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

The Correct Option is A

\(f(x) = log_e (x^2+1) - e^{-x} + 1\)
\(f‘(x) = \frac {2x}{x^2+1}+e^{−x}\)
\(f‘(x) = \frac {2}{x+\frac 1x }+ e^{-x}>0\) \(∀x∈R\)
\(g(x) = e^{−x}−2e^x\)
\(g‘(x) = −e^{−x}−2e^x<0\)  \(∀x∈R\)
⇒ 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): \((2,3)\)