Question

Which one of the following functions is monotonically increasing in its domain?

• A. \( f(x) = \log(1+x) - x + \frac{x^2}{2} \)
• B. \( g(x) = 2 \tan^{-1}x - x - 1 \)
• C. \( h(x) = 4\cos x + x \)
• D. \( u(x) = \log(1+x) - \frac{x}{x+1} \)

Hint:

A function \(f(x)\) is monotonically increasing in an interval if \(f'(x) \ge 0\) for all \(x\) in that interval. 1. Determine the domain of the function. 2. Calculate the first derivative \(f'(x)\). 3. Analyze the sign of \(f'(x)\) in the domain. Remember derivative rules: \( \frac{d}{dx}\log u = \frac{1}{u}\frac{du}{dx} \), \( \frac{d}{dx}\tan^{-1}u = \frac{1}{1+u^2}\frac{du}{dx} \), \( \frac{d}{dx}\cos u = -\sin u \frac{du}{dx} \).

Answer 1

The Correct Option is A

A function is monotonically increasing if its derivative is non-negative (\( f'(x) \ge 0 \)) in its domain.
The domain for \( \log(1+x) \) is \( 1+x>0 \implies x>-1 \).
% Option (1) \( f(x) = \log(1+x) - x + \frac{x^2}{2} \).
Domain: \( x>-1 \).
\( f'(x) = \frac{1}{1+x} - 1 + \frac{2x}{2} = \frac{1}{1+x} - 1 + x \) \( f'(x) = \frac{1 - (1+x) + x(1+x)}{1+x} = \frac{1-1-x+x+x^2}{1+x} = \frac{x^2}{1+x} \).
For \( x>-1 \), \( 1+x>0 \).
Also \( x^2 \ge 0 \).
So, \( f'(x) = \frac{x^2}{1+x} \ge 0 \) for all \( x>-1 \).
Thus, \(f(x)\) is monotonically increasing in its domain.
% Option (2) \( g(x) = 2 \tan^{-1}x - x - 1 \).
Domain: \( x \in \mathbb{R} \).
\( g'(x) = 2 \cdot \frac{1}{1+x^2} - 1 = \frac{2 - (1+x^2)}{1+x^2} = \frac{1-x^2}{1+x^2} \).
\( g'(x) \ge 0 \implies 1-x^2 \ge 0 \implies x^2 \le 1 \implies -1 \le x \le 1 \).
\(g(x)\) is not increasing for all \(x\) in its domain (e.
g.
, if \(x=2\), \(g'(2) = (1-4)/(1+4) = -3/5<0 \)).
% Option (3) \( h(x) = 4\cos x + x \).
Domain: \( x \in \mathbb{R} \).
\( h'(x) = -4\sin x + 1 \).
If \( h'(x) \ge 0 \), then \( 1 \ge 4\sin x \implies \sin x \le 1/4 \).
This is not true for all \(x\) (e.
g.
, if \(x=\pi/2\), \( \sin x = 1 \), which is not \( \le 1/4 \)).
So \(h(x)\) is not always increasing.
% Option (4) \( u(x) = \log(1+x) - \frac{x}{x+1} \).
Domain: \( x>-1 \).
\( u'(x) = \frac{1}{1+x} - \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{1+x} - \frac{x+1-x}{(x+1)^2} = \frac{1}{1+x} - \frac{1}{(x+1)^2} \) \( u'(x) = \frac{(1+x)-1}{(1+x)^2} = \frac{x}{(1+x)^2} \).
For \( u'(x) \ge 0 \), we need \( x \ge 0 \) (since \( (1+x)^2>0 \) for \( x>-1 \)).
\(u(x)\) is increasing for \(x \ge 0\), but decreasing for \( -1<x<0 \).
So not monotonically increasing in its entire domain.
Therefore, only \( f(x) = \log(1+x) - x + \frac{x^2}{2} \) is monotonically increasing in its domain.
This matches option (1).