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QJEEM24Ma5ae3: JEE Main 2024 Mathematics - Class 12 JEEM24Ma5ae3 Solution: The maximum area of a triangle whose one vertex is at and the…

The maximum area of a triangle whose one vertex is at (0,0)(0,0) and the other two vertices lie on the curve y=−2x2+54y=-2 x^2+54 at points (x,y)(x, y) and (−x,y)(-x, y) , where y>0 , is :

Multiple Choice Options

  • A 108
  • B 122
  • C 88
  • D 92

Step-by-Step Verified Solution

JEE Main 2024 (Online) 30th January Morning Shift Mathematics - Application of Derivatives Question 21 English Explanation

Area of Δ\Delta

amp;=12∣0amp;0amp;1xamp;yamp;1−xamp;yamp;1∣amp;⇒∣12(xy+xy)∣=∣xy∣amp;Area⁡(Δ)=∣xy∣=∣x(−2x2+54)∣amp;d(Δ)dx=∣(−6x2+54)∣⇒dΔdx=0 at x=3amp; Area =3(−2×9+54)=108\begin{aligned}& =\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\x & y & 1 \\-x & y & 1\end{array}\right| \\& \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \\& \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x^2+54\right)\right| \\& \frac{d(\Delta)}{d x}=\left|\left(-6 x^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text { at } x=3 \\& \text { Area }=3(-2 \times 9+54)=108\end{aligned}

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