Question

If the function $f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1$, where $a>0$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^2=q$, then $a$ equals

• A. $3$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$2$

Explanation for the correct option

Step 1: Solve for the extremities of the given function

Given that the function $f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1$ where $a>0$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^2=q$

For maxima and minima of the function $f\text{'}\left(x\right)=0$

Differentiating $f\left(x\right)$ with respect to $x$ we get

$f\text{'}\left(x\right)=2\times 3x^2-9a\times 2x+12a^2\times 1+0$

$\Rightarrow$$f\text{'}\left(x\right)=6x^2-18ax+12a^2$ $...\left(i\right)$

Differentiating $f\text{'}\left(x\right)$ with respect to $x$ we get

$f\text{'}\text{'}\left(x\right)=6\times 2x-18a\times 1$

$\Rightarrow$$f\text{'}\text{'}\left(x\right)=12x-18a$ $...\left(ii\right)$

Equating $f\text{'}\left(x\right)$ to $0$ we get,

$6x^2-18ax+12a^2=0$

$\Rightarrow$ $x^2-3ax+2a^2=0$

$\Rightarrow$$x^2-ax-2ax+2a^2=0$

$\Rightarrow$ $\left(x-a\right)\left(x-2a\right)=0$

$\Rightarrow$ $x=a$ and $x=2a$

Thus, the function has a maxima and a minima at either of $x=a$ or $x=2a$

Step 2: Solve for the required value

$f\text{'}\text{'}\left(a\right)=12a-18a$

$f\text{'}\text{'}\left(a\right)=-6a$

$\Rightarrow$$f\text{'}\text{'}\left(a\right)<0$ $\left[\because a>0\right]$

When $f\text{'}\text{'}\left(x\right)<0$ at a point , the function has a maxima at that point

Hence, $f\left(x\right)$ has a maxima at $x=a$

$\Rightarrow$$p=a$ $...\left(iii\right)$

$f\text{'}\text{'}\left(2a\right)=12\times 2a-18a$

$f\text{'}\text{'}\left(2a\right)=6a$

$\Rightarrow$$f\text{'}\text{'}\left(2a\right)>0$ $\left[\because a>0\right]$

When $f\text{'}\text{'}\left(x\right)>0$ at a point , the function has a minima at that point

Hence, $f\left(x\right)$ has a minima at $x=2a$

$\Rightarrow$$q=2a$ $...\left(iv\right)$

We know that, $p^2=q$

$\Rightarrow$$a^2=2a$

$\Rightarrow$ $a=0$ or $a=2$

As $a>0$, $a=2$

Thus, the required value of $a$ is $2$.

Hence, option (C) i.e. $2$ is the correct answer.