QJEEM22Ma8dc1: JEE Main 2022 Mathematics - Class 12 JEEM22Ma8dc1 Solution: If , then I equals

Given,

$\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }$

Now,

$L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx}$

$= \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} }$

$= \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx}$

$= \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx}$

[We know,

$\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C$ ]

$= {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2$

$= {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]$

$= {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)$

$= {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}$

$= {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}$

$= {8 \over 3} - {{2\pi } \over 4}$

$= {8 \over 3} - {\pi \over 2}$

Now in R.H.S.,

$\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$

$= \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } }$

$= \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1$

$= 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}$

$= 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}$

$= 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}$

$= {5 \over 6} - {\pi \over 4}$

Now,

$\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}$

$= \left[ {2y - {{{y^3}} \over 6}} \right]_1^2$

$= \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]$

$= 4 - {8 \over 6} - 2 + {1 \over 6}$

$= 2 - {7 \over 6}$

$= {5 \over 6}$

$\therefore$ $R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I$

As $L.H.S. = R.H.S.$

$\Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I$

$\Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}$

$\Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}$

$\Rightarrow I = 1 - {\pi \over 4}$

From option (c)

$\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy}$

$= \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy}$

$= 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1$

$= 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]$

$= 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]$

$= 1 - {\pi \over 4} = I$

$\therefore$ Option (c) is the right answer.

Note :

There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.