JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The points on the curve

y=12 x-x^3

at which the gradient is zero are

View answer and explanation

Correct option: D

(2,16),(-2,-16)

Option C (

(2,-16),(-2,16)

) is correct because setting the derivative of the curve equal to zero and solving for x gives x=

\pm

2, and substituting these values back into the original equation yields the corresponding y-values.

More Questions from TANGENTS & NORMALS

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z