The given function is f(x)=cosx+e−x Let α and β be any two roots of the equation excosx+1=0 such that α<β, then, cosα+e−α=0 and cosβ+e−β=0 This implies, f(x) is continuous at [α,β] and differentiable on (α,β) also f(α)−f(β) By Rolle's theorem there exists C∈(α,β) such that f′(c)=0 . This implies, −sinc−e−c=0 sinc+1=0 And, x=c is a root of exsinx+1=0