The least value of the function
`phi(x)=int_(5pi//4)^(x) (3sin t+4 cos t)dt`
on the integral `[5pi//4,4pi//3]`, is

Options

A. `sqrt(3)+(3)/(2)`
B. `-2sqrt(3)+(3)/(2)+(1)/(sqrt(2))`
C. `(3)/(2)+(1)/(sqrt(2))`
D. none of these

Answer

B

Explanation

We have,
`phi(x)=overset(x)underset(5pi//4)int (3 sin t+ 4cos t)dt`
`rArr phi'(x)=(3 sin x+4 os x)`
Clearly, `phi'(x) lt 0` for all `x in [5pi//4,4pi//3]`
`rArr phi(4pi//3)` is decreasing function on `[5pi//4,4pi//3]`
`rArr phi (4pi//3)` is the least value of `phi`(x) on `[5pi//4,4pi//3]`
Now,
`phi(4(pi)/(3))=overset(4pi//3)underset(5pi//4)int (3 sin t+4 cos t)dt`
`rArr phi(4(pi)/(3))=[-3 cos t+4 sin t]_(5pi//4)^(4pi//3)=(3)/(2)-2sqrt(3)+(1)/(sqrt(2))`