Q.3 For 0 ≤ 𝑥 ≤ 2𝜋, sin 𝑥 and cos 𝑥 are both decreasing functions in the interval ________.
(A) (0, π/2)
(B) (π/2, π)
(C) (π, 3π/2)
(D) (3π/2, 2π)
Answer: (B) (π/2, π)
Both sin 𝑥 and cos 𝑥 decrease throughout the interval (π/2, π) within 0 ≤ 𝑥 ≤ 2π. Their first derivatives confirm this behavior uniquely among the options.
Sine Function Analysis
Sin 𝑥 reaches its maximum at π/2 and decreases to 0 at π. The derivative cos 𝑥 stays negative from π/2 to 3π/2, so sin 𝑥 decreases on (π/2, π).
-
In (0, π/2): cos 𝑥 > 0, so sin 𝑥 increases.
-
In (π, 3π/2): cos 𝑥 < 0, so sin 𝑥 continues decreasing.
-
In (3π/2, 2π): cos 𝑥 > 0, so sin 𝑥 increases.
Cosine Function Analysis
Cos 𝑥 starts at 1 (𝑥=0), decreases to -1 at π, then increases. The derivative -sin 𝑥 proves cos 𝑥 decreases where sin 𝑥 > 0, which spans 0 to π.
-
In (π/2, π): sin 𝑥 > 0, so -sin 𝑥 < 0, confirming decrease.
-
In (0, π/2): sin 𝑥 > 0, so cos 𝑥 also decreases here.
-
In (π, 3π/2): sin 𝑥 > 0 early but derivative changes sign; overall increases after π.
-
In (3π/2, 2π): sin 𝑥 < 0, so -sin 𝑥 > 0, cos 𝑥 increases.
Option Breakdown
| Option | Sin 𝑥 Behavior | Cos 𝑥 Behavior | Both Decreasing? |
|---|---|---|---|
| (A) (0, π/2) | Increasing (cos 𝑥 > 0) | Decreasing (-sin 𝑥 < 0) | No |
| (B) (π/2, π) | Decreasing (cos 𝑥 < 0) | Decreasing (-sin 𝑥 < 0) | Yes |
| (C) (π, 3π/2) | Decreasing (cos 𝑥 < 0) | Increasing (-sin 𝑥 changes but net rise from -1) | No |
| (D) (3π/2, 2π) | Increasing (cos 𝑥 > 0) | Increasing (-sin 𝑥 > 0) | No |
Only (π/2, π) shows both derivatives negative simultaneously.
Sin x and cos x both decreasing in (π/2, π) is a key concept for calculus exams like JEE Main, GATE, and board tests. This sin x cos x decreasing interval analysis uses derivatives to check monotonicity across [0, 2π], helping students solve MCQs precisely.
Understanding Monotonicity with Derivatives
A function decreases where its derivative is negative. For sin 𝑥, derivative cos 𝑥 < 0 on (π/2, 3π/2). For cos 𝑥, derivative -sin 𝑥 < 0 when sin 𝑥 > 0 on (0, π). The overlap—where sin x and cos x are both decreasing—is (π/2, π).
Detailed Interval Check: Sin x Behavior
-
(0, π/2): cos 𝑥 > 0 → sin 𝑥 increasing from 0 to 1.
-
(π/2, π): cos 𝑥 < 0 → sin 𝑥 decreasing from 1 to 0.
-
(π, 3π/2): cos 𝑥 < 0 → sin 𝑥 decreasing from 0 to -1.
-
(3π/2, 2π): cos 𝑥 > 0 → sin 𝑥 increasing from -1 to 0.
Detailed Interval Check: Cos x Behavior
-
(0, π/2): -sin 𝑥 < 0 → cos 𝑥 decreasing from 1 to 0.
-
(π/2, π): -sin 𝑥 < 0 → cos 𝑥 decreasing from 0 to -1.
-
(π, 3π/2): sin 𝑥 shifts from positive to negative; cos 𝑥 rises overall from -1 toward 0.
-
(3π/2, 2π): -sin 𝑥 > 0 → cos 𝑥 increasing from 0 to 1.
Why Only (π/2, π) Works for Both
In (π/2, π), sin 𝑥 drops steadily while cos 𝑥 plunges to its minimum. Other intervals fail: (0, π/2) has sin 𝑥 rising; (π, 3π/2) sees cos 𝑥 rebounding; (3π/2, 2π) lifts both upward. Practice sign charts for quick exam solves.
Keywords: sin x cos x decreasing interval, both sin x and cos x decreasing, sin x decreasing 0 to 2π, cos x monotonicity intervals
