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

Question

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

Smart Achievers · Accepted Answer

(2)

Ans. $R - {\frac{1}{2}}$

Explanation:

Let $u (x) = \sin x$

$v (x) =$ $| x |$

$f (x) = v o u (x) = v [u (x)]$

$= v (\sin x)$

$=$ $| \sin x |$

$∵$ $u (x) = \sin x$ is a continuous function and $v (x) =$ $| x |$ is a continuous function.

$∴$ $f (x) = v \circ u (x)$ is also continuous everywhere but $v (x)$ is not differentiable at $x = 0$ .

$\Rightarrow f (x)$ is not differentiable where $\sin x = 0$

$x = n π, n \in Z$

Hence, $f (x)$ is continuous everywhere but not differentiable at $x = n π, n \in Z .$

$∵$ At $x = \frac{1}{2}$ , curve has two tangents at that point, therefore, from the graph it is clear that $y =$ $| 2 x - 1 |$ is not differentiable at $x = \frac{1}{2} .$

$∴$ $f (x) is differentiable in R - {\frac{1}{2}} .$

Solution

Q.
The set of points, where the function $f$ given by $f (x) =$ $| 2 x - 1 |$ $\sin x$ is differentiable is :

1 $R$

2 $R - {\frac{1}{2}}$

3 $(0, \infty)$

4 None of these

Want Us to Email you About Special Offers & Updates?