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Text Solution

`(0,1)``(-1/2,0)``(2,0)``(0,2)`

Answer :

BSolution :

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