JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

Consider

f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} } \right),\,\,x \in \left( {0,\frac{\pi }{2}} \right)

.
A normal to

y = f\left( x \right)

x = \frac{\pi }{6}

also passes through the point

View answer and explanation

Correct option: B

\left( {0,\,\frac{{2\pi }}{3}} \right)

The correct answer is B (

\left( {0,\\,\frac{{2\pi }}{3}} \right)

) because the normal to the curve at

x = \frac{\pi }{6}

passes through this point.

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}