A) \[x+y=1\]
B) \[x-y=1\]
C) \[x+y=-1\]
D) \[x-y=-1\]
Correct Answer: A
Solution :
[a] At \[x=0,\,\,\,y=1\]. Hence, the point at which normal is drawn is \[P(0,1)\]. Differentiating the given equation w.r.t. x, we have \[{{(1+x)}^{y}}\left\{ \log (1+x)\frac{dy}{dx}+\frac{y}{1+x} \right\}\] \[-\frac{dy}{dx}+\frac{1}{\sqrt{1-{{\sin }^{4}}x}}2\sin x\cos x=0\] or \[{{\left( \frac{dy}{dx} \right)}_{(0,1)}}=\frac{{{(1+0)}^{1}}\times \frac{1}{1+0}-\frac{2\sin 0}{\sqrt{1-{{\sin }^{2}}0}}}{1-{{(1+0)}^{1}}\log 1}=1\]. \[\therefore \] Slope of the normal \[=-1\]. Therefore, equation of the normal having slope \[-1\] at point \[P(0,1)\] is given by \[y-1=-(x-0)\] or \[x+y=1\]
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