JEE Advance - Mathematics (1993 - No. 23)
Find the equation of the normal to the curve
$$y = {\left( {1 + x} \right)^y} + {\sin ^{ - 1}}\left( {{{\sin }^2}x} \right)$$ at $$x=0$$
$$y = {\left( {1 + x} \right)^y} + {\sin ^{ - 1}}\left( {{{\sin }^2}x} \right)$$ at $$x=0$$
x - y = 1
x + y = 1
x - y = 0
x + y = 0
2x + y = 1
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