Question

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

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (C)

$2$

Given $f\left(x\right)=2x^3-9a{x}^2+12a^{2}x+1$ attains maximum and minimum at $p$ and $q$ respectively
$\therefore$ $f^{\prime }\left(p\right)=0,f^{\prime }\left(q\right)=0$
$f^{\prime \prime }\left(p\right)<0$ and $f^{\prime \prime }\left(q\right)>0$
Now, $f^{\prime }\left(p\right)=0$
and $f^{\prime }\left(q\right)=0$
$\Rightarrow$ $6p^2-18ap+12a^2=0$
and $6q^2-18aq+12a^2=0$
$\Rightarrow$ $p^2-3ap+2a^2=0$
and $q^2-3aq+2a^2=0$
$\Rightarrow$ $p=a,2a,q=a,2a.......\left(i\right)$
Now $f^{\prime \prime }\left(p\right)<0$
$\Rightarrow$ $12p-18a<0$
$\Rightarrow$ $p<\frac{3}{2}a........\left(ii\right)$
and $f^{\prime \prime }\left(q\right)>0$ $\Rightarrow$ $12q-18a>0$
$\Rightarrow$ $q>\frac{3}{2}a......\left(iii\right)$
From Eqs, $\left(i\right),\left(ii\right),\left(iii\right)$ we get
$p=a,q=2a$
Now $p^2=q$
$\Rightarrow$ $a^2=2a$
$\Rightarrow$ $a=0,2$
But for $a=0,f\left(x\right)=2x^3+1$ which does not attains a maximum or minimum fror any value of $x$. Hence $a=2$