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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, isA. `x+2y=0`B. `2x+y=0`C. `x-y=2`D. none of these |
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Answer» Correct Answer - A 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` |
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