f(x)=2x3−9ax2+12a2x+1 f′(x)=6x2−18ax+12a2 For maximum or minimum, f′(x)=0 ⇒6x2−18ax+12a2=0 x=
18a±√324a2−288a2
2×6
=
18a±√36a2
12
=
18a±6a
12
=2a,a
Now, f′′(x)=12x−18a At x=2a f′′(x)=24a−18a =6a>0, maxima
∴p=f(2a)=2×8a3−36a3+24a3+1
=4a3+1 At x=a f′′(x)=12×a−18a =−6a<a, minima ∴q=f(a) =2a3−9a3+12a3+1 =5a3+1 Also given p3=q ∴(4a3+1)3=(5a3+1) ⇒a=0, but a>0 (given) Hence, none of the given option is correct.