Question

The interval in which the \(f(x) = sinx-cosx, 0 ≤ x ≤ 2π\) is strictly decreasing is :

• A. \((\frac{π}{4},\frac{5π}{4})\)
• B. \((0,\frac{π}{4})∪(\frac{5π}{4},2π)\)
• C. \((\frac{3π}{4},\frac{7π}{4})\)
• D. \((0,\frac{3π}{4})∪(\frac{7π}{4},2π)\)

Answer 1

The Correct Option is C

To determine the interval where the function \( f(x) = \sin x - \cos x \) is strictly decreasing for \( 0 \leq x \leq 2\pi \), we begin by analyzing its derivative:

\( f'(x) = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x \)

The function is strictly decreasing where the derivative is negative:

\( \cos x + \sin x < 0 \)

To simplify this, divide both sides by \( \sqrt{2} \):

\( \frac{1}{\sqrt{2}}(\cos x + \sin x) < 0 \)

Using the identity \( \cos x + \sin x = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right) \), the inequality becomes:

\( \sqrt{2} \sin \left(x + \frac{\pi}{4} \right) < 0 \quad \Rightarrow \quad \sin \left(x + \frac{\pi}{4} \right) < 0 \)

The sine function is negative in the interval \( (\pi, 2\pi) \). Therefore:

\( x + \frac{\pi}{4} \in (\pi, 2\pi) \quad \Rightarrow \quad x \in \left( \pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4} \right) = \left( \frac{3\pi}{4}, \frac{7\pi}{4} \right) \)

Hence, the function \( f(x) = \sin x - \cos x \) is strictly decreasing on the interval:

\( \left( \frac{3\pi}{4}, \frac{7\pi}{4} \right) \)