A) \[x-y+1=0\]
B) \[x+y+1=0\]
C) \[2x-y+1=0\]
D) \[x+2y+2=0\]
E) \[2x+y-1=0\]
Correct Answer: A
Solution :
\[y={{(1+x)}^{y}}+{{\sin }^{-1}}({{\sin }^{2}}x)\] At \[x=0,y=1\] Let \[y=u+v\] where \[u={{(1+x)}^{y}},\] \[v={{\sin }^{-1}}({{\sin }^{2}}x)\] \[\Rightarrow \] \[\log u=y\log (1+x)\] \[\Rightarrow \] \[\frac{dy}{dx}\frac{1}{u}=\frac{y}{1+x}+\log (1+x)\frac{dy}{dx}\] and \[\frac{dv}{dx}=\frac{1}{\sqrt{1-{{\sin }^{2}}x}}.2\sin x\cos x\] \[\therefore \] \[\frac{dy}{dx}={{(1+x)}^{y}}\left[ \frac{y}{1+x}+\log (1+x)\frac{dy}{dx} \right]\] \[+\frac{\sin 2x}{\cos x}\] \[\frac{dy}{dx}[1-{{(1+x)}^{y}}\log (x+1)]\] \[=\frac{{{(1+x)}^{y}}y}{1+x}+\frac{\sin 2x}{\cos x}\] \[\therefore \] \[{{\left. \frac{dy}{dx} \right|}_{(0,1)}}=\frac{1}{1+0}=1\] \[\therefore \]Equation of tangent is \[(y-1)=1(x)\Rightarrow y-x-1=0\] \[\Rightarrow \] \[x-y+1=0\]
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