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

Definite Integrals — JEE Main · 2025 · 29 Jan 2025 · Shift 2 Question with Solution

24 \int_{0}^{\frac{π}{4}} (sin 4 x - \frac{π}{12} + [2 sin x]) d x = 2 π + α

, where [•] denotes the greatest integer function, then

α

is equal to ____.

Answer: 12

Split

∣4 x - \frac{π}{12} ∣

x = \frac{π}{48}

within

[0, \frac{π}{4}]

.

For

x \in [0, \frac{π}{48}]

4 x - \frac{π}{12} \leq 0

, so the term is

sin (\frac{π}{12} - 4 x)

; for

x \in [\frac{π}{48}, \frac{π}{4}]

it is

sin (4 x - \frac{π}{12})

\int_{0}^{π /48} sin (\frac{π}{12} - 4 x) d x = \frac{1}{4} [1 - cos \frac{π}{12}]

\int_{π /48}^{π /4} sin (4 x - \frac{π}{12}) d x = \frac{1}{4} [1 + cos \frac{π}{12}]

Adding these gives

\frac{1}{2}

, independent of

cos \frac{π}{12}

.

For

[2 sin x]

[0, \frac{π}{4}]

2 sin x < 1

(bracket

= 0

) while

x < \frac{π}{6}

, and

1 \leq 2 sin x < 2

(bracket

= 1

) for

\frac{π}{6} \leq x \leq \frac{π}{4}

. So

\int_{0}^{π /4} [2 sin x] d x = 1 \cdot (\frac{π}{4} - \frac{π}{6}) = \frac{π}{12}

Adding the two pieces, the whole integral equals

\frac{1}{2} + \frac{π}{12}

, and multiplying by

24

24 (\frac{1}{2} + \frac{π}{12}) = 12 + 2 π

Comparing with

2 π + α

gives

α = 12

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer