Question

The function $f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1=0$ has a local maximum at $x=\alpha$, and a local minimum at $x=\beta$ such that $\beta ={\alpha }^2$ then $a$ is equal to:

• A. $0$
• B. $\frac{1}{4}$
• C. $2$
• D. either $0$ or $2$

Answer 1

The correct option is C $2$

$f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1=0$
$f^{\prime }\left(x\right)=0$
$\Rightarrow 6(x^2-3ax+2a^2)=0$ or $6(x-a)(x-2a)=0$
$\therefore x=a,2a$
$f^{\prime \prime }\left(x\right)=6(2x-3a)=positive$ for $x=2a$ and hence local minimum at $x=2a=\beta$ say
$f^{\prime \prime }\left(x\right)=6(2x-3a)=negative$ for $x=a$ and hence local maximum at $x=a=\alpha$ say.
But $\beta ={\alpha }^2\Rightarrow 2a=a^2$ or $a(a-2)=0$
$\therefore a=2$
$a=0$ is ruled out as $f\left(x\right)=2x^3+1$ and $f^{\prime }\left(x\right)=6x^2$ which is always +ive and hence $f\left(x\right)$ is an increasing function in this case.