<s>[INST] The set of values of $p$ for which the points of extremum of the function $f(x)=x^3-3px^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ The set of values of $p$ for which the points of extremum of the function $f(x)=x^3-3px^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ is
$(A)(-3,5)$
$(B)(-3,3)$
$(C)(-1,3)$
$(D)(-1,5)$

$$f'(x)=3x^2-6px+3(p^2-1)=3[(p-x)^2-1]$$
For the points of extremum to lie between $(-2,4)$
The vertex of the parabola $f'(x)=3x^2-6px+3(p^2-1)$ should lie between $(-2,4)$
The vertex of the parabola $f'(x)=3x^2-6px+3(p^2-1)$ is $2p$
$$-2<2p<4\implies -1<p<2$$
But the answer is $(-1,3)$
 [/INST] Because you need $-3<p<3$ and $-1<p<5$, which gives $-1<p<3$
Derivative gives $(x-p)^2-1=0$ and $x_{max}=p-1$, $x_{min}=p+1$.
$-2<p+1<4$ with $-2<p-1<4$ gives $-1<p<3$.
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