Let S = {(x, y) ∈ R × R:x ≥ 0, y ≥ 0, y^2 ≤ 4x, y^2 ≤ 12 − 2x and 3y + √8x ≤ 5√8}.

Question

Let \(S=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x \geq 0, y \geq 0, \right.\)\(y^{2} \leq 4 x, y^{2} \leq 12-2 x\) 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) \(\frac{17}{5}\)

Answer 1

Correct option is (B) \(\frac{17}3\)

\(y^{2}=4 x, y^{2}=12-2 x \Rightarrow x=2, y=\sqrt{8}\)

\(A= \int\limits_{0}^{2} 2 \sqrt{x} d x+\frac{1}{2} \times 3 \times \sqrt{8} \)

\(=\left[2 \times \frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{2}+3 \sqrt{2}=\frac{4}{3} \times 2 \sqrt{2}+3 \sqrt{2}=\frac{17}{3} \sqrt{2}\)

\(\because A=\alpha \sqrt{2} \Rightarrow \alpha=\frac{17}{3}\)