Question

The set of points where the function f(x) = |2x – 1| sin x is differentiable, is

(a) R

(b) $R-\left\{\frac{1}{2}\right\}$

(c) (0, ∞)

(d) none of these

Answer 1

Let $g\left(x\right)=\left|2x-1\right|$ and $h\left(x\right)=\sin x$.

We know that, the trigonometric functions are differentiable in their respective domain.

So, $h\left(x\right)=\sin x$ is differentiable for all x ∈ R.

Now,

$g\left(x\right)=\left|2x-1\right|=\left\{\begin{array}{ll}2x-1,& x\ge \frac{1}{2}\\ -\left(2x-1\right),& x<\frac{1}{2}\end{array}\right.$

(2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all $x<\frac{1}{2}$ and for all $x>\frac{1}{2}$.

So, we need to check the differentiability of g(x) at $x=\frac{1}{2}$.

We have

$Lg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\lim }\frac{g\left({\displaystyle \frac{1}{2}}-h\right)-g\left({\displaystyle \frac{1}{2}}\right)}{-h}$

$\Rightarrow Lg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\lim }\frac{-\left[2\left({\displaystyle \frac{1}{2}}-h\right)-1\right]-\left(2\times {\displaystyle \frac{1}{2}}-1\right)}{-h}$

$\Rightarrow Lg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\lim }\frac{2h}{-h}$

$\Rightarrow Lg\text{'}\left(\frac{1}{2}\right)=-2$

And

$Rg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\lim }\frac{g\left({\displaystyle \frac{1}{2}}+h\right)-g\left({\displaystyle \frac{1}{2}}\right)}{h}$

$\Rightarrow Rg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\lim }\frac{\left[2\left({\displaystyle \frac{1}{2}}+h\right)-1\right]-\left(2\times {\displaystyle \frac{1}{2}}-1\right)}{h}$

$\Rightarrow Rg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\lim }\frac{2h}{h}$

$\Rightarrow Rg\text{'}\left(\frac{1}{2}\right)=2$

$\therefore Lg\text{'}\left(\frac{1}{2}\right)\ne Rg\text{'}\left(\frac{1}{2}\right)$

So, $g\left(x\right)=\left|2x-1\right|$ is not differentiable at $x=\frac{1}{2}$.

The function $g\left(x\right)=\left|2x-1\right|$ is differentiable for all $x\in \mathrm{R}-\left\{\frac{1}{2}\right\}$.

We know that, the product of two differentiable functions is differentiable.

$\therefore f\left(x\right)=g\left(x\right)\times h\left(x\right)=\left|2x-1\right|\sin x$ is differentiable for all $x\in \mathrm{R}-\left\{\frac{1}{2}\right\}$.

Thus, the set of points where the function $f\left(x\right)=\left|2x-1\right|\sin x$ is differentiable is $\mathrm{R}-\left\{\frac{1}{2}\right\}$.

Hence, the correct answer is option (b).