Question

Let $f :R→R$ and $g : R→R$ be two functions defined by $f(x) = log_e(x^2 \+ 1) – e^{–x} \+ 1$ and $g(x)=\frac {1−2e^{2x}}{e^x}$. Then, for which of the following range of α, the inequality $f(g((\frac {α−1)^2}{3}))>f(g(α−\frac 53))$ holds?

• A. (2, 3)
• B. (–2, –1)
• C. (1, 2)
• D. (–1, 1)

Answer 1

Answer: (A)

(2, 3)

Answer 2

$f(x) = log_e (x^2+1) - e^{-x} + 1$
$f‘(x) = \frac {2x}{x^2+1}+e^{−x}$
$f‘(x) = \frac {2}{x+\frac 1x }+ e^{-x}>0$ $∀x∈R$
$g(x) = e^{−x}−2e^x$
$g‘(x) = −e^{−x}−2e^x<0$ $∀x∈R$
⇒ f(x) is increasing and g(x) is decreasing function.
$f(g(\frac {(α−1)^2}{3})) > f(g(α−\frac 53))$
⇒ $\frac {(α−1)^2}{3} < α−\frac 53$
$= α^2 – 5α + 6 < 0$
= $(α – 2)(α – 3) < 0$
= $α ∈ (2, 3)$

So, the correct option is (A): $(2,3)$