JEE 2022 — Mathematics PYQ

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Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

If $\int_{0}^{1}(\sqrt{2x} - \sqrt{2x - x^2})dx = \int_{0}^{1}\left(1 - \sqrt{1 - y^2} - \frac{y^2}{2}\right)dy + \int_{1}^{2}\left(2 - \frac{y^2}{2}\right)dy + I$ , then $I$ equals:

Choose the correct answer:

Explanation

L.H.S.:

\text{L.H.S.} = \int_{0}^{2} \sqrt{2x} dx - \int_{0}^{2} \sqrt{2x - x^2} dx

= \sqrt{2} \left[ \frac{2}{3}x^{3/2} \right]_0^2 - \int_{0}^{2} \sqrt{1 - (x - 1)^2} dx

= \frac{8}{3} - 2\int_{0}^{1} \sqrt{1 - y^2} dy

R.H.S.:

\text{R.H.S.} = \int_{0}^{1} (1 - \frac{y^2}{2}) dy - \int_{0}^{1} \sqrt{1 - y^2} dy + \int_{1}^{2} (2 - \frac{y^2}{2}) dy + I

= \left[ y - \frac{y^3}{6} \right]_0^1 - \int_{0}^{1} \sqrt{1 - y^2} dy + \left[ 2y - \frac{y^3}{6} \right]_1^2 + I

= \frac{5}{6} - \int_{0}^{1} \sqrt{1 - y^2} dy + \frac{5}{6} + I = \frac{5}{3} - \int_{0}^{1} \sqrt{1 - y^2} dy + I

Equating L.H.S and R.H.S:

\frac{8}{3} - 2\int_{0}^{1} \sqrt{1 - y^2} dy = \frac{5}{3} - \int_{0}^{1} \sqrt{1 - y^2} dy + I

I = 1 - \int_{0}^{1} \sqrt{1 - y^2} dy

I = \int_{0}^{1} (1 - \sqrt{1 - y^2}) dy

Correct Option: (C)

Choose the correct answer:

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