The equation of the tangent to the curve `y=1-e^(x//2)` at the tangent to the curve `y=1-e^(x//2)` at the point of intersection with the y-axis, is

Options

A. `x+2y=0`
B. `2x+y=0`
C. `x-y=2`
D. none of these

Answer

A

Explanation

We have,
`y= 1-e^(x//2) rArr (dy)/(dx) = -(1)/(2) e^(x//2)`
The curves`y=1-e^(x//2)` meets y-axis at (0, 0).
`therefore((dy)/(dx))_((0","0))= -(1)/(2)`
The equation of the tangent at (0, 0) is
`y-0 = -(1)/(2) (x-0) rArr x+2y =0`