f(x)=2x3−9ax2+12a2x+1,a>0 f′(x)=6x2−18ax+12a2 ‌‌=6(x2−3ax+2a2)=0‌ for extreme values ‌ ‌‌=6(x−a)(x−2a)=0⇒x=a,2a f"(x)=12x−18a,f"(a)=−6a<0‌ Max at ‌x=a=p f"(2a)=6a>0‌ Min at ‌x=2a=q ‌ Given ‌p2=q⇒a2=2a⇒a=2