Question

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

Answer 1

Answer: 12

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