The value of `f(x)=3cos(sqrt ((pi^2)/(16)-x^2))` lie in the interval____

Answer

`[0, (3)/(sqrt(2))]`

Explanation

Given, `f(x)=3sin sqrt((pi^(2))/(16)-x^(2))`
` rArr" Domain " in[-(pi)/(4),(pi)/(4)]`
` therefore " For range,"f'(x)=3cos(sqrt((pi^(2))/(16)-x^(2)))*(1(-2x))/(2sqrt((pi^(2))/(16)-x^(2)))=0`
Where, `cos(sqrt((pi^(2))/(16)-x^(2)))=0 or x=0`
`[("neglecting "cos(sqrt((pi^(2))/(16)-x^(2)))=0 rArr (pi^(2))/(16)-x^(2)=(pi^(2))/(4)),(rArr x^(2)= -(3pi^(2))/(16)", never possible")]`
`implies x=0`
Thus, `f(0)=3"sin" (pi)/(4) = (3)/(sqrt(2))`
`and f(-(pi)/(4))=f((pi)/(4))=0`
Hence, range ` in [0,(3)/(sqrt(2))]`