Question

Let $f(x)=g(x)\left|(x-1)(x-2)(x-3)^{2}(x-4)^{3}\right|$, where $g(x) =x^{3}+b x^{2}+c x+d$. If $f(x)$ is differentiable for all $x \in R$ and $f'(3)+f'''(4)=0$, then the value of $g(5)$ is ____.

Answer 1

$f(x) =(x^{3}+bx^{2}+cx +d)|x-1||x-2|(x-3)^{2}|x-4|(x-4)^{2}$
$f(x)$ is differentiable for all real $x$, if $(x-1)$ and $(x-2)$ are factors of $g(x)$
Clearly , $f'(3) =0 $
So, $f'''(4)= 0$, for which (x - 4) must be factor of g(x)
$\therefore g(x)=x^{3}+b x^{2}+c x+d=(x-1)(x-2)(x-4)$
So, $g(5)=4 \times 3 \times 1=12$