The trigonometric equation `sin^(-1)x=2sin^(-1)a` has a solution for all real values (b) `|a|<1/a` `|a|lt=1/(sqrt(2))` (d) `1/2<|a|<1/(sqrt(2))`

Options

A. `|a| le (1)/sqrt(2)`
B. `1/2 lt |a| lt (1)/sqrt(2)`
C. all real vaues of a
D. `|a| lt 1/2`

Explanation

Let `alpha` a solution of `sin^(-1)x=2 sin^(-1) alpha` Then `sin^(-1) alpha =2 sin^(-1) alpha`
`rarr -(pi)/(2) le 2 sin^(-1) alpha le (pi)/(2)`
` rarr -(pi)/(4) le sin^(-1) a le (pi)/(4)`
`rarr sin((pi)/(4)) le alpha le sin(pi)/(4) rarr -(1)/sqrt(2) le alpha le (1)/sqrt(2) rarr |alpha| le (1)/sqrt(2)`