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

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

• $\sin 2x$: derivative $2\cos 2x$. On $(0,\pi/2)$, $2x \in (0,\pi)$, so $\cos 2x$ changes sign — not always negative.
• $\tan x$: derivative $\sec^2 x > 0$ always — increasing.
• $\cos x$: derivative $-\sin x$. On $(0,\pi/2)$, $\sin x > 0$, so $-\sin x < 0$ — decreasing. …

Concept and Intuition

To decide whether a function is decreasing on an interval, we look at its slope — the derivative. If the derivative is negative for every point in the interval, the function is strictly decreasing there. This is a direct application of the monotonicity of trigonometric functions.

For $\left(0, \frac{\pi}{2}\right)$, we know:

• $\sin x$ increases from $0$ to $1$.
• $\cos x$ decreases from $1$ to $0$.
• $\tan x$ increases from $0$ to $+\infty$.

But the options include composite functions like $\sin 2x$ and $\cos 3x$, so we must differentiate each and check the sign of the derivative on the given interval.

Step-by-step solution

1. Option (A): $\sin 2x$

Derivative: $\frac{d}{dx} \sin 2x = 2 \cos 2x$.

On $\left(0, \frac{\pi}{2}\right)$, the argument $2x$ runs from $0$ to $\pi$.

• For $2x \in (0, \frac{\pi}{2})$, $\cos 2x > 0$ → derivative positive.
• For $2x \in (\frac{\pi}{2}, \pi)$, $\cos 2x < 0$ → derivative negative. So the derivative changes sign — the function is not decreasing on the whole interval.

2. Option (B): $\tan x$

Derivative: $\frac{d}{dx} \tan x = \sec^2 x$.

On $\left(0, \frac{\pi}{2}\right)$, $\sec^2 x > 0$ always (since $\cos x > 0$).

Derivative is positive everywhere → $\tan x$ is increasing, not decreasing.

3. Option (C): $\cos x$

Derivative: $\frac{d}{dx} \cos x = -\sin x$.

On $\left(0, \frac{\pi}{2}\right)$, $\sin x > 0$, so $-\sin x < 0$.

Derivative is negative throughout → $\cos x$ is strictly decreasing on this interval.

4. Option (D): $\cos 3x$ …

Method: Comparing Monotonicity of Several Functions on a Given Interval

Use this when a question gives several different trig functions (possibly with different arguments like $2x$ or $3x$) and asks which one is increasing/decreasing on a common interval.

Steps

Step 1: Differentiate every option using the chain rule where needed.

For an argument like $kx$, remember $\frac{d}{dx}\sin(kx)=k\cos(kx)$ and $\frac{d}{dx}\cos(kx)=-k\sin(kx)$ — the constant $k$ never changes the sign, only the magnitude.

Step 2: Rescale the given interval for each option's argument.

If the argument is $kx$, multiply the endpoints of the given $x$-interval by $k$ to find the corresponding range of the argument — this is essential, since the sign of $\sin$ or $\cos$ depends on the argument, not on $x$ directly.

Step 3: Check whether the derivative keeps one sign across the whole rescaled range. …

Common Mistakes

Mistake 1: Assuming a function like $\cos 3x$ behaves the same as $\cos x$ over the same $x$-interval.

Why it's wrong: the coefficient inside the argument compresses the graph horizontally, so $\cos 3x$ completes three times as many oscillations as $\cos x$ over the same range of $x$ — its sign can change well within an interval where $\cos x$ itself would not. Correct approach: always rescale the interval to the actual argument ($3x$, not $x$) before judging the sign.

Mistake 2: Checking the derivative's sign at only one point instead of across the whole interval. …