Question

Let ƒ(x) = (log (1 + x))^–1 – x^–1, x > 0, then

• A. 1 < ƒ(x) < 2
• B. –1 < ƒ(x) < 0
• C. 0 < ƒ(x) < 1
• D. None of these

Answer 1

Answer: (C)

0 < ƒ(x) < 1

Answer 2

Consider g(x) = log (1 + x) on the interval [0, x].

Since g(x) is differentiable on (0, x), therefore by Lagrange’s mean value theorem, there exists c ∈ (0, x) such that

$\frac{\log(1 + x) - \log(1 + 0)}{x - 0}$ = g ′ (3) = $\frac{1}{1 + c}$

⇒  $\frac{\log(1 + x)}{x}$ = $\frac{1}{1 + c}$ < 1 … (1)

⇒ log (1 + x) < x ⇒ ƒ(x) > 0

Also, c ∈ (0, x) ⇒ c < x ⇒ $\frac{1}{1 + c}$ > $\frac{1}{1 + x}$

⇒ $\frac{\log(1 + x)}{x}$ > $\frac{1}{1 + x}$ [By (1)]

⇒ log (1 + x) > $\frac{x}{1 + x}$

⇒ $\frac{1}{\log(1 + x)}$ < $\frac{1 + x}{x}$

⇒ $\frac{1}{\log(1 + x)}$< 1 + $\frac{1}{x}$ ⇒ $\frac{1}{\log(1 + x)}$ – $\frac{1}{x}$ < 1

⇒ ƒ(x) < 1.