Question

Let $f\left(x\right)=(x^3+(a-1)x^2+(b-a)x-b)|x^2+8x+6|,$ where $a,b\in R$. If $f\left(x\right)$ is derivable in $(-\infty ,\infty )$, then the value of $\left(\frac{a+b}{4}\right)$ is

Answer 1

$f\left(x\right)=(x^3+(a-1)x^2+(b-a)x-b)|x^2+8x+6|$
$\Rightarrow f\left(x\right)=(x-1)(x^2+ax+b)|x^2+8x+6|$
For the function to be derivable both the equation
$x^2+ax+b=0$ and $x^2+8x+6=0$ must be same, so

$\Rightarrow a=8,b=6$
$\Rightarrow \left(\frac{a+b}{4}\right)=3.5$