If `a , b in [0,2pi]` and the equation `x^2+4+3sin(a x+b)-2x=0` has at least one solution, then the value of `(a+b)` can be (a)`(7pi)/2` (b) `(5pi)/2` (c) `(9pi)/2` (d) none of these

Options

A. `(7pi)/2`
B. `(5pi)/2`
C. `(9pi)/2`
D. none of these

Explanation

`x^(2)+4-2x+3 sin (ax+b)=0`
`(x-1)^(2)+3=-3 sin (ax+b)`
`L.H.S. ge 3` and `R.H.S. le 3`
`rArr L.H.S. = R.H.S.=3`
`(x-1)^(2)+3+3 sin (ax+b)=0`
`rArr x=1, sin (ax+b)=-1`
`rArr sin (a+b)=-1`
`rArr a+b=(4n-1) pi/2, n in I`
`rArr a+b=(7 pi)/2` (from the given options)