A) \[x+y=2\]
B) \[x+y=1\]
C) \[x-y=1\]
D) \[x-y=2\]
Correct Answer: B
Solution :
Given curve is \[y={{(1+x)}^{y}}+{{\sin }^{-1}}({{\sin }^{2}}x)\] At \[x=0,y=1\] Now, \[\frac{dy}{dx}={{(1+x)}^{y}}\left[ \frac{y}{1+x}+\frac{dy}{dx}\log (1+x) \right]\] \[+\frac{2\sin x\cos x}{\sqrt{1-{{\sin }^{4}}x}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(0,1)}}=1(1)+0=1\] Thus, equation of normal is \[y-1=-1(x-0)\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(0,1)}}=1(1)+0=1\]
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