QJEEM19Mae24c: JEE Main 2019 Mathematics - Class 12 JEEM19Mae24c Solution: If the area (in sq. units) of the region {(x, y) : y 2 4x, x +…

Question

Verified Solution • 2025–26 Session moderate

If the area (in sq. units) of the region {(x, y) : y²

\le

4x, x + y

\le

1, x

\ge

0, y

\ge

0} is a

\sqrt 2

+ b, then a – b is equal to :

Accepted Answer

Let P be the point common to x + y = 1 and y² = 4x

Then y² = 4(1-y)

\Rightarrow

y² - 4y - 4 = 0

\Rightarrow y = {{ - 4 \pm \sqrt {16 + 16} } \over 2}

\Rightarrow y = - 2 + 2\sqrt 2

Then P is (3 -

2\sqrt 2

, -2 +

2\sqrt 2

).

Hence shaded area = Area of region (OPN) + area of (

\Delta OPQ

)

\Rightarrow {\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt {xdx} + {1 \over 2}\left[ {1 - (3 - 2\sqrt 2 )} \right]} ^2}

\Rightarrow {2 \over 3}2\left( {\sqrt 2 - 1} \right)\left( {3 - 2\sqrt 2 } \right) + {1 \over 2}{\left[ {2(\sqrt 2 - 1)} \right]^2}

\Rightarrow {4 \over 3}\left\{ { - 7 + 5\sqrt 2 } \right\} + 2\left( {3 - 2\sqrt 2 } \right)

\Rightarrow \left( {{{20} \over 3} - 4} \right)\sqrt 2 + 6 - {{28} \over 3}

\Rightarrow {8 \over 3}\sqrt 2 - {{10} \over 3}

Then a =

{8 \over 3}

and b =

{-10 \over 3}

, so a - b = 6

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