If $f(x)=p x^3+q x^2+r x+t$ attains local minimum and local maximum values at $x=-2$ and $x=2$ respectively…
If $f(x)=p x^3+q x^2+r x+t$ attains local minimum and local maximum values at $x=-2$ and $x=2$ respectively and $p$ is a root of $9 x^2-1=0$, then $\mathrm{p}+\mathrm{q}+\mathrm{r}=$
$\frac{4}{3}$
4
$\frac{11}{3}$
$\frac{13}{3}$
Solution
$\because f^{\prime}(x)=3 p x^2+2 q x+r$ ...(i)
Since, $f(x)$ attain local minimum and maximum at $x=-2$ and $x=2$ respectively.
$\therefore f^{\prime}(x)=k(x+2)(x-2)=k\left(x^2-4\right)$ ...(ii)
From (i) and (ii)
$\begin{aligned} & k\left(x^2-4\right)=3 p x^2+2 q x+r \\ & \Rightarrow \quad k x^2-4 k=3 p x^2+2 q x+r\end{aligned}$
Comparing both sides, we get
$k=3 p, q=0$ and $r=-4 k$
Now, $f^{\prime \prime}(x)=6 p x+2 q=6 p x \quad\{\because q=0\}$
$\Rightarrow f^{\prime \prime}(-2)=-12 p>0 \quad\{\because x=-2$ is local minimum $\}$
$\Rightarrow p < 0$
$\because p$ is a root of $9 x^2-1=0 \Rightarrow x= \pm \frac{1}{3}$
$\therefore p < 0 \Rightarrow p=-\frac{1}{3}$
So, $k=3 \times\left(-\frac{1}{3}\right)=-1$
$\Rightarrow r=-4 k=-4(-1)=4$
$\therefore p+q+r=-\frac{1}{3}+0+4=\frac{11}{3}$.