Q

# Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-4

The set of points where  $\dpi{100} f\left ( x \right )= \frac{x}{1+\left | x \right |}$ is differentiable, is

• Option 1)

$\left ( -\infty ,0 \right )\cup \left ( 0,\infty \right )$

• Option 2)

$\left ( -\infty ,-1 \right )\cup \left ( -1,\infty \right )$

• Option 3)

$\left ( -\infty ,-\infty \right )$

• Option 4)

$\left ( 0 ,\infty \right )$

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As we learnt in

Condition for differentiable -

A function  f(x) is said to be differentiable at  $x=x_{\circ }$  if   $Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })$   both exist and are equal otherwise non differentiable

-

$f(x) =\frac{x}{1+\left | 1+x \right |}$

$f(x)=\left\{\begin{matrix} \frac{x}{1+x}, x>0 \\ 0,x=0 \\ \frac{x}{1-x}, x<0 \end{matrix}\right.$

$f'(x)=\left\{\begin{matrix} \frac{1}{({1+x})^2}, x>0 \\ \frac{1}{({1-x})^2}, x<0 \end{matrix}\right.$

So $f'(0^+)= f'(0)$

$\therefore$ f(x) is differentiable for each $x \epsilon R$

Option 1)

$\left ( -\infty ,0 \right )\cup \left ( 0,\infty \right )$

Incorrect

Option 2)

$\left ( -\infty ,-1 \right )\cup \left ( -1,\infty \right )$

Incorrect

Option 3)

$\left ( -\infty ,-\infty \right )$

Correct

Option 4)

$\left ( 0 ,\infty \right )$

Incorrect

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