Let f(x) be a a positive function and I1 = -(1/2)1 2xf(2x(1 - 2x)) , dx and I2 = -12 f(x(1

Q71 jee_main_2025_29_jan_evening Integration of Modulus and Greatest Integer Functions

If

24\int₀(pi)/(4)( |4x - (pi)/(12)| + [2 x])dx = 2π +α

, where

[\cdot ]

denotes the greatest integer function, then

α

is equal to

Numerical Answer. Answer: 12 to 12

Solution

Related Formula

Additivity property of definite integrals over split interval boundary points:

\intₐc g(x) dx = \intₐb g(x) dx + \intbc g(x) dx

Core Logic

Separate the integration tracking flow into two component operations:

I₁ = \int₀(pi)/(4) |4x - (pi)/(12)| dx

I₂ = \int₀(pi)/(4) [2 x] dx

Step 1: Solve Modulus Integral Term

The argument changes sign inside absolute value bounds when $4x - (pi)/(12) = 0 x = (pi)/(48)$ :

I₁ = \int₀(pi)/(48) - (4x - (pi)/(12)) dx + \int(pi)/(48)(pi)/(4) (4x - (pi)/(12)) dx

= (1)/(4) [ (4x - (pi)/(12)) ]₀(pi)/(48) - (1)/(4) [ (4x - (pi)/(12)) ](pi)/(48)(pi)/(4)

Evaluating across numerical boundaries gives:

24 \cdot I₁ = 24((1)/(2)) = 12

Step 2: Solve Greatest Integer Component and Combine

Analyze step limits inside greatest integer block $[2 x]$ $[2\sin x]$: For $x in [0, (pi)/(6))$ $x \in \left[0, \frac{pi}{6}\right)$, $0 \leq 2 x < 1 [2 x] = 0$ . For $x in [(pi)/(6), (pi)/(4)]$ , $1 \leq 2 x < \sqrt(2) [2 x] = 1$ .

I₂ = \int₀(pi)/(6) 0 dx + \int(pi)/(6)(pi)/(4) 1 dx = (pi)/(4) - (pi)/(6) = (pi)/(12)

Combine both sections aggregated by multiplier 24:

Total = 12 + 24((pi)/(12)) = 2π + 12

Comparing directly with expression statement parameters $2π + α$ $2\pi + \alpha$ isolates response value: $α = 12$ $\alpha = 12$

Pattern Recognition

Modulus arguments and step functions require isolating inflection transition numbers directly to break integrations up neatly.

Chapter Mix

Class 12 Mathematics: Definite Integration

Q jee_main_2025_28_jan_morning Properties of Definite Integrals (King's Property)

If

\int-(π)/(2)(π)/(2)(96x² ²x)/((1 + e^x)) dx = π (α π² +β),α ,β in Z,

then

(α + β)²

equals:

A. 144
B. 196
C. 100
D. 64

Solution

Related Formula

King's property for definite integration:

\intₐ^b f(x) dx = \intₐ^b f(a+b-x) dx

Core Logic

Apply the identity $x \to -x$ $x \to -x$ to the integral:

I = \int-(π)/(2)(π)/(2) (96x² ² x)/(1 + e^x) dx = \int-(π)/(2)(π)/(2) 96x² ² x1 + e-x dx

Step 1: Adding both integral variations

Adding the equations eliminates the exponential denominator term ( $1+e^x$ $1+e^x$):

2I = \int-(π)/(2)(π)/(2) 96x² ² x \cdot [(1)/(1+e^x) + (e^x)/(1+e^x)] dx

I = 48 \int₀(π)/(2) x² (1 + 2x) dx

Step 2: Evaluating the integrated components

Integrating by parts gives:

I = π (2π² - 12)

Matching coefficients with the template: $α = 2$ $\alpha = 2$ and \beta = -12.

(α + β)² = (2 - 12)² = (-10)² = 100

Pattern Recognition

Exponential denominators like $1+e^x$ $1+e^x$ in symmetric integral intervals are prime candidates for simplification using King's property.

Chapter Mix

Class 12 Maths: Definite Integrals

Q68 jee_main_2025_04_april_evening Properties of Definite Integrals

Let

f(x) + 2f((1)/(x)) = x² + 5

and

2 g (x) - 3 g ((1)/(2)) = x, x > 0

. If

α = \int_ 1 ^ 2 f (x) d x

, and

β = \int_ 1 ^ 2 g (x) d x

, then the value of

9α + β

is:

A. $1$ $1$
B. $0$ $0$
C. $10$ $10$
D. $11$ $11$

Solution

Core Logic

We have two functional equations to solve before integrating.

Equation 1: $f(x) + 2f((1)/(x)) = x² + 5$ Replace $x$ $x$ with $(1)/(x)$ $\frac{1}{x}$:

f((1)/(x)) + 2f(x) = (1)/(x²) + 5

Multiplying this new equation by $2$ $2$ and subtracting the original Equation 1 eliminates the $f((1)/(x))$ $f\left(\frac{1}{x}\right)$ term:

4f(x) + 2f((1)/(x)) - (f(x) + 2f((1)/(x))) = 2((1)/(x²) + 5) - (x² + 5)

3f(x) = (2)/(x²) - x² + 5 f(x) = (2)/(3x²) - (x²)/(3) + (5)/(3)

Step 1: Finding alpha

Integrate $f(x)$ $f(x)$ from $1$ $1$ to $2$ $2$:

α = \int₁² ( (2)/(3x²) - (x²)/(3) + (5)/(3) ) dx = [ -(2)/(3x) - (x³)/(9) + (5x)/(3) ]₁²

α = ( -(1)/(3) - (8)/(9) + (10)/(3) ) - ( -(2)/(3) - (1)/(9) + (5)/(3) ) = (19)/(9) - (8)/(9) = (11)/(9)

Thus, $9α = 11$ $9\alpha = 11$.

Step 2: Solving for g(x) and finding beta

We are given $2g(x) - 3g((1)/(2)) = x$ . Substitute $x = (1)/(2)$ $x = \frac{1}{2}$:

2g((1)/(2)) - 3g((1)/(2)) = (1)/(2) -g((1)/(2)) = (1)/(2) g((1)/(2)) = -(1)/(2)

Substitute this constant value back into the original equation:

2g(x) - 3(-(1)/(2)) = x 2g(x) + (3)/(2) = x g(x) = (x)/(2) - (3)/(4)

Now find $β$ $\beta$:

β = \int₁² ( (x)/(2) - (3)/(4) ) dx = [ (x²)/(4) - (3x)/(4) ]₁² = ( 1 - (3)/(2) ) - ( (1)/(4) - (3)/(4) ) = -(1)/(2) - (-(1)/(2)) = 0

Step 3: Calculating 9alpha + beta

Combining our values:

9α + β = 11 + 0 = 11

Pattern Recognition

Functional equations involving $x \to (1)/(x)$ $x \to \frac{1}{x}$ are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Functional Equations

Q66 jee_main_2025_04_april_morning Properties of Definite Integrals

The value of

\int₋₁¹ (1 + \sqrt(|x| - x))e^x + (\sqrt(|x| - x))e-xe^x + e-x dx

is equal to

A. $3 - 2\sqrt(2)3$ $3 - \frac{2\sqrt{2}}{3}$
B. $2 + 2\sqrt(2)3$ $2 + \frac{2\sqrt{2}}{3}$
C. $1 - 2\sqrt(2)3$ $1 - \frac{2\sqrt{2}}{3}$
D. $1 + 2\sqrt(2)3$ $1 + \frac{2\sqrt{2}}{3}$

Solution

Related Formula

King's property of definite integrals:

\intₐb f(x)dx = \intₐb f(a+b-x)dx

Core Logic

Let the given integral be $I$ $I$. Apply King's property by substituting $x \to -x$ $x \to -x$:

I = \int₋₁¹ (1 + \sqrt(|x| + x))e-x + (\sqrt(|x| + x))exe-x + ex dx

Add both integral expressions $2I = I + I$ $2I = I + I$:

2I = \int₋₁¹ (e^x + e-x) + (\sqrt(|x| - x) + \sqrt(|x| + x))(e^x + e-x)e^x + e-x dx

2I = \int₋₁¹ (1 + \sqrt(|x| - x) + \sqrt(|x| + x)) dx

Step 1: Apply Symmetry Properties

The integrand is completely even. Hence, convert intervals:

2I = 2\int₀¹ (1 + \sqrt(|x| - x) + \sqrt(|x| + x)) dx

For $x in [0,1]$ $x \in [0,1]$, $|x| = x \sqrt(|x| - x) = 0$ $|x| = x \implies \sqrt{|x| - x} = 0$ and $\sqrt(|x| + x) = \sqrt(2x)$ $\sqrt{|x| + x} = \sqrt{2x}$:

I = \int₀¹ (1 + \sqrt(2x)) dx

Step 2: Final Integration Execution

I = [ x + \sqrt(2) \cdot x3/23/2 ]₀¹ = [ x + 2\sqrt(2)3x3/2 ]₀¹

I = 1 + 2\sqrt(2)3

Pattern Recognition

When functions involve combinations of exponential components ( $e^x, e-x$ $e^x, e^{-x}$) over symmetric boundaries, adding the variable reflection eliminates exponential fractions instantly.

Chapter Mix

Class 12 Mathematics: Definite Integration

Q52 jee_main_2025_24_jan_morning Properties of Definite Integrals

If

I(m,n) = \int₀¹ xm-1 (1-x)ⁿ⁻¹ dx

where

m, n > 0

, then

I(9,14) + I(10,13)

is :

A. $I(9, 1)$ $I(9, 1)$
B. $I(19, 27)$ $I(19, 27)$
C. $I(1, 13)$ $I(1, 13)$
D. $I(9, 13)$ $I(9, 13)$

Solution

Related Formula

The beta function integral format satisfies:

I(m,n) = \int₀¹ xm-1 (1-x)ⁿ⁻¹ dx

Core Logic

Let's combine the terms of the requested sum directly by inserting their respective definitions:

I(9,14) = \int₀¹ x⁹⁻¹ (1-x)¹⁴⁻¹ dx = \int₀¹ x⁸ (1-x)¹³ dx

I(10,13) = \int₀¹ x¹⁰⁻¹ (1-x)¹³⁻¹ dx = \int₀¹ x⁹ (1-x)¹² dx

Step 1: Factoring out common algebraic terms

Summing the two components:

I(9,14) + I(10,13) = \int₀¹ [ x⁸ (1-x)¹³ + x⁹ (1-x)¹² ] dx

Factor out the common term $x⁸ (1-x)¹²$ $x^8 (1-x)^{12}$ inside the integrand:

= \int₀¹ x⁸ (1-x)¹² [ (1-x) + x ] dx

= \int₀¹ x⁸ (1-x)¹² (1) dx

= \int₀¹ x⁹⁻¹ (1-x)¹³⁻¹ dx = I(9,13)

Pattern Recognition

When dealing with linear combinations of beta functions with shifting parameter indices, directly writing down the definite integral expression often results in immediate algebraic cancellation or simplification via basic factoring.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Let f(x)$f(x)$ be a a positive function and I₁ = ∫-(1)/(2)¹ 2xf(2x(1 - 2x)) dx$I_1 = \int_{-\frac{1}{2}}^{1} 2xf(2x(1 - 2x)) \, dx$ and I₂ = ∫₋₁² f(x(1 - x)) dx$I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of (I₂)/(I₁)$\frac{I_2}{I_1}$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Perform Base Transformation Substitution

Step 2: Invoke Integral Mirror Properties

Step 3: Final Matrix Matching Evaluation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Definite Integrals Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Solve Modulus Integral Term

Step 2: Solve Greatest Integer Component and Combine

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Adding both integral variations

Step 2: Evaluating the integrated components

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Finding alpha

Step 2: Solving for g(x) and finding beta

Step 3: Calculating 9alpha + beta

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Symmetry Properties

Step 2: Final Integration Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Factoring out common algebraic terms

Pattern Recognition

Chapter Mix

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Let $f(x)$ $f(x)$ be a a positive function and $I₁ = \int-(1)/(2)¹ 2xf(2x(1 - 2x)) dx$ and $I₂ = \int₋₁² f(x(1 - x)) dx$ $I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of $(I₂)/(I₁)$ $\frac{I_2}{I_1}$ is equal to