The set of critical pionts of the fuction f(x) given by
`f(x)= x - log _(e)x+int_(2)^(x)((1)/(t)-2-2cos 4t)dt is `

Options

A. `{(pi)/(6)+(npi)/(2):n =0,1,2….}`
B. `{n pi : n in N }`
C. `{n pi+(pi)/(6):n in N} cup {(pi)/(2)}`
D. none of these

Answer

A

Explanation

We have `f(x)=x- log_ex+underset(2)overset(x)(1/t-2-2 cos 4t)dt`
`rArr f'(x) =1 -1/x+(1/x-2-2 cos 4 x)`
`rArr f(x)=-1-2 cos 4x `
Now
f(x)=0
`rArr cos4 x =-1/2=cos(2pi)/(3)`
`rArr 4x=2n pi pm (2pi)/(3), n in Z rArr x=(n pi)/(2)pm(pi)/(6),n in Z`
But , f(x) is defined for `x gt 0` Therefore
`f(x)=0rArr x=(npi)/(2)+(pi)/(6), n in N, x=(pi)/(6)`
Hence the set of critical points of f(x) is
`{(n pi)/(2)+pi/6: n= 0,1,2...}`