JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The slope of the normal at the point with abscissa

x=-2

of the graph of the function

f(x)=\left|x^{2}-x\right|

View answer and explanation

Correct option: D

1 / 3

Option D (1/3) is correct because it is the negative reciprocal of the derivative of the function evaluated at x = -2, considering the absolute value.

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The complete set of value of '

a

' for which there exists at least one line that is tangent to the graph of the curve

y=x^3-a x

at one point and normal to the graph at another point is given by The point of intersection of the tangents drawn to the curve

x^2 y=1-y

at the points where it is met by the curve

x y=1-y

, is given by Let

C

be the curve

y^3-3 x y+2=0

. If

H

is the set of points on the curve

\mathrm{C}

where the tangent is parallel to

x

-axis and

V

is the set of points on

C

where the tangent is parallel to

y

-axis, then The curve

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

has a vertical tangent at the point: If normal to the curve

y = f\left( x \right)

is parallel to

x

-axis, then The tangents to the curve

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

at the points

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

are parallel to: