Question

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

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (C)

$f(x)=2 x^3-3(2+p) x^2+12 p x+\ln \left(16-p^2\right)$
$f^{\prime}(x)=6 x^2-6(2+p) x+12 p=6\left(x^2-(2+p) x+2 p\right)=0$
$f^{\prime}(x)=0$ has $x=2, x=p$ roots.
But $f(x)$ has exactly one local maxima and one local minima so $p \neq 2$
Also $16- p ^2>0 \Rightarrow p ^2-16<0 \Rightarrow p \in(-4,4)$
Possible integral values of $p$ are $-3,-2,-1,0,1,3$
$\Rightarrow 6$ integral values