jee-main 2024 Q75

jee-main · India · session1_29jan_shift1 Integration by Substitution Substitution to Compute an Indefinite Integral with Initial Condition

For $x \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$, if $y ( x ) = \int \frac { \operatorname { cosec } x + \sin x } { \operatorname { cosec } x \sec x + \tan x \sin ^ { 2 } x } d x$ and $\lim _ { x \rightarrow \left( \frac { \pi } { 2 } \right) ^ { - } } y ( x ) = 0$ then $y \left( \frac { \pi } { 4 } \right)$ is equal to
(1) $\tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$
(2) $\frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$
(3) $- \frac { 1 } { \sqrt { 2 } } \tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$
(4) $\frac { 1 } { \sqrt { 2 } } \tan ^ { - 1 } \left( - \frac { 1 } { 2 } \right)$

For $x \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$, if $y ( x ) = \int \frac { \operatorname { cosec } x + \sin x } { \operatorname { cosec } x \sec x + \tan x \sin ^ { 2 } x } d x$ and $\lim _ { x \rightarrow \left( \frac { \pi } { 2 } \right) ^ { - } } y ( x ) = 0$ then $y \left( \frac { \pi } { 4 } \right)$ is equal to\\
(1) $\tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$\\
(2) $\frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$\\
(3) $- \frac { 1 } { \sqrt { 2 } } \tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$\\
(4) $\frac { 1 } { \sqrt { 2 } } \tan ^ { - 1 } \left( - \frac { 1 } { 2 } \right)$

Paper Questions

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