JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the area of the region ${(x, y) : ∣ x^{2} - 2∣ \leq x}$ is $A$ , then $6 A + 162$ is equal to ____.

Choose the correct answer:

Explanation

A = \int_{1}^{\sqrt{2}} [x - (2 - x^2)] \, dx + \int_{\sqrt{2}}^{2} [x - (x^2 - 2)] \, dx

A = \int_{1}^{\sqrt{2}} [x - (2 - x^2)] \, dx + \int_{\sqrt{2}}^{2} [x - (x^2 - 2)] \, dx

A = \int_{1}^{\sqrt{2}} [x - (2 - x^2)] \, dx + \int_{\sqrt{2}}^{2} [x - (x^2 - 2)] \, dx

= \left[ \frac{x^2}{2} - 2x + \frac{x^3}{3} \right]_{1}^{\sqrt{2}} + \left[ \frac{x^2}{2} - \frac{x^3}{3} + 2x \right]_{\sqrt{2}}^{2}

= \left( \frac{2}{2} - 2\sqrt{2} + \frac{2\sqrt{2}}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) + \left( \frac{4}{2} - \frac{8}{3} + 4 \right) - \left( \frac{2}{2} - \frac{2\sqrt{2}}{3} + 2\sqrt{2} \right)

= 1 - 2\sqrt{2} + \frac{2\sqrt{2}}{3} + \frac{3}{2} - \frac{1}{3} + 6 - \frac{8}{3} - 1 + \frac{2\sqrt{2}}{3} - 2\sqrt{2}

A = -\frac{8\sqrt{2}}{3} + \frac{9}{2}

\implies 6A + 16\sqrt{2} = 27