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  • Explain solution RD Sharma class 12 chapter 9 Differentiability exercise Very short answer type question 12 maths

Explain solution RD Sharma class 12 chapter 9 Differentiability exercise Very short answer type question  12 maths

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Hint:   If f(x) is differentiable at x=2 then \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=f^{\prime}(2)

Given:\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+|\mathrm{x}-3|

Explanation:

f(x)=\left\{\begin{array}{cc} -(x-1)-(x-3) & x<1 \\ x-1-(x-3) & 1<x<3 \\ (x-1)+(x-3) & x>3 \end{array}\right.

f(x)=\left\{\begin{array}{lr} -2 x+4 & x<1 \\ 2 & 1<x<3 \\ 2 x-4 & x>3 \end{array}\right.

Now, at x=2 , f(x)=2  so the function is differentiable at x=2 as it is a polynomial Function.

Hence , from the definition of differentiability

\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \text { exists. }

So,

\begin{aligned} &\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=f^{\prime} \\\\ &\lim _{x \rightarrow 2} \frac{2-2}{x-2}=f^{\prime} \\\\ &f^{\prime}(2)=0 \end{aligned}

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