Question

If \( f(x) = \cos x - \sin x \), and \( x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \), then \( f' \left( \frac{\pi}{3} \right) \) is equal to:

• A. \( \sqrt{3} + 1 \)
• B. \( \frac{\sqrt{3} + 1}{4} \)
• C. \( \frac{\sqrt{3} + 1}{2} \)
• D. \( \frac{\sqrt{3} - 1}{2} \)
• E. \( \frac{\sqrt{3} - 1}{4} \)

Hint:

For derivatives involving trigonometric functions, remember that the derivative of \( \cos x \) is \( -\sin x \), and the derivative of \( \sin x \) is \( \cos x \). Substitute these standard derivatives and evaluate using known trigonometric values.

Answer 1

The Correct Option is C

First, find the derivative of \( f(x) = \cos x - \sin x \): \[ f'(x) = -\sin x - \cos x \] Now, substitute \( x = \frac{\pi}{3} \) into the derivative: \[ f' \left( \frac{\pi}{3} \right) = -\sin \left( \frac{\pi}{3} \right) - \cos \left( \frac{\pi}{3} \right) \] Using the known values \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \) and \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \), we get: \[ f' \left( \frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2} - \frac{1}{2} \] \[ f' \left( \frac{\pi}{3} \right) = -\frac{\sqrt{3} + 1}{2} \] Thus, the correct answer is \( \boxed{\frac{\sqrt{3} + 1}{2}} \), as it matches the given options.