Question

Can someone answer the 4 th question​

Answer 1

Answer:

let f(x)=(2√3+3)sin x+2√3cos x

-√(2√3+3) square+(2√3)square≤f(x)≤√(2√3+3) square+(2√3) square

-√12+9+12√3+12≤f(x)≤√12+9+12√3+12

-√33+12√3≤f(x)≤√33+12√3

therefore,instead of -(2√3+√15) and (2√3+√15),it should be-√33+12√3 and √33+12√3