Exercise 6.2 · Q12 (MCQ): Which of $\cos x$⁠, $\cos 2x$⁠, $\cos 3x$⁠, $\tan x$ decrease on $\left(0,\frac\pi2\right)$⁠?

Basics you need

Solution

1. How each option will be decided

   For an option with $f'(x) \lt 0$ at every $x \in \left(0, \frac{\pi}{2}\right)$⁠: for any $x_1 \lt x_2$ in the interval, Theorem 1 on $[x_1, x_2]$ gives $f(x_1) \gt f(x_2)$⁠, so $f$ is decreasing (indeed strictly).

   For an option that is not decreasing, we exhibit one pair $x_1 \lt x_2$ in the interval with $f(x_1) \lt f(x_2)$⁠.

   "Decreasing on the interval" is a statement about every pair of points, so it needs a general argument (the sign of $f'$⁠), while its failure needs only one pair. All four functions are continuous and differentiable on $\left(0, \frac{\pi}{2}\right)$ ($\tan x$ because $\cos x \ne 0$ there), so Theorem 1 applies on every $[x_1, x_2]$ inside.
   Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

2. (A) $\cos x$⁠: derivative $-\sin x \lt 0$

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

   For $0 \lt x \lt \frac{\pi}{2}$⁠, the angle is in the first quadrant, so $\sin x \gt 0$ and $-\sin x \lt 0$⁠. By Step 1, $\cos x$ is (strictly) decreasing on $\left(0, \frac{\pi}{2}\right)$.

   This is part (a) of Example 9 restricted to a smaller interval: the book shows $\cos x$ is decreasing on all of $(0, \pi)$⁠.
   $\sin x \gt 0$ on $(0, \pi)$⁠. Basics: signs of sin and cos

3. (B) $\cos 2x$⁠: derivative $-2\sin 2x \lt 0$

   Chain rule with inner function $u = 2x$ (so $\frac{du}{dx} = 2$⁠) and outer function $\cos u$⁠:

   $$\frac{d}{dx}\cos 2x = -\sin 2x \cdot 2 = -2\sin 2x.$$

   For $0 \lt x \lt \frac{\pi}{2}$⁠, multiplying by 2 gives $0 \lt 2x \lt \pi$⁠, so $\sin 2x \gt 0$ and $-2\sin 2x \lt 0$⁠. By Step 1, $\cos 2x$ is (strictly) decreasing on $\left(0, \frac{\pi}{2}\right)$.

   We must find the range of the angle $2x$⁠, not of $x$⁠: the sign of $\sin 2x$ depends on where $2x$ is. Here $2x$ stays inside $(0, \pi)$⁠, the region where sine is positive.
   Chain rule (Theorem 4, Section 5.3.1). Basics: the chain rule

4. (C) $\cos 3x$⁠: the derivative changes sign at $\frac{\pi}{3}$

   Chain rule with inner function $u = 3x$⁠:

   $$\frac{d}{dx}\cos 3x = -3\sin 3x.$$

   For $0 \lt x \lt \frac{\pi}{2}$⁠, the angle $3x$ runs over $\left(0, \frac{3\pi}{2}\right)$⁠, which crosses $\pi$⁠:

   • $0 \lt x \lt \frac{\pi}{3}$⁠: $0 \lt 3x \lt \pi$⁠, so $\sin 3x \gt 0$ and the derivative is negative.
   • $\frac{\pi}{3} \lt x \lt \frac{\pi}{2}$⁠: $\pi \lt 3x \lt \frac{3\pi}{2}$⁠, so $\sin 3x \lt 0$ and the derivative is positive.

   So $\cos 3x$ falls until $x = \frac{\pi}{3}$ and then rises. An explicit pair: $x_1 = \frac{\pi}{3}$⁠, $x_2 = \frac{5\pi}{12}$⁠, both in $\left(0, \frac{\pi}{2}\right)$ because $\frac{5\pi}{12} \lt \frac{6\pi}{12}$⁠:

   $$\begin{aligned} \cos\left(3\cdot\tfrac{\pi}{3}\right) &= \cos\pi = -1,\\ \cos\left(3\cdot\tfrac{5\pi}{12}\right) &= \cos\tfrac{5\pi}{4} = -\tfrac{\sqrt2}{2}. \end{aligned}$$

   Since $-1 \lt -\frac{\sqrt2}{2} \approx -0.7071$⁠, we have $f(x_1) \lt f(x_2)$ with $x_1 \lt x_2$⁠. So $\cos 3x$ is not decreasing on $\left(0, \frac{\pi}{2}\right)$.

   Both points were chosen in the rising part $\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$⁠. At $x = \frac{\pi}{3}$⁠, $\cos 3x$ reaches its lowest possible value $-1$⁠, so any later point in the interval gives a larger value.
   $\cos\frac{5\pi}{4} = -\cos\frac{\pi}{4} = -\frac{\sqrt2}{2}$ (third quadrant, reference angle $\frac{\pi}{4}$⁠). Basics: signs of sin and cos

5. (D) $\tan x$⁠: derivative $\sec^2 x \gt 0$

   $$\frac{d}{dx}\tan x = \sec^2 x = \frac{1}{\cos^2 x}.$$

   On $\left(0, \frac{\pi}{2}\right)$⁠, $\cos x \ne 0$⁠, so $\cos^2 x \gt 0$ and $\sec^2 x \gt 0$⁠. So $\tan x$ is strictly increasing there. An explicit pair: $\tan\frac{\pi}{6} = \frac{1}{\sqrt3} \approx 0.5774 \lt 1 = \tan\frac{\pi}{4}$⁠, with $\frac{\pi}{6} \lt \frac{\pi}{4}$⁠. So $\tan x$ is not decreasing on $\left(0, \frac{\pi}{2}\right)$.

   A strictly increasing function on an interval is never decreasing there: any pair $x_1 \lt x_2$ has $f(x_1) \lt f(x_2)$⁠, the opposite of what "decreasing" needs. The explicit pair makes this concrete.
   $\frac{d}{dx}\tan x = \sec^2 x$ (Class XI, Limits and Derivatives).

6. The answer does not depend on the hidden brackets

   If the interval were $\left[0, \frac{\pi}{2}\right]$ instead: $\cos x$ and $\cos 2x$ are continuous on $\left[0, \frac{\pi}{2}\right]$ with negative derivative inside, so Theorem 1 (b) applies directly and they are still decreasing. The pair $\frac{\pi}{3}, \frac{5\pi}{12}$ still rules out $\cos 3x$⁠, and $\tan x$ is not even defined at $\frac{\pi}{2}$⁠. The answer is the same.

   Because the brackets are blank in this reprint's PDF, we check that the choice of bracket cannot change the answer, instead of guessing.

7. Examine every option

   Option	Derivative on $\left(0, \frac{\pi}{2}\right)$	Verdict
   (A)	$-\sin x \lt 0$	Correct. Decreasing (Step 2).
   (B)	$-2\sin 2x \lt 0$⁠, since $0 \lt 2x \lt \pi$	Correct. Decreasing (Step 3).
   (C)	$-3\sin 3x$⁠: negative before $\frac{\pi}{3}$⁠, positive after	Wrong. $\cos\pi = -1 \lt -\frac{\sqrt2}{2} = \cos\frac{5\pi}{4}$⁠, so it rises on $\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$ (Step 4).
   (D)	$\sec^2 x \gt 0$	Wrong. $\tan x$ is increasing: $\tan\frac{\pi}{6} \lt \tan\frac{\pi}{4}$ (Step 5).

   The question says "functions are", so more than one option may be right, and here two are. (C) is the trap: it starts off decreasing, like (A) and (B), but the angle $3x$ passes $\pi$ before $x$ reaches $\frac{\pi}{2}$⁠.

Every fact used, and where it comes from