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