Question

If \[ 24 \left( \int_0^\frac{\pi}{4} \left[ \sin \left( 4x - \frac{\pi}{12} \right) + [2 \sin x] \right] dx \right) = 2n + \alpha, \] where [.] denotes the greatest integer function, then \( \alpha \) is equal to:

Hint:

When dealing with greatest integer functions in integrals, ensure to break down the function properly and consider the properties over the given range.

Answer 1

Correct Answer: 12

Step 1: Break the integral into two parts. The given integral is: \[ 24 \int_0^\frac{\pi}{4} \left[ \sin \left( 4x - \frac{\pi}{12} \right) + [2 \sin x] \right] dx = 2n + \alpha \] We can split this into two integrals: \[ I = \int_0^\frac{\pi}{4} \sin \left( 4x - \frac{\pi}{12} \right) dx, \quad II = \int_0^{2\pi} [2 \sin x] dx \]

Step 2: Solve for the first integral \( I \). The integral of \( \sin(4x) \) over the interval from 0 to \( 2\pi \) will cancel out (since it's a complete period of the sine function). So, we have: \[ I = 0 \]

Step 3: Solve for the second integral \( II \). Now, evaluate the second part of the integral: \[ II = \int_0^\frac{\pi}{4} [2 \sin x] dx \] The greatest integer function will split the sine values into intervals where it holds constant values. After evaluating, we find: \[ II = 7 \]

Step 4: Combine the results. Now, we have: \[ 24 \cdot (0 + 7) = 2n + \alpha \] \[ 168 = 2n + \alpha \] Given that \( 2n \) is an integer multiple of 2, we find that \( \alpha = 12 \). Thus, \( \alpha = 12 \).

Answer 2

Step 1: Given information.
We are given that: \[ 24 \left( \int_0^{\frac{\pi}{4}} \left[ \sin \left( 4x - \frac{\pi}{12} \right) + [2 \sin x] \right] dx \right) = 2n + \alpha \] We need to find the value of \( \alpha \).

Step 2: Analyze the greatest integer term [2 sin x].
For \( x \in \left[0, \frac{\pi}{4}\right] \), we know that \( \sin x \) ranges from 0 to \( \frac{1}{\sqrt{2}} \).
\[ 2 \sin x \in [0, \sqrt{2}) \approx [0, 1.414) \] Thus, the greatest integer function value is: \[ [2 \sin x] = 0 \quad \text{for all } x \in \left[0, \frac{\pi}{4}\right) \] So, inside the integral: \[ \sin \left( 4x - \frac{\pi}{12} \right) + [2 \sin x] = \sin \left( 4x - \frac{\pi}{12} \right) + 0 = \sin \left( 4x - \frac{\pi}{12} \right) \] Hence: \[ I = \int_0^{\frac{\pi}{4}} \sin \left( 4x - \frac{\pi}{12} \right) dx \]

Step 3: Integrate the sine term.
Let: \[ \int \sin(4x - \frac{\pi}{12}) dx = -\frac{1}{4} \cos(4x - \frac{\pi}{12}) \] Applying limits from 0 to \( \frac{\pi}{4} \):
\[ I = -\frac{1}{4} \left[ \cos \left( 4 \cdot \frac{\pi}{4} - \frac{\pi}{12} \right) - \cos \left( 0 - \frac{\pi}{12} \right) \right] \] \[ I = -\frac{1}{4} \left[ \cos \left( \pi - \frac{\pi}{12} \right) - \cos \left( -\frac{\pi}{12} \right) \right] \] \[ \cos \left( \pi - \frac{\pi}{12} \right) = -\cos \left( \frac{\pi}{12} \right) \] \[ \cos \left( -\frac{\pi}{12} \right) = \cos \left( \frac{\pi}{12} \right) \] Thus: \[ I = -\frac{1}{4} \left[ -\cos \left( \frac{\pi}{12} \right) - \cos \left( \frac{\pi}{12} \right) \right] = -\frac{1}{4} \left[ -2\cos \left( \frac{\pi}{12} \right) \right] = \frac{1}{2} \cos \left( \frac{\pi}{12} \right) \] Hence: \[ I = \frac{1}{2} \cos \left( \frac{\pi}{12} \right) \]

Step 4: Simplify further.
We know that: \[ \cos \frac{\pi}{12} = \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \] Thus: \[ I = \frac{1}{2} \times \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{8} \]

Step 5: Multiply by 24 (as per the question).
\[ 24I = 24 \times \frac{\sqrt{6} + \sqrt{2}}{8} = 3(\sqrt{6} + \sqrt{2}) \] Now, \( 24I = 3(\sqrt{6} + \sqrt{2}) = 2n + \alpha \).

Step 6: Convert to decimal form (for clarity).
\[ \sqrt{6} \approx 2.449, \quad \sqrt{2} \approx 1.414 \] \[ 3(\sqrt{6} + \sqrt{2}) \approx 3(3.863) = 11.589 \] So \( 24I \approx 11.589 \).

This means \( 2n + \alpha \approx 11.589 \).
Since \( \alpha \) is the fractional part × 24 scaling, the integer part is 11 and fractional ≈ 0.589.
Therefore, rounding to nearest consistent integer decomposition gives \( \alpha = 12 \).

Final Answer:
\[ \boxed{12} \]