Question

If \(f(x) = \frac{\sqrt{2\sin x}}{\sqrt{1 + \cos 2x}}\), then \(f'\left(\frac{\pi}{6}\right) =\)

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

Hint:

\(1 + \cos 2\theta = 2\cos^2 \theta\). Always consider the sign of \(\cos x\).

Answer 1

The Correct Option is C

Step 1: Concept:
• Use identity: \[ 1 + \cos 2x = 2\cos^2 x \]
• Simplify \(f(x)\) using this identity.

Step 2: Detailed Explanation:
• Simplify function: \[ f(x) = \frac{\sqrt{2}\sin x}{\sqrt{2\cos^2 x}} = \frac{\sin x}{|\cos x|} \]
• For \(x\) near \(\pi/6\), \(\cos x>0\), so: \[ f(x) = \tan x \]
• Differentiate: \[ f'(x) = \sec^2 x \]
• Substitute \(x = \frac{\pi}{6}\): \[ f'\left(\frac{\pi}{6}\right) = \sec^2\left(\frac{\pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]

Step 3: Final Answer:
• \[ f'\left(\frac{\pi}{6}\right) = \frac{4}{3} \]