JEE Advanced · Mathematics · 28. Area Under Curves

Let \(S=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x \geq 0, y \geq 0, y^2 \leq 4 x, y^2 \leq 12-2 x\right.\) and \(\left.3 y+\sqrt{8} x \leq 5 \sqrt{8}\right\}\). If the area of the region \(S\) is \(\alpha \sqrt{2}\), then \(\alpha\) is equal to

A \(\frac{17}{2}\)
B \(\frac{17}{3}\)
C \(\frac{17}{4}\)
D \(\frac{17}{5}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{17}{3}\)

Step-by-step Solution

Detailed explanation

Point of intersection of all curves is \((2,2 \sqrt{2})\)
Area \(=\mathrm{A}_1+\mathrm{A}_2\)
\(\alpha \sqrt{2}=\int_0^2 2 \sqrt{\mathrm{x}} \mathrm{dx}+\frac{1}{2} \times 3 \times 2 \sqrt{2}\)
\(\begin{aligned} & \alpha \sqrt{2}=2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2+3 \sqrt{2} \\ & \alpha \sqrt{2}=\frac{17 \sqrt{2}}{3} \\ & \alpha=\frac{17}{3}\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Advanced

Explore more questions on app