Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

Let $f(x)=\left\{\begin{matrix}(x-2)^2\sin\left(\frac{1}{x-2}\right)-|x|&,&x≠2\\-2&,&x=2\end{matrix}\right.$ then the points where f (x) is not differentiable are

Options:

x = 0, 2 only

x = 0 only

x = 2 only

none of these

Correct Answer:

x = 0 only

Explanation:

$f'(2)=\underset{h→0}{\lim}\frac{h^2\sin\frac{1}{h}-(2+h)+2}{h}$

Since $(x-2)^2\sin\left(\frac{1}{x-2}\right)$ is differentiable at x = 0

and |x| is not differentiable at x = 0

∴ f (x) is not differentiable at x = 0