Consider the function `f(x) = sin^(-1)x`, having principal value branch `[(pi)/(2), (3pi)/(2)] and g(x) = cos^(-1)x`, having principal value brach `[0, pi]`
For `sin^(-1) x lt (3pi)/(4)`, solution set of x is

Options

A. `((1)/(sqrt2), 1]`
B. `(-(1)/(sqrt2), -1]`
C. `[-(1)/(sqrt2), (1)/(sqrt2)]`
D. `[-1, 1]`

Answer

A

Explanation

We have `f(x) = sin^(-1)x`, having range `[(pi)/(2), (3pi)/(2)]`
Consider the function `h(x) = (pi - sin^(-1) x)`, where
`sin^(-1) x in [-(pi)/(2), (pi)/(2)]`
Clearly, function `f(x) and h(x)` are identical.
Advantage with function `y = h(x)` is that we are well versed with `sin^(-1) x` if `(sin^(-1) x) in [-(pi)/(2), (pi)/(2)]`
`h(sin 10) = pi - sin^(-1) (sin 10)`
`= pi - sin^(-1) (sin (3pi - 10)))`
`= pi - (3pi - 10)`
`= 10 - 2pi`
Now, `f(x) lt (3pi)/(4)`
`:. h(x) lt (3pi)/(4)`
`rArr pi - sin^(-1) x lt (3pi)/(4)`
`rArr sin^(-1) x gt (pi)/(4)`
`rArr (pi)/(4) lt sin^(-1) x le (pi)/(2)`
`rArr (1)/(sqrt2) lt x le 1`