Definite Integrals JEE 2025: Shifted Sine and Floor Functions

How do you solve the shifted-sine and floor-function question in definite integrals JEE 2025?

The shifted sine contributes zero; the floor-function area determines the answer. This definite integrals JEE 2025 question asks you to find the unknown:

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

The floor function gives the greatest integer not exceeding its argument: $⌊ t ⌋ = greatest integer not exceeding t .$

The supplied bank tags this Hard and Subjective Answer Type; no answer options were supplied. Its year label is 2025, but its supplied date is 2023-01-24, so the exact year and shift attribution remain unresolved.

How do you split the integral and show that the sine term is zero?

Split the sum into two integrals, then substitute in the sine term. The endpoint cosine values cancel, giving zero. This follows the supplied solution’s method.

Define: $I = I_{1} + I_{2},$

I_{1} = \int_{0}^{π} \sin (4 x - \frac{π}{12}) d x, I_{2} = \int_{0}^{π} ⌊ 2 \sin x ⌋ d x .

For the first integral, set:

u = 4 x - \frac{π}{12}, d u = 4 d x, d x = \frac{d u}{4} .

Transform both limits before integrating:

x = 0 ⟹ u = - \frac{π}{12},

x = π ⟹ u = 4 π - \frac{π}{12} = \frac{47 π}{12} .

Therefore,

I_{1} = \frac{1}{4} \int_{- π / 12}^{47 π / 12} \sin u d u = \frac{1}{4} [- \cos u]_{- π / 12}^{47 π / 12},

I_{1} = \frac{1}{4} [- \cos (\frac{47 π}{12}) + \cos (- \frac{π}{12})] .

Cosine periodicity gives:

\frac{47 π}{12} = 4 π - \frac{π}{12}, \cos (\frac{47 π}{12}) = \cos (- \frac{π}{12}) .

Hence,

I_{1} = 0 .

The transformed interval has length:

\frac{47 π}{12} - (- \frac{π}{12}) = 4 π .

That is two complete sine periods. Oddness alone is not the justification, because the transformed limits are not symmetric about zero.

Where should you split the floor-function integral?

Split where the expression inside the floor reaches an integer. Here, the floor is one across the middle region, except at its isolated maximum, and zero across the two outer regions. Integer crossings determine the intervals, not arbitrary equal subdivisions.

First establish the range:

0 \leq 2 \sin x \leq 2 (0 \leq x \leq π) .

The possible floor values are zero, one and two. Find the crossings:

2 \sin x = 1 ⟹ x = \frac{π}{6}, \frac{5 π}{6},

2 \sin x = 2 ⟹ x = \frac{π}{2} .

The exact partition, including every boundary, is:

⌊ 2 \sin x ⌋ = {\begin{matrix} 0, & x \in [0, π / 6) \cup (5 π / 6, π], \\ 1, & x \in [π / 6, π / 2) \cup (π / 2, 5 π / 6], \\ 2, & x = π / 2 . \end{matrix}

At both outer crossing points, the input equals one exactly, so the floor is one, not zero. At the midpoint, the input reaches two, but it does not stay there over any interval.

Changing a bounded, integrable function at finitely many points leaves its definite integral unchanged. A single point has zero width, so the isolated height of two adds no area.

Thus the two height-one pieces can be combined for integration:

I_{2} = 0 \cdot \frac{π}{6} + 1 \cdot (\frac{5 π}{6} - \frac{π}{6}) + 0 \cdot \frac{π}{6} = \frac{2 π}{3} .

How do you restore the factor of 24 and calculate alpha?

Multiply the complete integral by 24, then subtract the constant on the right. The answer below is derived from the supplied expression, not quoted from a separately supplied official answer key.

Combine the two parts:

I = 0 + \frac{2 π}{3} = \frac{2 π}{3} .

Substitute into the original relation:

24 I = 24 \cdot \frac{2 π}{3} = 16 π = 2 π + α .

Therefore,

α = 16 π - 2 π = 14 π .

Check against the scaled integral:

2 π + 14 π = 16 π = 24 \cdot \frac{2 π}{3} .

Why can’t you integrate twice sine instead of its floor?

Removing the floor changes the integrand over intervals of positive length, so it changes the area. It is not comparable to ignoring one isolated point. Since no options were supplied, the following is an illustrative wrong result, not a listed distractor.

The invalid replacement is:

\int_{0}^{π} ⌊ 2 \sin x ⌋ d x \overset{incorrect}{\to} \int_{0}^{π} 2 \sin x d x .

The smooth sine integral evaluates to:

\int_{0}^{π} 2 \sin x d x = [- 2 \cos x]_{0}^{π} = 2 + 2 = 4 .

Using it would produce:

α_{wrong} = 24 \cdot 4 - 2 π = 96 - 2 π .

Compare the areas:

Floor-function area = \frac{2 π}{3}, Smooth sine area = 4 .

Between the outer threshold points, the floor stays at one except at the midpoint, while the smooth function keeps changing. Outside those points, the floor is zero even though the sine is positive in the interval’s interior. Write the floor intervals before attempting any antiderivative.

Which two related integrals can you solve with the same method?

The cosine and sine exercises below both use integer crossings followed by height-times-width calculations. They are original practice questions, not additional verified PYQs. Attempt each partition first, then compare your endpoint assignments with the worked answer.

How do you integrate the floor of twice cosine over the first quadrant?

The area comes from a single height-one interval. Cosine decreases throughout the first quadrant, and the maximum floor value occurs only at the left endpoint.

Evaluate:

J = \int_{0}^{π / 2} ⌊ 2 \cos x ⌋ d x .

The threshold is:

2 \cos x = 1 ⟹ x = \frac{π}{3} .

The exact values are:

⌊ 2 \cos x ⌋ = {\begin{matrix} 2, & x = 0, \\ 1, & x \in (0, π / 3], \\ 0, & x \in (π / 3, π / 2] . \end{matrix}

The isolated endpoint contributes no area. Therefore,

J = 1 \cdot \frac{π}{3} + 0 \cdot (\frac{π}{2} - \frac{π}{3}) = \frac{π}{3} .

How do you integrate the floor of three times sine from zero to pi?

Use two nonzero integer thresholds to locate the intervals of height one and two. The maximum height of three occurs only at the midpoint and contributes no area.

Evaluate and define:

K = \int_{0}^{π} ⌊ 3 \sin x ⌋ d x,

a = \arcsin (\frac{1}{3}), b = \arcsin (\frac{2}{3}), 0 < a < b < \frac{π}{2} .

The threshold equations give:

3 \sin x = 1 ⟹ x = a, π - a,

3 \sin x = 2 ⟹ x = b, π - b .

The complete partition is:

⌊ 3 \sin x ⌋ = {\begin{matrix} 0, & x \in [0, a) \cup (π - a, π], \\ 1, & x \in [a, b) \cup (π - b, π - a], \\ 2, & x \in [b, π - b] ⧵ {π / 2}, \\ 3, & x = π / 2 . \end{matrix}

Sum the two height-one strips and the central height-two region:

K = 2 (b - a) + 2 (π - 2 b) = 2 π - 2 a - 2 b .

For your next floor-function integral:

Find the integer thresholds within the input’s range.
Order every crossing point.
Assign the floor height on each interval and check isolated values.
Sum height multiplied by interval length.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

Why is the integral of sin(4x − π/12) from 0 to π zero?

With u = 4x − π/12, the transformed interval has length 4π, covering two complete sine periods. The endpoint cosine values therefore cancel, giving zero. Oddness alone is not a valid justification because the transformed limits are not symmetric about zero.

How do you integrate floor(2 sin x) from 0 to π?

The integer crossings occur at π/6, π/2 and 5π/6. The floor is one between π/6 and 5π/6, except for an isolated value of two at π/2, and zero outside that region. The isolated value adds no area, so the integral is 5π/6 − π/6 = 2π/3.

Why can you ignore a single point in a floor-function integral?

Changing a bounded, integrable function at finitely many points does not change its definite integral. For floor(2 sin x), the value two occurs only at x = π/2 and has zero width, so it contributes no area.

What is alpha in the shifted-sine and floor-function question?

For 24 times the integral of sin(4x − π/12) + floor(2 sin x) from 0 to π equal to 2π + alpha, the integral is 2π/3. Thus 24 × 2π/3 = 16π, giving alpha = 14π. This result is derived from the supplied expression, not a separately supplied official answer key.

Is this floor-function question a verified JEE 2025 PYQ?

The supplied question bank labels it 2025 but gives the date 2023-01-24. These details conflict, so the exact year and shift attribution remain unresolved.