jee-main 2025 Q12

jee-main · India · session1_23jan_shift2 Differential equations Solving Separable DEs with Initial Conditions

Let $x = x ( y )$ be the solution of the differential equation $y = \left( x - y \frac { \mathrm {~d} x } { \mathrm {~d} y } \right) \sin \left( \frac { x } { y } \right) , y > 0$ and $x ( 1 ) = \frac { \pi } { 2 }$. Then $\cos ( x ( 2 ) )$ is equal to :
(1) $1 - 2 \left( \log _ { e } 2 \right) ^ { 2 }$
(2) $1 - 2 \left( \log _ { \mathrm { e } } 2 \right)$
(3) $2 \left( \log _ { e } 2 \right) - 1$
(4) $2 \left( \log _ { e } 2 \right) ^ { 2 } - 1$

Let $x = x ( y )$ be the solution of the differential equation $y = \left( x - y \frac { \mathrm {~d} x } { \mathrm {~d} y } \right) \sin \left( \frac { x } { y } \right) , y > 0$ and $x ( 1 ) = \frac { \pi } { 2 }$. Then $\cos ( x ( 2 ) )$ is equal to :\\
(1) $1 - 2 \left( \log _ { e } 2 \right) ^ { 2 }$\\
(2) $1 - 2 \left( \log _ { \mathrm { e } } 2 \right)$\\
(3) $2 \left( \log _ { e } 2 \right) - 1$\\
(4) $2 \left( \log _ { e } 2 \right) ^ { 2 } - 1$

Paper Questions

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