Question

Let $f\left(x\right)=2x^3-3(2+p)x^2+12px+\ln (16-p^2).$ If $f\left(x\right)$ has exactly one local maximum and one local minimum, then the number of possible integral values of $p$ is

Answer 1

Given : $f\left(x\right)=2x^3-3(2+p)x^2+12px+\ln (16-p^2)$
Differentiating w.r.t. $x,$ we get
$f^{\prime }\left(x\right)=6x^2-6(2+p)x+12p$
$\Rightarrow f^{\prime }\left(x\right)=6(x^2-(2+p)x+2p)$
For maximum or minimum, $f^{\prime }\left(x\right)=0$
$\Rightarrow x^2-(2+p)x+2p=0$
$\Rightarrow x=2,p$
Since $f\left(x\right)$ has exactly one local maximum and one local minimum, therefore $p\ne 2.$

Also, $16-p^2>0$
$\Rightarrow p^2-16<0$
$\Rightarrow p\in (-4,4)$
Possible integral values of $p$ are $-3,-2,-1,0,1,3$
Hence, number of possible integral values of $p$ is $6.$