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sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x ...

`sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x` , then roots of the equation are- a.`0` b. `1` c.`-1\ ` d. `-2`

A

0

B

1

C

2

D

3

Text Solution

Verified by Experts

The correct Answer is:
D
`"sin"^(-1)(3x)/(5)+"sin"^(-1)(4x)/(5)=sin^(-1)x`
`rArr sin^(-1)((3x)/(5)sqrt(1-(16x^(2))/(25))+(4x)/(5)sqrt(1-(9x^(2))/(25)))=sin^(-1)x`
`rarr (3x)/(5)(sqrt(25-16x^(2)))/(5)+(4x)/(5)(sqrt(25-9x^(2)))/(5)=x`
`rArr x = 0` or `3sqrt(25-16x^(2))+4sqrt(25-9x^(2))=25`
Now `sqrt(225-144x^(2))+sqrt(400-144x^(2))=25` .....(1)
`rArr (175)/(sqrt(400-144x^(2))-sqrt(225-144x^(2)))=25`
`rArr sqrt(400-144x^(2))-sqrt(225-144x^(2))=7` ....(2)
From (1) and (2), `sqrt(400-144x^(2))=16`
`rArr 400-144x^(2)=256`
`rArr x^(2)=1`
`rArr x = pm 1`
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