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If the equation x^(2)+12+3sin(a+bx)+6x=0...

If the equation `x^(2)+12+3sin(a+bx)+6x=0` has atleast one real solution, where `a, b in [0,2pi]`, then the value of a - 3b is `(n in Z)`

A

`2n pi`

B

`(2n+1)pi`

C

`(4n-1)(pi)/(2)`

D

`(4n+1)(pi)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`x^(2)+12+3 sin (a+bx)+6x =0`
`rArr (x+3)^(2)+3+3sin (a+bx)=0`
`rArr (x+3)^(2)+3=-3 sin (a+ bx)`
`L.H.S. ge 3` but `R.H.S. le 3`
`L.H.S. = R.H.S. = 3`
`therefore x=-3` and `sin (a + bx) =-1`
`rArr sin (a-3b)=-1`
or `a-3b=(4n-1)(pi)/(2), n in Z`
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