The function $f(x)=2 x^3-9 a x^2+12 a^2 x+1(a>0)$ attains its maximum and minimum at $p$ and $q$…
The function $f(x)=2 x^3-9 a x^2+12 a^2 x+1(a>0)$ attains its maximum and minimum at $p$ and $q$ respectively and $p^2=q$. Then, $a=$
$1$
$2$
$1 / 2$
$3$
Solution
$\begin{aligned} & f(x)=2 x^3-9 a x^2+12 a^2 x+1, a>0 \\ & f^{\prime}(x)=6 x^2-18 a x+12 a^2\end{aligned}$
$=6\left(x^2-3 a x+2 a^2\right)=6(x-a)(x-2 a)$
For maxima or minima
$f^{\prime}(x)=0 \Rightarrow x=a, 2 a$
$\therefore \quad p=a$ and $q=2 a$
According to question,
$p^2=q$ and $a^2=2 a$
$\Rightarrow a(a-2)=0 \Rightarrow a=0,2$
$\therefore \quad a=2$ $[\because a>0]$
