JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 13)
If $$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$$, then
$$g'\left( { - {\pi \over 2}} \right) = 0$$
$$g'\left( { - {\pi \over 2}} \right) = - 2\pi $$
$$g'\left( {{\pi \over 2}} \right) = 2\pi $$
$$g'\left( {{\pi \over 2}} \right) = 0$$
Explanation
$$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$$
g'(x) = 2cos 2x sin$$-$$1(sin2 x) $$-$$ cos x sin$$-$$1(sin x)
$$g'\left( {{\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$$
$$g'\left( { - {\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$$
No option is matching.
g'(x) = 2cos 2x sin$$-$$1(sin2 x) $$-$$ cos x sin$$-$$1(sin x)
$$g'\left( {{\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$$
$$g'\left( { - {\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$$
No option is matching.
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