jee-main 2022 Q75

jee-main · India · session2_27jul_shift1 Areas by integration

The area of the smaller region enclosed by the curves $y ^ { 2 } = 8 x + 4$ and $x ^ { 2 } + y ^ { 2 } + 4 \sqrt { 3 } x - 4 = 0$ is equal to
(1) $\frac { 1 } { 3 } ( 2 - 12 \sqrt { 3 } + 8 \pi )$
(2) $\frac { 1 } { 3 } ( 2 - 12 \sqrt { 3 } + 6 \pi )$
(3) $\frac { 1 } { 3 } ( 4 - 12 \sqrt { 3 } + 8 \pi )$
(4) $\frac { 1 } { 3 } ( 4 - 12 \sqrt { 3 } + 6 \pi )$

The area of the smaller region enclosed by the curves $y ^ { 2 } = 8 x + 4$ and $x ^ { 2 } + y ^ { 2 } + 4 \sqrt { 3 } x - 4 = 0$ is equal to\\
(1) $\frac { 1 } { 3 } ( 2 - 12 \sqrt { 3 } + 8 \pi )$\\
(2) $\frac { 1 } { 3 } ( 2 - 12 \sqrt { 3 } + 6 \pi )$\\
(3) $\frac { 1 } { 3 } ( 4 - 12 \sqrt { 3 } + 8 \pi )$\\
(4) $\frac { 1 } { 3 } ( 4 - 12 \sqrt { 3 } + 6 \pi )$

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