For the functions `f(x)= int_(0)^(x) (sin t)/t dt ` where ` x gt 0`. At `x=n pi ` f(x) attains

Options

A. maximum or minimum according as n is odd or even respectively .
B. minimum or maximum according as n is odd or even respectively
C. maximum at x= n `pi`
D. minimum at x= n `pi`

Answer

A

Explanation

We have `f(x)=oversetxunderset0int(sin)/tdt`
`rArrf(x)=(sinx)/x and f''(x)=(xcosx-sinx)/x^2`
For maximum/minimum, we must have
f'(x)=0
`rArr(sinx)=0 rArr sinx=0 rArr x=npi,n in N " " [thereforexgt0]`
At x=n`pi`, we have
`f''(x)=(npicosnpi-sinnpi)/(n^2pi^2)=(-1)^n/(npi)`
`f''(x)lt0` , if n is an odd natural number
and , `f''(x)lt0` if n is even natural number
Hence f(x) attains local maximum or minimum at `x=npi` according as n is an odd natural number or even natural number