Q.Which of the following functions are decreasing on $\left(0, \frac{\pi}{2}\right)$? (A) $\cos x$ (B) $\cos 2x$ (C) $\cos 3x$ (D) $\tan x$

Concept: Monotonicity of Trigonometric Functions — a function is decreasing on an interval if its derivative is negative throughout that interval.

Step 1: Check each function’s derivative on $\left(0, \frac{\pi}{2}\right)$.

• (A) $\frac{d}{dx}(\cos x) = -\sin x$. On $(0, \pi/2)$, $\sin x > 0$, so $-\sin x < 0$. Hence $\cos x$ is decreasing.
• (B) $\frac{d}{dx}(\cos 2x) = -2\sin 2x$. On $(0, \pi/2)$, $2x \in (0, \pi)$, so $\sin 2x > 0$ for $2x \in (0, \pi)$ (i.e., $x \in (0, \pi/2)$). Thus $-2\sin 2x < 0$. $\cos 2x$ is decreasing. …

We need to check monotonicity — whether each function strictly decreases over the entire open interval $\left(0, \frac{\pi}{2}\right)$. The cleanest way is to examine the sign of the first derivative. If $f'(x) < 0$ for every $x$ in the interval, then $f$ is strictly decreasing there. If $f'(x)$ changes sign, the function is not monotonic (it may increase in parts).

Let’s go function by function.

1. Function (A): $f(x) = \cos x$

   $f'(x) = -\sin x$.

   On $\left(0, \frac{\pi}{2}\right)$, $\sin x > 0$ (positive). Therefore $-\sin x < 0$ everywhere in the interval.

   So $\cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$. This is a classic fact — the cosine curve falls from $1$ to $0$ over this quadrant.

2. Function (B): $f(x) = \cos 2x$

   $f'(x) = -2\sin 2x$.

   For $x \in \left(0, \frac{\pi}{2}\right)$, $2x \in (0, \pi)$.

   $\sin 2x$ is positive on $(0, \pi)$ except at the endpoints where it is zero. So $\sin 2x > 0$ for all $x$ in the open interval.

   Hence $-2\sin 2x < 0$ throughout.

   So $\cos 2x$ is also strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.

3. Function (C): $f(x) = \cos 3x$

   $f'(x) = -3\sin 3x$.

   Here $3x \in (0, \frac{3\pi}{2})$.

   $\sin \theta$ is positive on $(0, \pi)$ and negative on $(\pi, \frac{3\pi}{2})$.

   So there is a point inside the interval where $3x = \pi$, i.e. $x = \frac{\pi}{3}$, at which $\sin 3x = 0$.

   For $x < \frac{\pi}{3}$, $\sin 3x > 0$ so $f'(x) < 0$ (decreasing).

   For $x > \frac{\pi}{3}$, $\sin 3x < 0$ so $f'(x) > 0$ (increasing).

   Since the derivative changes sign, $\cos 3x$ is not monotonic on the whole interval — it decreases then increases. …

Method: Testing Several Candidate Functions for Monotonicity on the Same Interval

This method applies to "which of the following is decreasing/increasing" style questions where several trigonometric functions with different scaled arguments must each be checked against one fixed interval.

Steps

Step 1: Differentiate each candidate separately.

Use the standard derivative rules, including the chain rule for a scaled argument, e.g. $\dfrac{d}{dx}\cos kx = -k\sin kx$.

Step 2: Rewrite the given interval in terms of the actual argument of each derivative.

If the argument is scaled (like $2x$ or $3x$ instead of $x$), convert the interval bounds accordingly — for $x \in (0, \tfrac{\pi}{2})$, the argument $kx$ ranges over $(0, \tfrac{k\pi}{2})$. This is where sign mistakes usually happen, because a scaled argument can sweep through more than one "sign region" even though $x$ itself stays inside a single quadrant. …

Common Mistakes

Mistake 1: Applying the sign pattern of $\sin x$ directly to a scaled argument like $3x$.

Why it's wrong: as $x$ ranges over $(0,\tfrac{\pi}{2})$, the argument $3x$ actually sweeps through $(0,\tfrac{3\pi}{2})$ — well past $\pi$, where $\sin$ changes sign — so the derivative of $\cos 3x$ does not keep one sign throughout, unlike $\cos x$ itself. Correct approach: always convert the interval to the scaled argument's own range before judging the sign.

Mistake 2: Assuming $\tan x$ behaves like $\cos x$ because both "look similar" near $x=0$. …