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Let f(x)=int(0)^(x)(e^(t))/(t)dt(xgt0), ...

Let `f(x)=int_(0)^(x)(e^(t))/(t)dt(xgt0),`
then `e^(-a)[f(x+1)-f(1+a)]=`

A

a) `int_(0)^(x)=(e^(t))/((t+a))dt`

B

b) `int_(1)^(x)(e^(t))/(t+a)dt`

C

c) `e^(-a)int_(1+a)^(x+a)(e^(t))/(t)dt`

D

d) `int_(0)^(x)(e^(t-a))/((t+a))dt`

Text Solution

Verified by Experts

The correct Answer is:
B, C
`e^(-a)[f(x+a)-f(1+a)]`
`" "=e^(-a)[int_(0)^(x+a)(e^(t).dt)/(t)-int_(0)^(1+a)(e^(t))/(t)dt]`
`" "=e^(-a)[int_(0)^(x+a)(e^(t).dt)/(t)+int_(1+a)^(0)(e^(t))/(t)dt]`
`" "=e^(-a)[int_(1+a)^(x+a)(d^t.dt)/(t)]`
`" "=e^(-a)int_(1)^(x)(e^(y+a))/(y+a).dy" "("Put, t"=y+a, dt=dy)`
`" "=int_(1)^(x)(e^(y).dy)/(t+a)`
`" "=int_(1)^(x)(e^(t).dt)/(t+a)`
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