Question

$f (x)= \begin{cases} e^{x^{2}}+x, & \text{ $x>\,0$} \\[2ex] ax+b, & \text{ $x \leq 0$ } \end{cases}$
is differentiable at x = 0 then

• A. $a=1, b=-1$
• B. $a=-1, b=1$
• C. $a=1, b=1$
• D. $a=-1, b=-1$

Answer 1

Answer: (C)

First $f(x)$ must be continuous at $x=0$
For which $f\left(0^{-}\right)=f\left(0^{+}\right)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0}\left(e^{x^{2}}+x\right)=\displaystyle\lim _{x \rightarrow 0}(a x+b)$
$\Rightarrow b=1$
Also $f'(x) = \begin{cases} 2 x e^{x^{2}}+1, & \text{if $x>\,0$ is even} \\[2ex] a, & \text{if $x<\,0$ } \end{cases}$
$f(x)$ is differentiable at $x=0$ then $f'\left(0^{+}\right)=f'\left(0^{-}\right)$
$\Rightarrow a=1$