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← JEE Main 2022 · 25 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 13 of 90

JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 13

Let f:R→Rf:\mathbb R\to\mathbb R and g:R→Rg:\mathbb R\to\mathbb R be two functions defined by f(x)=log⁡e(x2+1)−e−x+1f(x)=\log_e\left(x^2+1\right)-e^{-x}+1 and g(x)=1−2e2xexg(x)=\dfrac{1-2e^{2x}}{e^x}. Then, for which of the following range of α\alpha, the inequality
f(g((α−1)23))>f(g(α−53))f\left(g\left(\frac{(\alpha-1)^2}3\right)\right)>f\left(g\left(\alpha-\frac53\right)\right)
holds ?

Choose one option
Answer and worked solution

Official answer: A

ff is increasing. f′(x)=2xx2+1+e−xf'(x)=\dfrac{2x}{x^2+1}+e^{-x}; for x≥0x\ge0 both terms are ≥0\ge0, and for x<0x<0, e−x>1≥∣2xx2+1∣e^{-x}>1\ge\left|\dfrac{2x}{x^2+1}\right|.

gg is decreasing. g(x)=e−x−2exg(x)=e^{-x}-2e^x.

So f∘gf\circ g is decreasing, and the inequality means (α−1)23<α−53\dfrac{(\alpha-1)^2}3<\alpha-\dfrac53, i.e. α2−5α+6<0\alpha^2-5\alpha+6<0: α∈(2,3)\alpha\in(2,3).

(A)