JEE Mains · Maths · STD 12 - 6. Application of derivatives
The maximum area of a triangle whose one vertex is at \((0,0)\) and the other two vertices lie on the curve \(y=-2 x^2+54\) at points \((x, y)\) and \((-x, y)\) where \(\mathrm{y}>0\) is :
- A \(88\)
- B \(122\)
- C \(92\)
- D \(108\)
Answer & Solution
Correct Answer
(D) \(108\)
Step-by-step Solution
Detailed explanation
\(=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ \mathrm{x} & \mathrm{y} & 1 \\ -\mathrm{x} & \mathrm{y} & 1\end{array}\right|\) \(\Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y|\)…
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