jee-main 2022 Q76

jee-main · India · session1_24jun_shift2 Differential equations Solving Separable DEs with Initial Conditions

The slope of normal at any point $( x , y ) , x > 0 , y > 0$ on the curve $y = y ( x )$ is given by $\frac { x ^ { 2 } } { x y - x ^ { 2 } y ^ { 2 } - 1 }$. If the curve passes through the point $( 1,1 )$, then $e \cdot y ( e )$ is equal to
(1) $\frac { 1 - \tan ( 1 ) } { 1 + \tan ( 1 ) }$
(2) $\tan ( 1 )$
(3) 1
(4) $\frac { 1 + \tan ( 1 ) } { 1 - \tan ( 1 ) }$

The slope of normal at any point $( x , y ) , x > 0 , y > 0$ on the curve $y = y ( x )$ is given by $\frac { x ^ { 2 } } { x y - x ^ { 2 } y ^ { 2 } - 1 }$. If the curve passes through the point $( 1,1 )$, then $e \cdot y ( e )$ is equal to\\
(1) $\frac { 1 - \tan ( 1 ) } { 1 + \tan ( 1 ) }$\\
(2) $\tan ( 1 )$\\
(3) 1\\
(4) $\frac { 1 + \tan ( 1 ) } { 1 - \tan ( 1 ) }$

Paper Questions

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