The maximum area of a triangle whose one vertex is at (0,0) and the other two vertices lie on the curve y=−2x2+54 at points (x,y) and (−x,y) where y>0 is :
(A)88
(B)122
(C)92
(D)108✓ Correct
Answer: (D)
Solution
Vertices (0,0),(x,y),(−x,y) with y=−2x2+54 give area =∣xy∣=x(−2x2+54) for x>0.
A(x)=−2x3+54x,A′(x)=−6x2+54=0⇒x=3.
A(3)=−2(27)+54(3)=108.
JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option