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Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 2  maths textbook solution

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Answer:  Not applicable

Hint: Using Lagrange’s mean theorem and limits.

Given: f(x)=|x| \text { on }[-1,1]


Lagrange’s mean value theorem states if a functionf(x) is continuous on a closed interval    [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that

Differentiability at x = 0:


\begin{aligned} &\lim _{x \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h} \\ &\lim _{x \rightarrow 0^{-}} \frac{-(0-h)-0}{-h} \end{aligned}

=\lim _{x \rightarrow 0^{-}} \frac{h-0}{-h}

\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{h}{-h} \\ =&-1 \end{aligned}

RHD = \lim _{x \rightarrow 0^{+}} \frac{f(0-h)-f(0)}{-h}

\begin{aligned} &=\lim _{x \rightarrow 0^{-}} \frac{(0-h)-0}{-h} \\ &=\lim _{x \rightarrow 0^{-}} \frac{-h-0}{-h} \\ &=\lim _{x \rightarrow 0^{-}} \frac{-h}{-h} \\ &=1 \end{aligned}

\mathrm{LHS} \neq \mathrm{RHS}

\therefore f(x) is not differentiable at x = 0

\thereforeLagrange’s mean value theorem is not applicable for function f (x) = |x| on [-1, 1]



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