Q.Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 2x + \cos x$, then $f$:
(A) has a minimum at $x = \pi$
(B) has a maximum at $x = 0$
(C) is a decreasing function
(D) is an increasing function

The key idea is monotonicity of trigonometric functions: check the sign of $f'(x)$ over $\mathbb{R}$.

1. Differentiate: $f'(x) = 2 - \sin x$.
2. Since $-1 \leq \sin x \leq 1$, we have $2 - \sin x \geq 2 - 1 = 1 > 0$ for all $x \in \mathbb{R}$. …

The key to this problem lies in monotonicity — whether a function is increasing or decreasing. For a differentiable function, the sign of the derivative tells us everything: if $f'(x) > 0$ for all $x$, the function is strictly increasing; if $f'(x) < 0$ for all $x$, it is strictly decreasing. Here, the presence of $\cos x$ might tempt you to think about oscillations, but the linear term $2x$ dominates.

Let’s work through it.

1. Find the derivative. Differentiate term by term:

2. Analyse the range of $f'(x)$. We know that $\sin x$ oscillates between $-1$ and $1$ for all real $x$. So the smallest possible value of $2 - \sin x$ occurs when $\sin x$ is largest, i.e. $\sin x = 1$:

The largest possible value occurs when $\sin x = -1$:

Hence $f'(x) \in [1, 3]$ for all $x \in \mathbb{R}$.

3. Interpret the sign. Since $1 > 0$, we have $f'(x) > 0$ for every real $x$. A function whose derivative is positive everywhere is strictly increasing on its entire domain. …

Method: Proving Global Monotonicity Using Bounds on sin x / cos x

This technique applies whenever a function combines a steadily growing (or shrinking) term with a bounded oscillating trig term, and you must decide if it is increasing/decreasing everywhere — without being given a specific point.

Steps

Step 1: Differentiate the whole function.

Differentiate term by term using the standard rules, including $\frac{d}{dx}\sin x=\cos x$ and $\frac{d}{dx}\cos x=-\sin x$.

Step 2: Isolate the bounded trig piece and apply its known range.

Every occurrence of $\sin x$ or $\cos x$ in the derivative satisfies

Substitute the extreme values to find the smallest and largest values $f'(x)$ can possibly take.

Step 3: Compare that worst-case bound to zero. …

Common Mistakes

Mistake 1: Assuming the oscillating trig term creates turning points.

Why it's wrong: seeing $\cos x$ inside a function tempts students to expect rises and falls like the trig function itself. Correct approach: what matters is the sign of the whole derivative, not the trig term alone — if the non-trig part of the derivative dominates the bounded trig part everywhere, the wiggle never actually flips the overall sign.

Mistake 2: Using a non-strict bound and losing track of whether it's actually strict.

Why it's wrong: some students loosely write $f'(x) \ge 0$ instead of working out the exact minimum, and miss that the bound is strictly greater than zero. Correct approach: work out the exact minimum of the derivative (here $2-\sin x \ge 2-1=1$) and confirm it is strictly greater than zero before declaring strict monotonicity. …