VITEEE 2015 Paper

Let f (x) = log (1 + x) - $\frac{x}{1+x}$ ∴ f ' (x) = $\frac{1}{1+x}$ - $\frac{(1+x).1 - x.1}{(1+x)^2}$ = $\frac{1}{1+x}$ - $\frac{1}{(1+x)^2}$ = $\frac{x}{(1+x)^2}$ which is positive. [Since x > 0] ∴ f(x) is monotonic increasing, when x > 0. ⇒ f (x) > f (0) Now, f(0) = log 1 - 0 = 0 ∴ f (x) > 0 ⇒ log (1 + x) - $\frac{x}{x+1}$ > 0 ⇒ $\frac{x}{x+1}$ < log (1 + x) ... (i) Also, for x > 0, $x^2$ > 0 ⇒ $x^2$ x > x ⇒ x (x + 1) > x ⇒ x > $\frac{x}{x+1}$ ... (ii) From eqs. (i) and (ii), we get $\frac{x}{x+1}$ < log (1 + x) < x [Since log (1 + x) < x for x > 0]