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{cases} -(2x-1)sin\,x, & \text{ $x < 1/2$} \\[2ex] (2x-1)sin\,x, & \text{ $x \ge 1/2$} \end{cases}$
$Lf '\left(\frac{1}{2}\right)=$ $\displaystyle \lim_{h \to 0^-} $ $\frac{f \left(1/2+h\right)-f \left(1/2\right)}{h}$
$=\displaystyle \lim_{h \to 0^-}$ $\frac{-\left[2\left(1/2+h\right)-1\right]sin\left(1/2+h\right)-0}{h}$
$=\displaystyle \lim_{h \to 0^-} $$\frac{-2h\,sin\left(1/2+h\right)}{h}=-2\,sin\left(1/2\right)$
$Rf '\left(1/2\right)=$ $\displaystyle \lim_{h \to 0^+} $ $\frac{f \left(1/2+h\right)-f \left(1/2\right)}{h}$
$=\displaystyle \lim_{h \to 0^+}$ $\frac{\left[2\left(1/2+h\right)-1\right]sin\left(1/2+h\right)-0}{h}$
$=\displaystyle \lim_{h \to 0^+}$ $\frac{2h\,sin\left(1/2+h\right)}{h}=2sin(1/2)$
$Lf'(1/2) \ne Rf'(1/2)$
So, $f(x)$ is not differentiable at $x = 1/2$
Hence $f(x)$ is differentiable $\forall \,x \in R-\left\{\frac{1}{2}\right\}$