JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Analyzing +4 | -1

A curve is given by the equations

x = a\cos \theta + \frac{1}{2}b\cos 2\theta ,y = a\sin \theta + \frac{1}{2}b\sin 2\theta

. Then the points for which

\frac{{{d^2}y}}{{d{x^2}}} = 0

are given by

View answer and explanation

Correct option: C

\cos \theta = - \frac{{{a^2} + 2{b^2}}}{{3ab}}

Option C (

\cos \theta = - \frac{{{{a^2} + 2{{b^2}}}}}{{3ab}}

) is correct because it is the result of setting the second derivative to zero and solving for

\cos \theta

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →

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}}