Question

$f(x)=\{\begin{array}{ll}{e}^{x^2}+x,& \text{ x>0}\\ ax+b,& \text{ x≤0 }\end{array}$
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 $f\left(0^{-}\right)=f\left(0^{+}\right)$
$\Rightarrow \underset{x\to 0}{\lim }\left(e^{x^2}+x\right)=\underset{x\to 0}{\lim }(ax+b)$
$\Rightarrow b=1$
Also $f^{\prime }(x)=\{\begin{array}{ll}2x{e}^{x^2}+1,& \text{if x>0 is even}\\ a,& \text{if x<0 }\end{array}$
$f(x)$ is differentiable at $x=0$ then $f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)$
$\Rightarrow a=1$