The tangent to the curve `y=e^(2x)` at the point (0,1) meets X-axis at

Options

A. `((1)/(2),0)`
B. `(-(1)/(2),0)`
C. `(2,0)`
D. (0,0)

Answer

B

Explanation

Given curves is `y=e^(2x)`
On differentiating, we get
`(dy)/(dx)=2e^(2x)implies((dy)/(dx))_((0,1))=2e^(0)=2`
Equation of tangent at (0,1) and slope 2 is
y-1=2(x-0)`implies`y=2x+1
this tangent, meets X-axis,
`thereforey=0`
`implies0=2x+1impliesx=-(1)/(2)`
`therefore` Coordinate of the point on X-axis is `(-(1)/(2),0)`.