Let `f(x)=4 cos sqrt(x^(2)-pi^(2)/9)`. Then, the range of f(x) is :

Options

A. `[-1, 1] `
B. `[-4, 4]`
C. `[0, 1 ]`
D. none of these

Answer

B

Explanation

Clearly, f(x) is defined for
`x^(2) - (pi^(2))/(9) ge 0`
`rArr x le -(pi)/(3) or x ge (pi)/(3)`
`rArr x in (-oo, -pi//3]uu[pi//3, oo)`
Let `y = 4 cos sqrtx^(2) - (pi)/(2))`
`rArr cos^(-1) (y)/(4) = sqrt(x^(2) - (pi^(2))/(9))`
`rArr x^(2) - (pi)/(9) =(cos^(-1) .(y)/(4))^(2) rArr x = pm sqrt((pi^(2))/(9) + (cos^(-1).(y)/(4))^(2))`
clearly, x is real all y for which `cos^(-1).(y)/(4)`is meaningful.
Now, `cos^(-1)((y)/(4))` is defined, if
`-1 le (y)/(4) lt 1 rArr - 4 lt y lt 4rArr y in [-4,4]`
Hence, range of `f =[-4,4]`