Applications-Of-Derivatives Question 434
Question: The equation of normal to the curve $ y={{(1+x)}^{y}}+{{\sin }^{-1}}({{\sin }^{2}}x) $ at $ x=0 $ is
Options:
A) $ x+y=1 $
B) $ x-y=1 $
C) $ x+y=-1 $
D) $ x-y=-1 $
Show Answer
Answer:
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}}{ \log (1+x)\frac{dy}{dx}+\frac{y}{1+x} } $ $ -\frac{dy}{dx}+\frac{1}{\sqrt{1-{{\sin }^{4}}x}}2\sin x\cos x=0 $ or $ {{( \frac{dy}{dx} )}_{(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 $
BETA
