Question

Let \(f(x)=2\sqrt{\,2\sin x+2\sqrt{2}\cos x\,},\; x\in\mathbb{R}\). Then the value of \(f\!\left(\frac{\pi}{12}\right)\) is equal to

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

Hint:

Converting the form \((\sin x + \cos x)\) to \(\sqrt{2} \sin(x + 45^\circ)\) is often much faster than trying to calculate values for unusual angles like \(15^\circ\) directly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We can simplify the function using the identity \(a\sin x + b\cos x = \sqrt{a^2+b^2}\sin(x+\phi)\) where \(\tan \phi = b/a\).
Alternatively, we can directly substitute the values of sine and cosine for \(\pi/12\).

Step 2: Key Formula or Approach:
1. \(f(x) = 2\sqrt{2}(\sin x + \cos x)\).
2. \(\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})\).

Step 3: Detailed Explanation:
Simplify \(f(x)\):
\[ f(x) = 2\sqrt{2} \cdot \sqrt{2} \sin(x + \frac{\pi}{4}) = 4 \sin(x + \frac{\pi}{4}) \]
Now evaluate at \(x = \frac{\pi}{12}\):
\[ f(\frac{\pi}{12}) = 4 \sin(\frac{\pi}{12} + \frac{\pi}{4}) \]
Find the sum of angles:
\[ \frac{\pi}{12} + \frac{3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} \]
Substitute back:
\[ f(\frac{\pi}{12}) = 4 \sin(\frac{\pi}{3}) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

Step 4: Final Answer:
The value is \(2\sqrt{3}\).