If the equation x^2+4+3sin⁡ax+b-2x=0 has atleast one real solution, where a, b∈[0, 2π], then one possible value of (a+b) can be equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) (7π)/2. x-1^2+3+3sin⁡ax+b=0 x-1^2+3=-3sin⁡(ax+b) L.H.S ≥3, RHS ∈[ −3,3 ] now sin⁡(ax+b)=-1 ∴x=1 sin⁡b+a=-1

If the equation $x^{2} + 4 + 3 \sin (a x + b) - 2 x = 0$ has atleast one real solution, where $a$ , $b \in [0, 2 π]$ , then one possible value of $(a + b)$ can be equal to