jee-main 2021 Q71

jee-main · India · session4_27aug_shift2 Differentiating Transcendental Functions Evaluate derivative at a point or find tangent slope

If $y ( x ) = \cot ^ { - 1 } \left( \frac { \sqrt { 1 + \sin x } + \sqrt { 1 - \sin x } } { \sqrt { 1 + \sin x } - \sqrt { 1 - \sin x } } \right) , x \in \left( \frac { \pi } { 2 } , \pi \right)$, then $\frac { d y } { d x }$ at $x = \frac { 5 \pi } { 6 }$ is: (1) 0 (2) - 1 (3) $\frac { - 1 } { 2 }$ (4) $\frac { 1 } { 2 }$

If $y ( x ) = \cot ^ { - 1 } \left( \frac { \sqrt { 1 + \sin x } + \sqrt { 1 - \sin x } } { \sqrt { 1 + \sin x } - \sqrt { 1 - \sin x } } \right) , x \in \left( \frac { \pi } { 2 } , \pi \right)$, then $\frac { d y } { d x }$ at $x = \frac { 5 \pi } { 6 }$ is:
(1) 0
(2) - 1
(3) $\frac { - 1 } { 2 }$
(4) $\frac { 1 } { 2 }$

Paper Questions

Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q70 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q86 Q87 Q88 Q89 Q90