Question

Let $f\left(x\right)=\{\begin{array}{cc}x+e^{2x}-1,& x<0\\ x^2+2\lambda x,& x\ge 0\end{array}.$ If $f\left(x\right)$ is differentiable at $x=0,$ then the value of $2\lambda$ is

Answer 1

Given : $f\left(x\right)=\{\begin{array}{cc}x+e^{2x}-1,& x<0\\ x^2+2\lambda x,& x\ge 0\end{array}$
$R.H.D.=f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{h^2+2\lambda h-0}{h}$
$\Rightarrow f^{\prime }\left(0^{+}\right)=2\lambda$

$L.H.D.=f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{(-h+e^{-2h}-1)-0}{-h}\left(\frac{0}{0}\right)\text{form}$
By L'Hospital's rule,
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{-2e^{-2h}-1}{-1}$
$\Rightarrow f^{\prime }\left(0^{-}\right)=3$
Since $f\left(x\right)$ is differentiable at $x=0,$
$\Rightarrow L.H.D.=R.H.D.$
$\therefore$ $2\lambda =3$