Given, f(x)=2x3−9ax2+12a2x+1 f′(x)=6x2−18ax+12a2 Equate f′(x)=0 ⇒6x2−18ax+12a2=0 ⇒x2−3ax+2a2=0 ⇒(x−a)(x−2a)=0 ⇒x=a,2a Now, f"(x)=12x−18a ⇒f"(a)=12a−18a<0 ⇒f"(2a)=24−18a>0 ∴ Minimum value attained at x=2a Maximum value attained at x=a ∴p=a and q=2a⇒p2=q gives, a2=2a ⇒a(a−2)=0⇒a=0 and a=2 Since, a≠0,a=2.