Test Index

VITEEE 2014 Paper

Section: Mathematics

© examsnet.com

Question : 91 of 120

Marks: +1, -0

The equation of normal to the curve y =

(1 + x)^y

\sin^{-1}(\sin^2 x)

at x = 0 is

x + y = 1
x - y = 1
x + y = - 1
x - y = -1

Solution:

Given curve is

y =

(1 + x)^y

+

\sin^{-1}(\sin^2 x)

On differentiating w.r.t x, we get

\frac{dy}{dx}

=

(1 + x)^y

\left[\frac{y}{1+x} + \log(1+x)\frac{dy}{dx}\right]

+

\frac{2\sin x \cos x}{\sqrt{1-\sin^4 x}}

⇒

\left(\frac{dy}{dx}\right)_{\text{at }(0,1)}

= 1 [Since at x = 0 , y = 1]

Slope of normal at (x = 0) = -1

∴ Equation of normal at x = 0 and y = 1 is

y - 1 = - 1 (x - 0)

⇒ y - 1 = - x ⇒ x + y = 1

© examsnet.com

Go to Question:

Prev Question

Next Question