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  • Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 28

Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 28

Answers (1)

best_answer

(b)

Hint: put \mathrm{LHD}=\mathrm{RHD} \text { at } \mathrm{x}=1 / 2

Given: f(x)=|2 x-1| \sin x

Explanation:

f(x)= \begin{cases}(2 x-1) \sin x & x>1 / 2 \\ -(2 x-1) \sin x & x<1 / 2\end{cases}

f(x) is differentiable at x=\frac{1}{2}

LHD at x=\frac{1}{2}

\begin{aligned} &\lim _{x \rightarrow \frac{1^{-}}{2}} \frac{f(x)-f\left(\frac{1}{2}\right)}{x-\frac{1}{2}}=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\left(2\left(\frac{1}{2}-h\right)-1\right) \sin \left(\frac{1}{2}-h\right)-0}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(-1+2 h+1) \sin \left(\frac{1}{2}-h\right)}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0}-2 \sin \left(\frac{1}{2}-h\right)=-2 \sin \frac{1}{2} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(x)-f\left(\frac{1}{2}\right)}{x \rightarrow \frac{1^{+}}{2}}=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{x-\frac{1}{2}} \\ \end{aligned}

\begin{aligned}&=\lim _{h \rightarrow 0} \frac{2 h \sin \left(\frac{1}{2}+h\right)}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} 2 \sin \left(\frac{1}{2}+h\right)=2 \sin \frac{1}{2} \\ &L H D=R H D\end{aligned}

Therefore f(x) is not differentiable at x=\frac{1}{2}

\mathrm{R}-{1 / 2}

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