For `x epsilon(0,(5pi)/2)`, definite `f(x)=int_(0)^(x)sqrt(t) sin t dt`. Then `f` has

Options

A. local maximum at ` pi and 2 pi `
B. local manimum at `pi and 2 pi `
C. local minimum at `pi` and maximum at` 2pi `
D. local maximum at `pi` and minimum at `2 pi `

Answer

D

Explanation

`f(x)=underset(0)overset(x)int sqrtt t dt rArr f (x) sqrt(x) sin x `
At points local maximum or minimum , we have
`f(x)=0 rArr sqrt(x)=0 rArr x = pi, 2pi " " [because x in (0,(pi)/(2))]`
Changes in singns of f(x) in the neighourhoods of `pi and 2 pi ` are as shown in fig .30

Clearly, f(x) chages its sing from positve to negative in the neighbourhood of `x = pi` and negative to positve in the neighbourhood of `x = 2 pi` . Thus f(x) has a local maximum at `x=pi` and a local minimum at `x=2pi`