Definite Integration

∵‌‌f(x)=

x

∫

−x

(|t|−t2)e−t2dt,x>0
∴‌‌f(x)=2‌

x

∫

0

(t−t2)e−t2dt
f′(x)=2(x−x2)e−x2.....(i)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.