Find the range of `f(x) = (sin^(-1) x)^(2) + 2pi cos^(-1) x + pi^(2)`

Options

A. `cos.(pi^(2))/(8)`
B. `sin.(pi^(2))/(4)`
C. `cos.(pi^(2))/(2)`
D. none of these

Answer

D

Explanation

Let `cos^(-1) x = a rArr a in [0, pi]`
and `sin^(-1) y = b rArr b in [-pi//2, pi//2]`
We have `a + b^(2) = (p pi^(2))/(4)`...(i)
and `ab^(2) = (pi^(4))/(16)`..(ii)
since `b^(2) in [0, pi^(2)//4]`, we get `a + b^(2) in [0, pi + pi^(2)//4]`
So, from Eq. (i) we get `0 le (p pi^(2))/(4) le pi + (pi^(2))/(4)`
i.e., `0 le p le (4)/(pi) + 1`
Since `p in Z, " so " p = 0, 1 " or " 2`
Substituting the value of `b^(2)` from Eq. (i) Eq. (ii), we get
`a((p pi^(2))/(4) -a) = (pi^(4))/(16)`
`rArr 16a^(2) - 4p pi^(2) a + pi^(4) = 0`...(iii)
Since `a in R, " we have " D ge 0`
i.e., `16^(2) ge 4 - 64 pi^(4) ge 0`
or `p^(2) ge 4 " or " p ge 2 " or " p =2`
Substituting `p =2` in Eq. (iii), we get
`16a^(2) -8pi^(2) a + pi^(4) = 0`
or `(4a - pi^(2))^(2) = 0`
or `a = (pi^(2))/(4) = cos^(-1) x " or " x = cos.(pi^(2))/(4)`
From Eq. (ii), we get `(pi^(2))/(4) b^(2) = (pi^(4))/(16)`
or `b = +- (pi)/(2) = sin^(-1) y " or " y = +- 1`