Exercise 6.2 · Q13 (MCQ): On which interval is $x^{100}+\sin x-1$ decreasing?

Basics you need

Solution

1. Differentiate $f$

   Power rule for $x^{100}$⁠, $\frac{d}{dx}\sin x = \cos x$⁠, and the constant $-1$ has derivative 0:

   $$f'(x) = 100x^{99} + \cos x.$$

   $f$ is continuous and differentiable at every real number, so Theorem 1 can be used on any closed interval.

   Each option is an interval, and the sign of $f'$ on it decides whether $f$ can be decreasing there.
   $\frac{d}{dx}x^n = nx^{n-1}$⁠, $\frac{d}{dx}\sin x = \cos x$ (Class XI, Limits and Derivatives).

2. Option (A): $f' \gt 0$ on $(0, 1)$

   Let $0 \lt x \lt 1$⁠. Then $x^{99} \gt 0$⁠, so $100x^{99} \gt 0$⁠. Also $0 \lt x \lt 1 \lt \frac{\pi}{2}$⁠, so $x$ is a first-quadrant angle and $\cos x \gt 0$⁠. A sum of two positive numbers is positive:

   $$f'(x) = \underbrace{100x^{99}}_{\gt 0} + \underbrace{\cos x}_{\gt 0} \gt 0.$$
   On $(0, 1)$ the power term is tiny ($0.5^{99}$ is about $10^{-30}$⁠), so it is $\cos x$ that keeps $f'$ positive here. That is why we need $1 \lt \frac{\pi}{2}$⁠.
   $\cos x \gt 0$ for $0 \lt x \lt \frac{\pi}{2}$⁠. Basics: signs of sin and cos

3. Option (C): $f' \gt 0$ on $\left(0, \frac{\pi}{2}\right)$

   Let $0 \lt x \lt \frac{\pi}{2}$⁠. Again $100x^{99} \gt 0$ (as $x \gt 0$⁠) and $\cos x \gt 0$ (first quadrant), so $f'(x) \gt 0$⁠.

   The argument of Step 2 never used $x \lt 1$ except to place $x$ in the first quadrant, so it works on the larger interval $\left(0, \frac{\pi}{2}\right)$ too.
   Both terms are positive, so their sum is positive. Basics: sign of a product

4. Option (B): $f' \gt 0$ on $\left(\frac{\pi}{2}, \pi\right)$

   Here $\cos x \lt 0$⁠, so we cannot just add two positives; we compare sizes. Let $\frac{\pi}{2} \lt x \lt \pi$⁠. Since $x \gt \frac{\pi}{2} \gt 1$⁠, we have $x^{99} \gt 1$⁠, so $100x^{99} \gt 100$⁠. And $\cos x \ge -1$⁠. Hence

   $$\begin{aligned} f'(x) &= 100x^{99} + \cos x\\ &\gt 100 + (-1) = 99 \gt 0. \end{aligned}$$
   The negative term $\cos x$ can never be more negative than $-1$⁠, while the power term exceeds 100. The bound $\cos x \ge -1$ is all we need; we do not have to know the exact value of $\cos x$⁠.
   $-1 \le \cos x \le 1$ for every real $x$⁠. Basics: signs of sin and cos

5. So $f$ is increasing on each interval, not decreasing

   By Steps 2–4 and Theorem 1, $f$ is strictly increasing on each of $(0, 1)$⁠, $\left(\frac{\pi}{2}, \pi\right)$ and $\left(0, \frac{\pi}{2}\right)$⁠: every pair $x_1 \lt x_2$ in one of them has $f(x_1) \lt f(x_2)$⁠, the opposite of what "decreasing" needs. Explicit pairs (values rounded):

   • (A): $f(0.2) \approx -0.801331$ and $f(0.8) \approx -0.282644$⁠.
   • (B): $f(2) \approx 1.2677\times10^{30}$ and $f(3) \approx 5.1538\times10^{47}$⁠.
   • (C): $f(0.5) \approx -0.520574$ and $f(1.5) \approx 4.0656\times10^{17}$⁠.

   In each case the later point has the larger value, so $f$ is not decreasing on any of the three intervals.

   In fact Steps 2 and 4 together show $f'(x) \gt 0$ for every $x \gt 0$ (for $0 \lt x \lt 1$ use Step 2; for $x \ge 1$⁠, $100x^{99} \ge 100 \gt 1 \ge -\cos x$⁠). All three options lie inside $(0, \infty)$⁠, so they all fail together. "None of these" does not mean $f$ never decreases: for $x \le -1$⁠, $f'(x) \le -100 + 1 \lt 0$⁠, so $f$ is decreasing on $(-\infty, -1]$⁠, an interval not offered.
   Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

6. The answer does not depend on the hidden brackets

   With square brackets the intervals gain end points. At those end points $f'(0) = 0 + 1 = 1$⁠, $f'(1) = 100 + \cos 1 \gt 0$⁠, $f'\left(\frac{\pi}{2}\right) = 100\left(\frac{\pi}{2}\right)^{99} \gt 0$ and $f'(\pi) = 100\pi^{99} - 1 \gt 0$⁠. So $f' \gt 0$ on the closed intervals too, $f$ is still increasing on each, and the answer is unchanged.

   The brackets are blank in this reprint's PDF, so we make sure the reading cannot affect the answer instead of guessing.

7. Examine every option

   Option	Sign of $f'$	Verdict
   (A)	$100x^{99} \gt 0$⁠, $\cos x \gt 0$⁠: $f' \gt 0$	Wrong. $f$ is increasing on $(0, 1)$⁠: $f(0.2) \lt f(0.8)$⁠.
   (B)	$f' \gt 100 - 1 = 99 \gt 0$	Wrong. $f$ is increasing on $\left(\frac{\pi}{2}, \pi\right)$⁠: $f(2) \lt f(3)$⁠.
   (C)	$100x^{99} \gt 0$⁠, $\cos x \gt 0$⁠: $f' \gt 0$	Wrong. $f$ is increasing on $\left(0, \frac{\pi}{2}\right)$⁠: $f(0.5) \lt f(1.5)$⁠.
   (D)	(none of the intervals works)	Correct.

   Option (B) is the tempting one: $\sin x$ is decreasing on $\left(\frac{\pi}{2}, \pi\right)$⁠, and a student who forgets the $x^{100}$ term would choose it. But $x^{100}$ grows enormously there and swamps the small drop in $\sin x$⁠.

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