If the equation x2+5+4cos⁡(αx+β)=2x has at least one solution where α,β∈[2,5] then the value of α+β equal to

x2+5+4cos⁡(αx+β)=2x⇒(x−1)2+4=−4cos⁡(αx+β)Here (x−1)2+4≥4,−4cos⁡(αx+β)≤4But, the given equation has at least one solution, then (x−1)2+4=4=−4cos⁡(αx+β)∴ x=1⇒cos⁡(α+β)=−1⇒α+β=3π=9.42(∵α+β∈[4,10])