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Q.
If the function $ f(x ) = 2x^3 - 9ax^2 + 12a^2x + 1 $ attains its maximum and minimum at $ p $ and $ q $ respectively such that $ p^2 = q $ , then $ a $ equals
Given, $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ attains
maximum and minimum at $p$ and $q$ respectively.
$\therefore \,\, f'(p)=0, f'(q)=0$
$f''(p)<0 $ and $ f''(q)>0 $
Now$\,\,\,f '(p)=0$
and $\,\,\,f'(q)=0$
$\Rightarrow \,\,\, 6 p^{2}-18 a p+12 a^{2}=0 $
and $6 q^{2}-18 a q+12 a^{2}=0$
$\Rightarrow \,\,\, p^{2}-3 a p+2 a^{2}=0$
and $\,\,\, q^{2}-3 a q+2 a^{2}=0$
$\Rightarrow \,\,\, p=a, 2 a, q=a, 2 a \,\,\,\,\,\, ...(i)$
Now,$\,\,\, f''(p)<0 $
$\Rightarrow \,\,\, 12 p-18 a<0 $
$\Rightarrow \,\,\, p<\frac{3}{2} a \,\,\,\,\,...(ii)$
and $\,\,\, f''(q)>0 \Rightarrow 12 q-18 a>0 $
$\Rightarrow \,\,\, q>\frac{3}{2} a \,\,\,\,\,\,\, ...(iii)$
and $\,\,\,f''(q)>0 \Rightarrow 12 q-18 a>0$
$\Rightarrow \,\,\, ' q>\frac{3}{2} a$
From Eqs. (i), (ii) and (iii), we get $p=a, q=2 a$
Now, $\,\,\, p^{2}=q$
$\Rightarrow \,\,\, a^{2}=2 a $
$\Rightarrow \,\,\, a=0,2$
But for $a=0, f(x)=2 x^{3}+1$ which does not
attains a maximum or minimum for any value of $x$.
Hence, $a=2$.