Question

The equation of the tangent to the curve $ y={{(1+x)}^{y}}+{{\sin }^{-1}}({{\sin }^{2}}x) $ at $ x=0 $ is:

• A. $ x-y+1=0 $
• B. $ x+y+1=0 $
• C. $ 2x-y+1=0 $
• D. $ x+2y+2=0 $

Answer 1

The Correct Option is A

$ 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 $