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  • State True or False for the statements Rolle’s theorem is applicable for the function f(x) = |x - 1| in [0, 2].

State True or False for the statements
Rolle’s theorem is applicable for the function f(x) = |x - 1| in [0, 2].

 

Answers (1)

False

As per Rolle’s Theorem, Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers. Then there exists some c in (a, b) such that f’(c) = 0.

We have, f(x) = |x - 1| in [0, 2].

Since, polynomial and modulus functions are continuous everywhere f(x) is continuous

Now, x-1=0

⇒ x=1

We need to check if f(x) is differentiable at x=1 or not.

We have,

\begin{aligned} &f(x)=\left\{\begin{array}{l} -(x-1), \text { if } x<1 \\ (x-1), \text { if } x>1 \end{array}\right.\\ &\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\\ &\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{(1-h)-1}\\ &\lim _{h \rightarrow 0} \frac{[1-(1-h)-0]}{(1-h)-1} \quad(\because f(x)=1-x, \text { if } x<1)\\ &\lim _{n \rightarrow 0} \frac{[\mathrm{h}]}{-\mathrm{h}}\\ &\lim _{=h \rightarrow 0} \frac{[\mathrm{h}]}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}-1=-1\\ &\lim _{\mathrm{Rf}^{\prime}(1)}=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\\ &\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{(1+h)-1}\\ &\lim _{h \rightarrow 0} \frac{[(1+h)-1-0]}{(1+h)-1}(\because f(x)=x-1, \text { if } 1<x)\\ &\lim _{=h \rightarrow 0} \frac{[\mathrm{h}]}{\mathrm{h}}=\lim _{h \rightarrow 0} 1=1 \end{aligned}

⇒ Lf’(1) ≠ Rf’(1)

⇒ f(x) is not differentiable at x=1.

Hence, Rolle’s theorem is not applicable on f(x) since it is not differentiable at x=1 ∈ (0,2).

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