The maximum value of f(x)`=int_(0)^(1) t sin (x+pi t)dt` is

Options

A. `(1)/(pi)sqrt(pi^(2)+4)`
B. `(1)/(pi^(2))sqrt(pi^(2)+4)`
C. `sqrt(pi^(2)+4)`
D. `(1)/(2pi^(2))sqrt(pi^(2)+4)`

Answer

B

Explanation

We have,
`f(x)=underset(0)overset(1)int t sin(xunderset(II)+pi t)dt`
`rArr f(x)=[-(t)/(pi)cos(x+pit)]_(0)^(1)+(1)/(pi)underset(0)overset(1)int cos (x+pit)dt`
`rArr f(x)=(1)/(pi)cos x+(1)/(pi^(2))[sin(x+pit)]_(0)^(1)dt`
`rArr f(x)=(1)/(pi)cos x-(2)/(pi^(2))sin x`
We know that
`-sqrt(a^(2)+b^(2))le a cos t+b sin t le sqrt(a^(2)+b^(2))` for all t
`:. -sqrt((1)/(pi^(2))+(4)/(pi^(4))) le (1)/(pi)cos x-(2)/(pi^(2))sin x le sqrt((1)/(pi^(2))+(4)/(pi^(4)))` for all x
`rArr -(sqrt(pi^(2)+4)/(pi^(2))) le f(x) le (sqrt(pi^(2)+4)/(pi^(2)))` for all x
`rArr f_("max")(x)=(1)/(pi^(2))sqrt(pi^(2)+4)`