JEE MAIN - Mathematics (2015 (Offline) - No. 7)
The area (in sq. units) of the region described by
$$\left\{ {\left( {x,y} \right):{y^2} \le 2x} \right.$$ and $$\left. {y \ge 4x - 1} \right\}$$ is :
$$\left\{ {\left( {x,y} \right):{y^2} \le 2x} \right.$$ and $$\left. {y \ge 4x - 1} \right\}$$ is :
$${{15} \over {64}}$$
$${{9} \over {32}}$$
$${{7} \over {32}}$$
$${{5} \over {64}}$$
Explanation
Required area
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$$ = $$ Area of $$ABCD$$ $$-$$ $$ar$$ $$(ABOCD)$$
$$ = {1 \over 4}\left[ {{{{y^2}} \over 2} + y} \right]_{ - 1/2}^1\,\, - {1 \over 2}\left[ {{{{y^3}} \over 3}} \right]_{ - 1}^1$$
$$ = {1 \over 4}\left[ {{3 \over 2} + {3 \over 8}} \right] - {9 \over {48}}$$
$$ = {{15} \over {32}} - {9 \over {48}} = {{27} \over {96}} = {9 \over {32}}$$
$$ = $$ Area of $$ABCD$$ $$-$$ $$ar$$ $$(ABOCD)$$
$$ = {1 \over 4}\left[ {{{{y^2}} \over 2} + y} \right]_{ - 1/2}^1\,\, - {1 \over 2}\left[ {{{{y^3}} \over 3}} \right]_{ - 1}^1$$
$$ = {1 \over 4}\left[ {{3 \over 2} + {3 \over 8}} \right] - {9 \over {48}}$$
$$ = {{15} \over {32}} - {9 \over {48}} = {{27} \over {96}} = {9 \over {32}}$$
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