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  • The set of points where the function f given by f(x) = |2x - 1| sinx differentiable is A. R

The set of points where the function f given by f(x) = |2x - 1| \sin x differentiable is
A. R

B. R-\left \{ \frac{1}{2} \right \}
C. \left ( 0,\infty \right )
D. None of these

Answers (1)

B)

We have, f(x) = |2x - 1| \sin x

Now, 2 x-1=0
\Rightarrow x=\frac{1}{2}
Now we will check the differentiability of f(x) \ at \ 1 / 2.
\mathrm{Lf}^{\prime}\left(\frac{1}{2}\right)=\lim _{\mathrm{x} \rightarrow \frac{1}{2}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}\left(\frac{1}{2}\right)}{\mathrm{x}-\frac{1}{2}}

\\ =\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{\left(\frac{1}{2}-h\right)-\frac{1}{2}} \\ =\lim _{h \rightarrow 0} \frac{\left|2\left(\frac{1}{2}-h\right)-1\right| \sin \left(\frac{1}{2}-h\right) -0}{-h} \quad(\because f(x)=|2 x-1| \sin x) \\ =\lim _{h \rightarrow 0} \frac{|-2 h| \sin \left(\frac{1}{2}-h\right) }{-h} \\ =\lim _{h \rightarrow 0} \frac{2 h \sin \left(\frac{1}{2}-h\right) }{-h} \\ \quad \lim _{h \rightarrow 0}-2 \sin \left(\frac{1}{2}-h\right) =-2 \sin \left(\frac{1}{2}\right)

\begin{aligned} &\operatorname{Rf}^{\prime}\left(\frac{1}{2}\right)=\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)}{\left(\frac{1}{2}+h\right)-\frac{1}{2}}\\ &\lim _{=h \rightarrow 0} \frac{\left|2\left(\frac{1}{2}+h\right)-1\right| \sin \left(\frac{1}{2}+h\right) -0}{h} \quad(\because f(x)=|2 x-1| \sin x)\\ &\lim _{=h \rightarrow 0} \frac{|2 h| \sin \left(\frac{1}{2}+h\right) }{h}\\ &\lim _{h \rightarrow 0} \frac{2 h \sin \left(\frac{1}{2}+h\right) }{h}\\ &\lim _{=h \rightarrow 0} 2 \sin \left(\frac{1}{2}+h\right) =2 \sin \left(\frac{1}{2}\right) \\ &\therefore \mathrm{Lf}^{\prime}\left(\frac{1}{2}\right) \neq \mathrm{Rf}^{\prime}\left(\frac{1}{2}\right)\\ &\text { Hence, } f(x) \text { is not differ entiable }\\ &x=\frac{1}{2} \end{aligned}

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