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 ` (a) `[-1/2,0]` (b) `[-1,-1/2]` `[0,1]` (d) `[1/2,1]`

Options

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

Answer

B

Explanation

the equation of curve is `y=e^(2x)`
Since, it passes through the point (0,1).
`therefore (dy)/(dx)=e^(2x).2=2.e^(2x)`
`rArr (dy)/(dx)_(0,1) = 2.e^(2.0)=2`= Slope of the tangent to the curve.
`therefore` Equation of tangent is `y-1=2(x-0)`
`rArr y=2x+1`
Since, tangent to the curve `y=e^(2x)` at the point (0,1) meets X-axis i.e., y=0.
`therefore 0=2x+1 rArr x=-1/2`
So, the required point is `(-1/2,0)`.