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 $(-2,4)$

Question

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

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$$f'(x)=3x^2-6px+3(p^2-1)=3[(p-x)^2-1]$$
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 $(-1,3)$

Answer 1

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 $-1<p<3$.

Answer 2

The x-coordinate of vertex of the quadratic equation is $-\dfrac{b}{2a}$, not $-\dfrac{b}{a}$.

Here is where you went wrong.