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Equation of the tangent to the curve y=`e^(-abs(x))` at the point where it cuts the line x=1-A. is ey+x=2B. is x+y=eC. is ex+y=1D. does not exist |
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Answer» Correct Answer - 1 The point of intersection is (1,`e^(-1)`) `because` x=1, so equation of the curve is =`e^(-x)` `rArr(dy)/(dx)=-e^(-x)` `[(dy)/(dx)]_(x=1)=-e^(-1)`. Hnce equation of tangent is `y-e^(-1)=-e^(-1)(x-1)or,ey+x=2` |
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