JEE MAIN - Mathematics Hindi (2015 (Offline) - No. 13)
माना $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$, जहाँ $$|x|<\frac{1}{\sqrt{3}}$$ है, तो $$y$$ का एक मान है :
$${{3x - {x^3}} \over {1 + 3{x^2}}}$$
$${{3x + {x^3}} \over {1 + 3{x^2}}}$$
$${{3x - {x^3}} \over {1 - 3{x^2}}}$$
$${{3x + {x^3}} \over {1 - 3{x^2}}}$$
Comments (0)
