Q.On which of the following intervals is the function $f$ given by $f(x) = x^{100} + \sin x - 1$ decreasing ? (A) $(0, 1)$ (B) $\left(\frac{\pi}{2}, \pi\right)$ (C) $\left(0, \frac{\pi}{2}\right)$ (D) None of these

The key idea is that a function is decreasing where its derivative is negative.

Step 1: Differentiate.

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

Step 2: Check sign on each interval.

• On $(0,1)$: $x^{99} > 0$, so $100x^{99} > 0$; $\cos x > 0$ here. Sum is positive → increasing.
• On $\left(0, \frac{\pi}{2}\right)$: same reasoning — both terms positive → increasing. …

The core idea is simple: a function decreases only where its derivative is negative. So the entire problem reduces to checking the sign of $f'(x) = 100x^{99} + \cos x$ on each interval.

Why this approach? Because monotonicity (increasing/decreasing) is defined by the sign of the first derivative. If $f'(x) > 0$ on an interval, $f$ is increasing there; if $f'(x) < 0$, it is decreasing. No need to graph the function or solve complicated equations — just test the sign of the derivative.

Now, let’s examine each interval carefully.

1. Interval $(0, 1)$

   Here $x$ is positive and less than 1. So $x^{99} > 0$, and $100x^{99} > 0$. Also, $\cos x$ is positive for $x \in (0, 1)$ because $\cos 0 = 1$ and $\cos 1 \approx 0.54$. So both terms are positive, hence $f'(x) > 0$ everywhere on $(0,1)$. The function is increasing, not decreasing.

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

   On $(0, \pi/2)$, $x^{99} > 0$ (since $x>0$), so $100x^{99} > 0$. And $\cos x > 0$ for all $x$ in this interval (cosine is positive in the first quadrant). Again, $f'(x) > 0$. So $f$ is increasing here too.

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

   Here $x$ is between about 1.57 and 3.14. $x^{99}$ is still positive (since $x>0$), so $100x^{99} > 0$. But $\cos x$ becomes negative in the second quadrant — for example, $\cos \pi = -1$. So the derivative is $f'(x) = \text{positive large number} + \text{negative number}$. Could it become negative? …

Method: Determining the Sign of a Derivative That Sums a Dominant Term and a Bounded Term

Use this method when the derivative is a sum $f'(x) = (\text{large/unbounded term}) + (\text{bounded oscillating term})$, such as a high power of $x$ added to a sine or cosine — a classic trap in "which interval is the function decreasing on" style questions.

Steps

Step 1: Differentiate the function and identify the two kinds of terms in $f'(x)$.

Separate a polynomial term, whose magnitude can grow very large, from a trigonometric term, which is always bounded between $-1$ and $1$.

Step 2: On each candidate interval, determine the sign and rough size of the unbounded term.

For $x>0$, a term like $100x^{99}$ is positive, and for $x$ not extremely close to $0$, it is very large in magnitude.

Step 3: Compare that size against the maximum possible size of the bounded term. …

Common Mistakes

Mistake 1: Assuming the sign of $\cos x$ alone determines the sign of $f'(x)$ on $\left(\tfrac{\pi}{2},\pi\right)$.

Why it's wrong: $\cos x$ is indeed negative there, but it is added to $100x^{99}$, which is far larger in magnitude for any $x$ near $\pi/2$ or beyond — the sum stays positive. Correct approach: always compare the actual sizes of both terms before deciding which one "wins," rather than reading off the sign of the more familiar bounded term.

Mistake 2: Assuming an MCQ must have a "clean" matching option and forcing a decreasing interval to exist.

Why it's wrong: "none of these" can legitimately be correct — the derivative can stay positive on every listed interval. Correct approach: test all options honestly rather than forcing a match to one of the numbered choices. …