jee-main 2025 Q74

jee-main · India · session2_08apr_shift1 Areas by integration
Q74. The parabola $y ^ { 2 } = 4 x$ divides the area of the circle $x ^ { 2 } + y ^ { 2 } = 5$ in two parts. The area of the smaller part is equal to:
(1) $\frac { 1 } { 3 } + 5 \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$
(2) $\frac { 1 } { 3 } + \sqrt { 5 } \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$
(3) $\frac { 2 } { 3 } + 5 \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$
(4) $\frac { 2 } { 3 } + \sqrt { 5 } \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$
Q74. The parabola $y ^ { 2 } = 4 x$ divides the area of the circle $x ^ { 2 } + y ^ { 2 } = 5$ in two parts. The area of the smaller part is equal to:\\
(1) $\frac { 1 } { 3 } + 5 \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$\\
(2) $\frac { 1 } { 3 } + \sqrt { 5 } \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$\\
(3) $\frac { 2 } { 3 } + 5 \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$\\
(4) $\frac { 2 } { 3 } + \sqrt { 5 } \sin ^ { - 1 } \left( \frac { 2 } { \sqrt { 5 } } \right)$