Question

Consider the function $f\left(x\right)=x^3-p{x}^2+qx$ defined on $[1,3].$ If $f$ satisfies the hypothesis of Rolle's theorem such that $c=\frac{3}{2},$ then the value of $4p+q+16$ is

Answer 1

$f\left(x\right)=x^3-p{x}^2+qx$
$f$ satisfies the hypothesis of Rolle's theorem.
$\Rightarrow f\left(1\right)=f\left(3\right)$
$\Rightarrow 1-p+q=27-9p+3q$
$\Rightarrow 8p-2q=26$
$\Rightarrow 4p-q=13\cdots \left(1\right)$

Also, $f^{\prime }\left(c\right)=0$
$\Rightarrow 3c^2-2pc+q=0$
$\Rightarrow 3{\left(\frac{3}{2}\right)}^2-2p\left(\frac{3}{2}\right)+q=0$
$\Rightarrow 27-12p+4q=0\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right),$ we get
$p=\frac{25}{4},q=12$

$\therefore 4p+q+16=25+12+16=53$