JEE Advance - Mathematics Hindi (2017 - Paper 2 Offline - No. 13)
यदि $$ग(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$$, तो
$$g'\left( { - {\pi \over 2}} \right) = 0$$
$$g'\left( { - {\pi \over 2}} \right) = - 2\pi $$
$$g'\left( {{\pi \over 2}} \right) = 2\pi $$
$$g'\left( {{\pi \over 2}} \right) = 0$$
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