f(x)=loge(x2+1)−e−x+1⇒f′(x)=x2+12x+e−x>0,∀x∈R
So f(x) increasing
g(x)=e−x−2ex⇒g′(x)=−(e−x+2ex)<0∀x∈R
So g(x) decreasing
⇒f(g(x)) is decreasing
⇒f(g(3(α−1)2))>f(g(α−35))
⇒3(α−1)2<α−35
⇒(α−2)(α−3)<0⇒α∈(2,3)
JEE Main 2022 — Mathematics Calculus
Let f:R→R and g:R→R be two functions defined by f(x)=loge(x2+1)−e−x+1 and g(x)=ex1−2e2x⋅ Then, for which of the following range of α, the inequality f(g(3(α−1)2))>f(g(α−35)) holds?
Held on 25 Jun 2022 · Verified 6 Jul 2026.
(−2,−1)
(2,3)
(1,2)
(−1,1)
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