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If tan^(-1)(x^2+3|x|-4)+cot^(-1)(4pi+sin...

If `tan^(-1)(x^2+3|x|-4)+cot^(-1)(4pi+sin^(-1)s in 14)=pi/2, t h e n` the value of `sin^(-1)2x` is `6-2pi` (b) `2pi-6` `pi-3` (d) `3-pi`

A

`6 - 2pi`

B

`2pi - 6`

C

`pi - 3`

D

`3 -pi`

Text Solution

Verified by Experts

The correct Answer is:
A, B
`tan^(-1) (x^(2) + 3 |x| -4) + cot^(-1) (4 pi + sin^(-1) sin 14) = (pi)/(2)`
`rArr tan^(-1) (x^(2) + 3 |x| -4) + cot^(-1) (4pi + 14 - 4pi) = (pi)/(2)`
`rArr x^(2) + 3 |x| -4 = 14`
`rArr x^(2) + 6 |x| -3 |x| -18 = 0`
`rArr x = +- 3`
So, `sin^(-1) sin 2x = sin^(-1) sin 6 - 2pi`
or `sin^(-1) sin (-6) = 2pi - 6`
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