Question

If $f\left(x\right)=\{\begin{array}{c}{e}^x,x<2\\ a+bx,x\ge 2\end{array}$ is differentiable for all $xϵR$ then

• A. $a+b=0$
• B. $a+2b=e^2$
• C. $b=e^2$
• D. $Allofthese$

Answer 1

The correct option is C $b=e^2$

R.H.D.

$\underset{h\to 0}{lim}\frac{f\left(2+h\right)-f\left(2\right)}{h}$

$=\underset{h\to 0}{lim}\frac{a+b\left(2+h\right)-a-2b}{h}=b$

L.H.D.

$\underset{h\to 0}{lim}\frac{f\left(2-h\right)-f\left(2\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{e^{2-h}-e^2}{-h}$

$=\underset{h\to 0}{lim}\frac{e^2.e^{-h}-e^2}{-h}$

$=\underset{h\to 0}{lim}\frac{e^2\left(\frac{1}{e^h}-1\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{e^2\left(1-e^h\right)}{-e^{h}h}$

$=\underset{h\to 0}{lim}{e}^2\left(\frac{e^h-1}{e^h}\times \frac{1}{h}\right)$

$=\underset{h\to 0}{lim}{e}^2\left(\frac{e^h-1}{h}\times \frac{1}{e^h}\right)$

$=e^2\left[1\right]=e^2$ $\left[\underset{h\to 0}{lim}\frac{e^h-1}{h}=1\right]$

For $f\left(x\right)$ to be differential for all $x\in R$

R.H.D.=L.H.D.

$b=e^2$