If `cos^(-1)x gt sin^(-1) x`, then

Options

A. `1/(sqrt(2)) lt x le 1`
B. `0 le x le 1/(sqrt(2))`
C. `-1 le x lt 1/(sqrt(2))`
D. `x gt 0`

Answer

C

Explanation

We have, ` cos^(1)x gt sin^(-1)x`, where `x in [-1,1]`
`rArr x lt cos(sin^(-1)x)`
`rArr x lt cos[cos^(-1)sqrt(1-x^(2))] , ["let" sin^(-1)x = theta rArr sin theta = x/1]` v
`[:' cos theta = sqrt(1-sin^(2)theta) = sqrt(1-x^(2)) rArr cos^(-1)sqrt(1-x^(2))]`
`rArr x lt sqrt(1-x^(2))`
`rArr x^(2) lt 1 - x^(2) rArr 2x^(2) lt 1`
`rArr x^(2) lt 1/2 rArr x +-(1/(sqrt(2)))"....."(i)`
Also, `-1 le x le 1"....."(ii)`
`:. -1 le x le 1/(sqrt(2))`

Alternate Method
`pi/2 - sin^(-1)x gt sin^(-1)x`
`pi/2 gt 2 sin^(-1) x rArr pi/4 gt sin^(-1)x`
`1/(sqrt(2)) gt x rArr (1)/(sqrt(2)) lt x le 1`
We know that, `sin^(-1)x in[(-pi)/(2),(pi)/(2)]`