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

Step-by-step solution

x - 1 2 + 3 + 3 sin ⁡ a x + b = 0 x - 1 2 + 3 = - 3 sin ⁡ ( a x + b ) L . H . S ≥ 3 , R H S ∈ [ − 3 , 3 ] now s i n ⁡ ( a x + b ) = - 1 ∴ x = 1 sin ⁡ b + a = - 1