If `f(x)=sqrt(4-x^2)+sqrt(x^2-1)` , then the maximum value of `(f(x))^2` is ____________

Answer

6

Explanation

Let `x^(2)=4cos^(2) theta+sin^(2) theta.` Then
`(4-x^(2))=3 sin^(2) theta and (x^(2)-1)=3 cos^(2)theta`
` :. f(x)=sqrt(3)|sin theta|+sqrt(3)|cos theta|`
`or y_(min)=sqrt(3) and y_(max)=sqrt(3)((1)/(sqrt(2))+(1)/(sqrt(2)))=sqrt(6)`
Hence, range of `f(x) " is " [sqrt(3),sqrt(6)].`
Hence, maximum value of `(f(x))^(2) ` is 6.