JEE MAIN - Mathematics Hindi (2019 - 8th April Morning Slot - No. 21)
यदि $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$,
x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ तो $$dy \over dx$$ के बराबर होगा:
x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ तो $$dy \over dx$$ के बराबर होगा:
$$2x - {\pi \over 3}$$
$${\pi \over 6} - x$$
$${\pi \over 3} - x$$
$$x - {\pi \over 6}$$
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