Let f(x) + 2f((1)/(x)) = x² + 5 and 2 g (x) - 3 g ((1)/(2)) = x, x > 0. If α = ∫_ 1 ^ 2 f (x) d x, and β = ∫_ 1 ^ 2 g (x) d x, then the value of 9α + β is:

Solution & Explanation

Core Logic

We have two functional equations to solve before integrating.

Equation 1: f(x) + 2f((1)/(x)) = x² + 5 Replace x with (1)/(x):

f((1)/(x)) + 2f(x) = (1)/(x²) + 5

Multiplying this new equation by 2 and subtracting the original Equation 1 eliminates the f((1)/(x)) 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) from 1 to 2:

α = ∫₁² ( (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.

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

We are given 2g(x) - 3g((1)/(2)) = x. Substitute x = (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:

β = ∫₁² ( (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 → (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

Reference Study Guides

More Definite Integrals Previous-Year Questions — Page 10

Q28 jee_main_2024_01_february_morning Properties of Definite Integrals
If ∫-π/2π/2 8√(2) x dx(1+ex)(1+ ⁴x)=απ+β ₑ(3+2√(2)), where α, β are integers, then α²+β² equals
Numerical Answer. Answer: 8 to 8

Solution

Related Formula

King's Property for symmetric integration limits:

∫₋ₐa f(x) dx = ∫₋ₐa f(-x) dx
Core Logic

Let the given definite integral be I:

I = ∫-π/2π/2 8√(2) x dx(1+ex)(1+ ⁴x) (1)

Apply King's property by replacing x with -x (since -π/2 + π/2 = 0):

I = ∫-π/2π/2 8√(2) (-x) dx(1+e(-x))(1+ ⁴(-x)) I = ∫-π/2π/2 8√(2) x dx(1+e- x)(1+ ⁴x) = ∫-π/2π/2 8√(2) x · ex dx(ex+1)(1+ ⁴x) (2)
Step 1: Simplify by Adding Expressions

Adding equations (1) and (2):

2I = ∫-π/2π/2 8√(2) x(1 + ex) dx(1+ex)(1+ ⁴x) 2I = ∫-π/2π/2 8√(2) x dx1+ ⁴x

Since the integrand is even, we can change the limits from 0 to π/2:

2I = 2 ∫₀π/2 8√(2) x dx1+ ⁴x I = ∫₀π/2 8√(2) x dx1+ ⁴x
Step 2: Substitution and Algebraic Deconstruction

Let x = t x dx = dt. Limits change from 0 to 1:

I = ∫₀¹ 8√(2) dt1+t⁴ = 4√(2) ∫₀¹ (2 dt)/(1+t⁴)

Dividing the numerator and denominator by t², we write it as two distinct expressions:

I = 4√(2) [ ∫₀¹ (1+(1)/(t²))/(t²+(1)/(t²)) dt - ∫₀¹ (1-(1)/(t²))/(t²+(1)/(t²)) dt ] I = 4√(2) [ ∫-∞⁰ (dz)/(z²+2) - ∫∞² (dk)/(k²-2) ]

where z = t - (1)/(t) and k = t + (1)/(t).

Step 3: Integrate and Evaluate Parameters

Evaluating standard anti-derivatives:

I = 4√(2) [ 1√(2) ⁻¹( z√(2)) ]-∞⁰ - 4√(2) [ 12√(2)ln| k-√(2)k+√(2)| ]∞² I = 4√(2)( π2√(2)) - 2 ln| 2-√(2)2+√(2)| = 2π - 2ln(√(2)-1)² I = 2π + 2ln(3+2√(2))

Matching parameters yields α = 2 and β = 2. Thus:

α² + β² = 2² + 2² = 8
Pattern Recognition

Sees: Exponential variables causing asymmetry in symmetric integration bounds. Shortcut: Using King's rule completely eliminates the confusing ex factor, leaving behind a straightforward rational trigonometric configuration.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q10 jee_main_2024_29_january_evening Indefinite Integration
If ∫ (3)/(2) x + (3)/(2) x√( ³ x ³ x (x - θ)) d x = A √( θ x - θ) + B √( θ - θ x) + C, where C is the integration constant, then AB is equal to
  • A. 4sec(2θ)
  • B. 4 θ
  • C. 2 θ
  • D. 8secec(2θ)

Solution

Related Formula
(x - θ) = x θ - x θ
Core Logic

Let us partition the given integral I into two clean parts I₁ and I₂:

I = ∫ 3/2 x√( ³ x ³ x ( x θ - x θ)) dx + ∫ 3/2 x√( ³ x ³ x ( x θ - x θ)) dx

Factoring out appropriate powers of x and x from inside the roots:

I₁ = ∫ ² x√( x θ - θ) dx I₂ = ∫ ² x√( θ - x θ) dx
Step 1: Evaluating the Integrals

For I₁, substitute x θ - θ = t² ² x dx = (2t dt)/( θ):

I₁ = ∫ (2t dt)/(t θ) = (2t)/( θ) = 2 θ √( x θ - θ)

For I₂, substitute θ - x θ = z² ² x dx = (2z dz)/( θ):

I₂ = ∫ (2z dz)/(z θ) = (2z)/( θ) = 2 θ √( θ - x θ)
Step 2: Combining Coefficients

Comparing with the given expression, we find the coefficients:

A = 2 θ, B = 2 θ

Multiplying them together:

AB = 4 θ θ = (4)/( θ θ) = (8)/(2 θ θ) = 8 (2θ)
Pattern Recognition

When integrating roots of trigonometric functions involving (x - θ), try dividing/multiplying fields by ⁿ x or ⁿ x to force x / ² x templates.

Chapter Mix

Class 12 Mathematics: Integrals

Q24 jee_main_2024_29_january_evening Definite Integration
If ∫(π)/(6)(π)/(3)√(1 - 2x) dx = α +β √(2) +γ √(3) where α ,β and γ are rational numbers, then 3α + 4β - γ is equal to
Numerical Answer. Answer: 6 to 6

Solution

Related Formula
√(1 - 2x) = √(( x - x)²) = | x - x|
Core Logic

The absolute function | x - x| flips sign at x = (π)/(4) within the boundary range:

  • For x in [(π)/(6), (π)/(4)]: x ≥ x | x - x| = x - x
  • For x in [(π)/(4), (π)/(3)]: x ≥ x | x - x| = x - x
Step 1: Boundary Splitting Integration
I = ∫(π)/(6)(π)/(4) ( x - x) dx + ∫(π)/(4)(π)/(3) ( x - x) dx I = [ x + x ](π)/(6)(π)/(4) + [ - x - x ](π)/(4)(π)/(3) I = ( 1√(2) + 1√(2) - ((1)/(2) + √(3)2) ) + ( -(1)/(2) - √(3)2 - (- 1√(2) - 1√(2)) ) I = ( √(2) - 1 + √(3)2 ) + ( √(2) - 1 + √(3)2 ) = 2√(2) - 1 - √(3) = -1 + 2√(2) - √(3)
Step 2: Coefficient Matrix Matching

Comparing with the given baseline layout:

α = -1, β = 2, γ = -1

Evaluating the required target metrics:

3α + 4β - γ = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6
Pattern Recognition

Never forget that √(f(x)²) = |f(x)|. Skipping modulus checks inside definite root boundaries leads to incorrect answers.

Chapter Mix

Class 12 Mathematics: Integrals

Q jee_main_2024_27_jan_morning Properties of Definite Integrals
If (a, b) be the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (3, 1), and I₁=∫ₐbx~sin(4x-x²)dx, I₂=∫ₐbsin(4x-x²)dx, then 36 I₁I₂ is equal to:
  • A. 72
  • B. 88
  • C. 80
  • D. 66

Solution

Related Formula
∫ₐ^b f(x) dx = ∫ₐ^b f(a+b-x) dx (King's Rule)
Core Logic

First, find the orthocentre (a,b) of Δ ABC with vertices A(1, 2), B(2, 3), and C(3, 1). Slope of AB = (3-2)/(2-1) = 1. The altitude from C onto AB must be perpendicular to AB, so its slope is -1. Equation of altitude from C(3,1):

y - 1 = -1(x - 3) ⇒ x + y = 4

The orthocentre (a, b) lies on all altitudes, including this one. Thus, it satisfies a + b = 4.

Step 1: Applying Definite Integral Properties

Given I₁ = ∫ₐ^b x (4x-x²) dx, let's rewrite the argument of sine:

4x - x² = x(4-x)

Apply King's Rule replacing x with (a+b-x). Since we proved a+b = 4, substitute x with (4-x):

I₁ = ∫ₐ^b (4-x) ((4-x)(4 - (4-x))) dx I₁ = ∫ₐ^b (4-x) ((4-x)x) dx I₁ = ∫ₐ^b (4-x) (4x-x²) dx
Step 2: Evaluating the Integral Ratio

Expand the newly formed integral:

I₁ = 4 ∫ₐ^b (4x-x²) dx - ∫ₐ^b x (4x-x²) dx

Notice that the second term is I₁ and the first integral is I₂:

I₁ = 4I₂ - I₁ ⇒ 2I₁ = 4I₂ ⇒ (I₁)/(I₂) = 2
Step 3: Final Output Evaluation

We need the value of 36 (I₁)/(I₂):

36 × 2 = 72
Pattern Recognition

Whenever you see ∫ₐ^b x · f(x(a+b-x)) dx, immediately apply King's Rule to factor out x. You rarely need the individual values of the integration bounds, only their sum.

Chapter Mix

Class 11 Maths: Straight Lines Class 12 Maths: Definite Integration

Q9 jee_main_2024_27_jan_morning Integration of Irrational Functions
If ∫₀¹ 1√(3+x)+√(1+x)dx=a+b√(2)+c√(3), where a, b, c are rational numbers, then 2a+3b-4c is equal to:
  • A. 4
  • B. 10
  • C. 7
  • D. 8

Solution

Related Formula
∫ xⁿ dx = xⁿ⁺¹n+1
Core Logic

To evaluate integrals with sum of square roots in the denominator, multiply and divide by the conjugate to rationalize it.

I = ∫₀¹ √(3+x)-√(1+x)(√(3+x)+√(1+x))(√(3+x)-√(1+x))dx I = ∫₀¹ √(3+x)-√(1+x)(3+x) - (1+x)dx I = (1)/(2) ∫₀¹ (√(3+x) - √(1+x)) dx
Step 1: Integration and Bounds Setup

Integrate the resulting expression:

I = (1)/(2) [ (3+x)3/23/2 - (1+x)3/23/2 ]₀¹ I = (1)/(2) · (2)/(3) [ (3+x)3/2 - (1+x)3/2 ]₀¹ I = (1)/(3) [ ((4)3/2 - (2)3/2) - ((3)3/2 - (1)3/2) ]
Step 2: Term Simplification

Evaluate the boundary powers: 43/2 = 8 23/2 = 2√(2) 33/2 = 3√(3) 13/2 = 1

Substitute back into the expression:

I = (1)/(3) [ 8 - 2√(2) - 3√(3) + 1 ] = (1)/(3) [ 9 - 2√(2) - 3√(3) ] I = 3 - (2)/(3)√(2) - √(3)
Step 3: Finding Co-efficients

Comparing with a+b√(2)+c√(3) yields: a = 3, b = -(2)/(3), c = -1

Compute 2a+3b-4c:

2(3) + 3(-(2)/(3)) - 4(-1)

6 - 2 + 4 = 8

Pattern Recognition

Whenever you see a sum of square roots in the denominator of an integrand, the immediate algorithmic next step is rationalization.

Chapter Mix

Class 12 Maths: Definite Integration

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