JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The tangent to the curve

y=e^{k x}

at a point

(0,1)

meets the

x

-axis at

(a, 0)

where

a \in[-2,-1]

, then

k \in

View answer and explanation

Correct option: D

\left[\frac{1}{2}, 1\right]

Option D ([

\frac{1}{2}, 1

]) is correct because finding the tangent equation at (0,1) and using the given range for the x-intercept 'a' allows us to determine the range for k.

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