If 24int_0^fracpi4left(sin left|4x - fracpi12right| + [2sin x]right)mathrmdx = 2pi +alpha$24\int_{0}^{\frac{pi}{4}}\left(\sin \left|4x - \frac{pi}{12}\right| + [2\sin x]\right)\mathrm{d}x = 2\pi +\alpha$, where [cdot ]$[\cdot ]$ denotes the greatest integer function, then alpha$\alpha$ is equal to

### Related Formula Additivity property of definite integrals over split interval boundary points: int_a^c g(x) \, dx = int_a^b g(x) \, dx + int_b^c g(x) \, dx$$\int_{a}^{c} g(x) \, dx = \int_{a}^{b} g(x) \, dx + \int_{b}^{c} g(x) \, dx$$ ### Core Logic Separate the integration tracking flow into two component operations: I_1 = int_0^fracpi4 sinleft|4x - fracpi12 ight| \, dx$$I_1 = \int_{0}^{\frac{pi}{4}} \sin\left|4x - \frac{pi}{12} ight| \, dx$$ I_2 = int_0^fracpi4 [2sin x] \, dx$$I_2 = \int_{0}^{\frac{pi}{4}} [2\sin x] \, dx$$ ### Step 1: Solve Modulus Integral Term The argument changes sign inside absolute value bounds when 4x - fracpi12 = 0 implies x = fracpi48$4x - \frac{pi}{12} = 0 \implies x = \frac{pi}{48}$: I_1 = int_0^fracpi48 -sinleft(4x - fracpi12 ight) \, dx + int_fracpi48^fracpi4 sinleft(4x - fracpi12 ight) \, dx$$I_1 = \int_{0}^{\frac{pi}{48}} -\sin\left(4x - \frac{pi}{12} ight) \, dx + \int_{\frac{pi}{48}}^{\frac{pi}{4}} \sin\left(4x - \frac{pi}{12} ight) \, dx$$ = frac14 left[ cosleft(4x - fracpi12 ight) right]_0^fracpi48 - frac14 left[ cosleft(4x - fracpi12 ight) right]_fracpi48^fracpi4$$= \frac{1}{4} \left[ \cos\left(4x - \frac{pi}{12} ight) \right]_{0}^{\frac{pi}{48}} - \frac{1}{4} \left[ \cos\left(4x - \frac{pi}{12} ight) \right]_{\frac{pi}{48}}^{\frac{pi}{4}}$$ Evaluating across numerical boundaries gives: 24 cdot I_1 = 24left(frac12right) = 12$$24 \cdot I_1 = 24\left(\frac{1}{2}\right) = 12$$ ### Step 2: Solve Greatest Integer Component and Combine Analyze step limits inside greatest integer block [2sin x]$[2\sin x]$: For x in left[0, fracpi6right)$x \in \left[0, \frac{pi}{6}\right)$, 0 le 2sin x < 1 implies [2sin x] = 0$0 \le 2\sin x < 1 \implies [2\sin x] = 0$. For x in left[fracpi6, fracpi4right]$x \in \left[\frac{pi}{6}, \frac{pi}{4}\right]$, 1 le 2sin x < sqrt2 implies [2sin x] = 1$1 \le 2\sin x < \sqrt{2} \implies [2\sin x] = 1$. I_2 = int_0^fracpi6 0 \, dx + int_fracpi6^fracpi4 1 \, dx = fracpi4 - fracpi6 = fracpi12$$I_2 = \int_{0}^{\frac{pi}{6}} 0 \, dx + \int_{\frac{pi}{6}}^{\frac{pi}{4}} 1 \, dx = \frac{pi}{4} - \frac{pi}{6} = \frac{pi}{12}$$ Combine both sections aggregated by multiplier 24: textTotal = 12 + 24left(fracpi12 ight) = 2pi + 12$$\text{Total} = 12 + 24\left(\frac{pi}{12} ight) = 2\pi + 12$$ Comparing directly with expression statement parameters 2pi + alpha$2\pi + \alpha$ isolates response value: alpha = 12$\alpha = 12$ ### Pattern Recognition Modulus arguments and step functions require isolating inflection transition numbers directly to break integrations up neatly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration