The curve x+y−loge⁡(x+y)=2x+5 has a vertical tangent at the point (α,β) . Then α+β=

x+y−loge⁡(x+y)=2x+5Differentiate both sides with respect to x⇒1+y'−1x+y⋅(1+y')=2⇒(1+y')(x+y)−1+y'=2x+2y⇒x+y+y'(x+y−1)−1=2x+2y y′=x+y+1x+y−1 since slope of vertical tangent is undefined, denominator=0x+y-1=0 ⇒α+β-1=0⇒α+β=1