Question

If $f\left(x\right)=\{\begin{array}{cc}a{e}^{-ax},& x\le 0\\ x^3-4x+2b,& x>0\end{array}$

is differentiable at $x=0$, then the value of $|a|+\sqrt{b^2-1}$ is

• A. $\sqrt{3}$
• B. $2$
• C. $0$
• D. None of the above

Answer 1

The correct option is B $2$

At $x=0$
L.H.D. $=-a^2{e}^{-ax}{|}_{x=0}=-a^2$
R.H.D $=3x^2-4{|}_{x=0}=-4$
L.H.D. = R.H.D.
$\Rightarrow -a^2=-4\Rightarrow a^2=4\Rightarrow |a|=2$

Function is differentiable at $x=0$
$\Rightarrow$ Function is continuous at $x=0$
L.H.L. $=\underset{x\to 0^{-}}{lim}a{e}^{-ax}$
R.H.L. $=\underset{x\to 0^{+}}{lim}{x}^3-4x+2b$
L.H.L. = R.H.L.
$\Rightarrow a=2b$
$\Rightarrow a^2=4b^2$
$\Rightarrow b^2=\frac{a^2}{4}=\frac{4}{4}=1$
$\therefore |a|+\sqrt{b^2-1}=2+\sqrt{1-1}=2$