Question

The set of values of $p$ for which the points of extremum of the function, $f\left(x\right)=x^3-p{x}^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ is

• A. $(-3,5)$
• B. $(-3,3)$
• C. $(-1,3)$
• D. $(-1,5)$

Answer 1

The correct option is C $(-1,3)$

$f\left(x\right)=x^3-3p{x}^2+3(p^2-1)x+1$

$f^{\prime }\left(x\right)=3\{x^2-2px+(p-1)(p+1)\}$

$f^{\prime }\left(x\right)=3\{x-(p-1)\}\{x-(p+1)\}$

Using given condition,

$-2<p-1<4$ and $-2<p+1<4$

$\Rightarrow -1<p<5$ and $-3<p<3$

$\therefore p\in (-1,3)$