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Equation of the tangent to the curve y=e...

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=2

B

is x+y=e

C

is ex+y=1

D

does not exist

Text Solution

Verified by Experts

The correct Answer is:
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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