Question

Let \(S=\left\{(x,y)\in\R\times\R:x\ge0,y\ge0,y^2\le12-2x\ \text{and}\ 3y+\sqrt8\ x\le5\sqrt8\right\}\). If the area of the region S is \(\alpha\sqrt2\), then α is equal to

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

Answer 1

The Correct Option is B

Step 1: Setting Up the Integral 

The required area is given by: \[ \text{Required Area} = \int_0^2 \sqrt{x} \,dx + \frac{1}{2} \times 3 \times \sqrt{8} \]

Step 2: Solving for Limits

Given equations: \[ y^2 = 4x, \quad y^2 = 12 - 2x \] Solving for \( x \): \[ x = 2, \quad y = \sqrt{8} \]

Step 3: Evaluating the Integral

\[ A = \int_0^2 2\sqrt{x} \,dx + \frac{1}{2} \times 3 \times \sqrt{8} \] Evaluating the integral: \[ 2 \times \frac{2}{3} x^{3/2} \Big|_0^2 + 3\sqrt{2} \] \[ = \frac{4}{3} \times 2\sqrt{2} + 3\sqrt{2} \] \[ = \frac{17}{3} \sqrt{2} \]

Step 4: Finding \( \alpha \)

Since the total area is expressed as: \[ \alpha \sqrt{2}, \quad \text{we have} \quad \alpha = \frac{17}{3} \]

Final Answer:

\[ \alpha = \frac{17}{3} \]