Hence option (d) is wrong. g′(x)=xe−x22x......(ii) Add (i) and (ii), f′(x)+g′(x)=2xe−x2 f(x)+g(x)=−e−x2+c ∵f(0)=g(0)=0 f(x)+g(x)=−e−x2+1 f(√ℓn3)+g(√ℓn3)=1−
1
3
=
2
3
So, option (a) is wrong. H(x)=ψ1(x)−1−αx=e−x+−1−αx x≥1α∈(1,x) H(1)=e−1+1−1−α<0 H′(x)=−e−x+1−α<0⇒H(x) is decreasing So option (b) is wrong. (c) ψ2(x)=2(ψ1(β)−1) Applying L.M.V.T to ψ2(x) in [0,x] ψ2′(β)=
ψ2(x)−ψ2(0)
x
2β−2+2e−β=
ψ2(x)−0
x
⇒ψ2(x)=2x(ψ1(β−1)) has one solution So, option (c) is correct.