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Let xgt0 be a fixed reacl number. Then t...

Let `xgt0` be a fixed reacl number. Then the integral `underset(0)overset(oo)(f)e^(-1)|x-t|dt` is equal to -

A

`x+2e^(-x)-1`

B

`x-2e^(-x)+1`

C

`x+2e^(-x)+1`

D

`-x-2e^(-x)+1`

Text Solution

Verified by Experts

The correct Answer is:
A
`f(x)=underset(0)overset(oo)(f)e^(-t)|x-t|dt`
`f(x)=underset(0)overset(oo)(f)e^(-t)(x-t)dt+f(x)=underset(x)overset(oo)(f)e^(-t)(t-x)dt`
`f^(')(x)=e^(-x)(x-x)-e^(-0).0+f(x)=underset(0)overset(x)(f)e^(-t)(1)dt+0-e^(-x)(x-x).1+f(x)=underset(0)overset(x)(f)e^(-t)(1)dt+0-e^(x)(x-x).1+`f(x)=underset(x)overset(oo)(f)-e^(-t)dt`
`=[-e^(-t)]_(0)^(x)+[e^(-t)]_(x)^(oo)`
`=-e^(-x)+1+0-e^(-x)`
`f^(')(x)=1-2e^(-x)`
`dy=1-2e^(-x)dx`
`y=x+2e^(-x)+c`
`f(x)=x+2e^(-x)+c`
`f(0)=underset(0)overset(oo)(f)e^(-t)t` dt
`=[(-e^(-t))]_(0)^(oo)+=underset(0)overset(oo)(f)e^(-t)t` dt
`=[0-e^(-t)]_(0)^(oo)`
`=0+1
f(0)=1
`f(0)=1=10+23^(-0)+c`
c=-1
`f(x)=x+2e^(x)-1`
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