle's Theorem from these functions?
✓

Answer

By Rolle’s theorem, for a function $\text{f}:[\text{a},\text{b}]\rightarrow\text{R},$ if
  1. f is continuous on [a, b],
  2. f is differentiable on (a, b) and
  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 conditions of the hypothesis.
  1. $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
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.
Thus, f(x) is not continuous on [5, 9].
Also, f(5) = [5] = 5 and f(9) = [9] = 9
$\therefore\ \text{f}(5)\neq\text{f}(9)$
The differentiability of f on (5, 9) is checked in the following way.
Let n be an integer such that $\text{n}\in(5,9).$
The left hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}0=0$
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, f is not differentiable on (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 on $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[5,9]$
  1. ​​​​​​​$\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$
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.
Thus, f (x) is not continuous on [−2, 2].
Also, f(-2) = [-2] = -2 and f(2) = [2] = 2
$\therefore\ \text{f}(-2)\neq\text{f}(2)$
The differentiability of f on (-2, 2) is checked in the following way.
Let n be an integer such that $\text{n}\in(-2,2).$
The left hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{f}(\text{n}+\text{h})-\text{f}(\text{n})}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim_\limits{\text{h}\rightarrow0^+}0=0$
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, f is not differentiable on (-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 on $\text{f}(\text{x})=[\text{x}]\text{ on }\text{x}\in[-2,2]$​​​​​​​

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