Question

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

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

Answer 1

Answer: (B)

$f\left(x\right)=\left|2x-1\right|sinx$
$f(x)=\{\begin{array}{ll}-(2x-1)sinx,& \text{ x<1/2}\\ (2x-1)sinx,& \text{ x≥1/2}\end{array}$
$L{f}^{\prime }\left(\frac{1}{2}\right)=$ $\underset{h\to 0^{-}}{\lim }$ $\frac{f\left(1/2+h\right)-f\left(1/2\right)}{h}$
$=\underset{h\to 0^{-}}{\lim }$ $\frac{-\left[2\left(1/2+h\right)-1\right]sin\left(1/2+h\right)-0}{h}$
$=\underset{h\to 0^{-}}{\lim }$$\frac{-2hsin\left(1/2+h\right)}{h}=-2sin\left(1/2\right)$
$R{f}^{\prime }\left(1/2\right)=$ $\underset{h\to 0^{+}}{\lim }$ $\frac{f\left(1/2+h\right)-f\left(1/2\right)}{h}$
$=\underset{h\to 0^{+}}{\lim }$ $\frac{\left[2\left(1/2+h\right)-1\right]sin\left(1/2+h\right)-0}{h}$
$=\underset{h\to 0^{+}}{\lim }$ $\frac{2hsin\left(1/2+h\right)}{h}=2sin(1/2)$
$L{f}^{\prime }(1/2)\ne R{f}^{\prime }(1/2)$
So, $f(x)$ is not differentiable at $x=1/2$
Hence $f(x)$ is differentiable $\forall x\in R-\left\{\frac{1}{2}\right\}$