JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

For the curves

x=a \sqrt{\cos 2 \theta} \cos \theta, y=a \sqrt{\cos 2 \theta} \sin \theta

, the tangent at

\theta=\frac{\pi}{6}

is ......

View answer and explanation

Correct option: A

Parallel to

\mathrm{X}

-axis or

\mathrm{X}

-axis itself.

Option A (Parallel to X-axis or X-axis itself) is correct because at

\theta = \frac{\pi}{6}

, the derivative

\frac{dy}{dx}

is 0, indicating a horizontal tangent.

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}}