Question

Given $$\begin{gathered} f(x)=\{x^2{e}^{2(x-1)},\text{ 0}\le x\le 1. \\ asgn(x+1)cos(2x-2)+b{x}^2,1<x\le 2 \end{gathered}$$

If $f(x)$ is differentiable at $x=1,$ then the value of $|a-b|$ is

Answer 1

Answer: 3

$f\left(1^{-}\right)=1=f\left(1\right)$ and $f\left(1^{+}\right)=a+b$
For continuity at $x=1,a+b=1\dots \dots \left(i\right)$
In $0<x<1,f^{{}^{\prime }}\left(x\right)=2x{e}^{2\left(x-1\right)}+2x^2{e}^{2\left(x-1\right)}$
$\therefore f^{{}^{\prime }}\left(1^{-}\right)=4$
In $1<x<2,f^{{}^{\prime }}\left(x\right)=-2asin\left(2x-2\right)+2bx$
$\therefore f^{{}^{\prime }}\left(1^{+}\right)=2b$
For differentiability at $x=1,2b=4$ . This with $(i)$ gives the values $a=-1,b=2$
then $|b-a|=3$