Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

If $f(x)= \begin{cases}\frac{x}{1+|x|}, & |x| \geq 1 \\ \frac{x}{1-|x|}, & |x|<1\end{cases}$ then f(x) is

Options:

discontinuous and non-differentiable at x = -1, 1, 0

discontinuous and non-differentiable at x = -1, whereas continuous and differentiable at x = 0, 1

discontinuous and non-differentiable at x = -1, 1, whereas continuous and differentiable at x = 0

none of these

Correct Answer:

discontinuous and non-differentiable at x = -1, 1, whereas continuous and differentiable at x = 0

Explanation:

We have,

$f(x)= \begin{cases}\frac{x}{1+|x|}, & |x| \geq 1 \\ \frac{x}{1-|x|}, & |x|<1\end{cases}$

$\Rightarrow f(x)= \begin{cases}\frac{x}{1-x}, & x \leq-1 \\ \frac{x}{1+x}, & -1<x<0 \\ \frac{x}{1-x}, & 0<x<1 \\ \frac{x}{1+x}, & x \geq 1\end{cases}$

Clearly, f(x) is discontinuous and hence non-differentiable at x = ±1.

It can be easily seen that f(x) is continuous and differentiable at x = 0.