A) \[x+y=1\]
B) \[x+y+1=0\]
C) \[2x-y+1=0\]
D) \[x+2y+2=0\]
Correct Answer: A
Solution :
We have, \[y={{(1+x)}^{y}}+{{\sin }^{-1}}\left( {{\sin }^{2}}x \right)\] ...(i) when \[x=0,\]we have \[y=1\] Differentiating (i) w.r.t. x we get \[\frac{dy}{dx}={{\left( 1+x \right)}^{y}}\left\{ \frac{dy}{dx}\log (1+x)+\frac{y}{1+x} \right\}+\frac{\sin 2x}{\sqrt{1-{{\sin }^{4}}x}}\]\[\Rightarrow \,\,{{\left( \frac{dy}{dx} \right)}_{(0,1)}}=1\Rightarrow -{{\left( \frac{dx}{dy} \right)}_{(0,1)}}=-1.\] So the equation of the normal at \[(0,1)\] is \[y-1=-1\left( x-0 \right)\Rightarrow x+y=1\]
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