Answer:
Option B,C
Explanation:
We have,
f′(x)=e(f(x)−g(x))g′(x)∀x∈R
⇒ f′(x)=ef(x)eg(x)g′(x)
⇒ f′(x)ef(x)=g′(x)eg(x)
⇒ e−f(x)f′(x)=e−g(x)g′(x)
On integrating both sides, we get
e−f(x)=e−g(x)+c
At x=1,
e−f(1)=e−g(1)+c
e−1=e−g(1)+c [ ∵ f(1)= 1] ......(i)
At x=2,
e−f(2)=e−g(2)+c
⇒ e−f(2)=e−1+C [ ∵ g(2)= 1] ..........(ii)
From Eqs. (i) and (ii)
e−f(2)=2e−1−e−g(1) ............(iii)
⇒ e−f(2)>2e−1
We know that , e-x is decreasing
∵ −f(2)<loge2−1
f(2)>1−loge2
⇒ e−g(1)+e−f(2)=2e−1 [from Eq,(iii)]
⇒ e−g(1)<2e−1
−g(1)<loge2−1
⇒ g(1)>−loge2