JEE MAIN - Mathematics Hindi (2020 - 8th January Morning Slot - No. 17)
ƒ(x) = (sin(tan–1x) + sin(cot–1x))2 – 1 को मानें, |x| > 1.
यदि $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ और $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$, तो y($${ - \sqrt 3 }$$) का मान है :
यदि $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ और $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$, तो y($${ - \sqrt 3 }$$) का मान है :
$${{5\pi } \over 6}$$
$$ - {\pi \over 6}$$
$${\pi \over 3}$$
$${{2\pi } \over 3}$$
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