Question

Given $f\left(\right.x\left.\right)=\left\{\right. x^{2}e^{2 \left(\right. x - 1 \left.\right)},\text{ 0}\leq x\leq 1. \\ asgn\left(\right.x+1\left.\right)cos\left(\right.2x-2\left.\right)+bx^{2}, \, 1 < x\leq 2$

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

Answer 1

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