JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The slope of the tangent to the curve

\mathrm{y}=\int_0^{\mathrm{x}} \frac{\mathrm{dx}}{1+\mathrm{x}^3}

at the point where

\mathrm{x}=1

is -

View answer and explanation

Correct option: A

1 / 2

Option A (1/2) is correct because the derivative of the integral gives

\frac{{1}}{{1+x^3}}

, and evaluating at x=1 yields 1/2.

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