Home
Class 12
MATHS
Equation of tangent of the curve y = 1 -...

Equation of tangent of the curve `y = 1 - e^(x//2)` at that point at which the curve crosses the y-axis, is :

A

`x+y=1`

B

`2x+y = 1`

C

`x=-2y`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • APPLICATIONS OF DERIVATIVES

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise 6.1|18 Videos
  • APPLICATIONS OF DERIVATIVES

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise 6.2|19 Videos
  • APPLICATIONS OF DERIVATIVES

    NAGEEN PRAKASHAN ENGLISH|Exercise Exercise 6h (multiple Choice Questions)|10 Videos
  • APPLICATIONS OF INTEGRALS

    NAGEEN PRAKASHAN ENGLISH|Exercise Miscellaneous Exercise|19 Videos

Similar Questions

Explore conceptually related problems

Find the equation of tangent of tangent of the curve y = b * e^(-x//a) at that point at which the curve meets the Y-axis.

Find the equation of tangent of the curve yx^(2)+x^(2)-5x+6=0 at that point at which curve crosses the X-axis.

The equation of the tangent to the curve y=b e^(-x//a) at the point where it crosses the y-axis is (a)x/a-y/b=1 (b) a x+b y=1 (c)a x-b y=1 (d) x/a+y/b=1

The equation of the tangent to the curve y = e^(2x) at (0,1) is

The equation of tangent to the curve y=be^(-x//a) at the point where it crosses Y-axis is

Find the equation of the tangent to the curve y=(x^3-1)(x-2) at the points where the curve cuts the x-axis.

Find the equation of the tangent to the curve y=(x^3-1)(x-2) at the points where the curve cuts the x-axis.

The equation of the tangent to the curve y=e^(-|x|) at the point where the curve cuts the line x = 1, is

Write the equation of the tangent to the curve y=x^2-x+2 at the point where it crosses the y-axis.

Find the equation of tangent of the curve x=at^(2), y = 2at at point 't'.