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The tangent to the curve `y=e^(2x)` at the point (0,1) meets X-axis atA. `(0,1)`B. `(-1/2,0)`C. `(2,0)`D. `(0,2)` |
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Answer» Correct Answer - B 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)`. |
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