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 9

Q jee_main_2025_28_jan_evening Integration by Parts
Let f: Rarrow R be a twice differentiable function such that f(2)=1. If F(x)=xf(x) for all xin R, ∫₀²xF(x)dx=6 and ∫₀²x²F(x)dx=40, then F(2)+∫₀²F(x)dx is equal to:
  • A. 11
  • B. 15
  • C. 6
  • D. 13

Solution

Related Formula

Integration by Parts formula:

∫ u · v dx = u ∫ v dx - ∫ ( u' ∫ v dx ) dx
Core Logic

Given F(x) = xf(x) and f(2) = 1 F(2) = 2f(2) = 2.

Let's apply Integration by Parts to the first given integral:

∫₀² x F'(x) dx = 6 [ xF(x) ]₀² - ∫₀² F(x) dx = 6 2F(2) - 0 - ∫₀² F(x) dx = 6 2(2) - ∫₀² F(x) dx = 6 ∫₀² F(x) dx = 4 - 6 = -2
Step 1: Simplify Second Integral via Integration by Parts

Now look at the second integral:

∫₀² x² F''(x) dx = 40

Applying Integration by parts (taking u = x² and v = F''(x)):

[ x² F'(x) ]₀² - ∫₀² 2x F'(x) dx = 40 4 F'(2) - 0 - 2 ∫₀² x F'(x) dx = 40

We already know ∫₀² x F'(x) dx = 6:

4 F'(2) - 2(6) = 40 4 F'(2) - 12 = 40 4 F'(2) = 52 F'(2) = 13
Step 2: Sum the Values

We need to find F'(2) + ∫₀² F(x) dx: 13 + (-2) = 11

Pattern Recognition

Notice how the definition of f(x) is mostly a distraction to find F(2)=2. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures.

Chapter Mix

Class 12 Mathematics: Definite Integration

Q57 jee_main_2025_28_jan_evening Leibnitz Rule and Definite Integral
Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫₀xtf(t)dt. If g(x³)=x⁶+x⁷ then value of Σr=1¹⁵f(r³) is:
  • A. 320
  • B. 340
  • C. 270
  • D. 310

Solution

Related Formula

Newton-Leibnitz Theorem for differentiation under integral sign:

(d)/(dx) ( ∫₀x tf(t) dt ) = xf(x)
Core Logic

Given:

g(x) = ∫₀x tf(t) dt

Differentiating both sides with respect to x:

g'(x) = xf(x) f(x) = (g'(x))/(x)

We are given g(x³) = x⁶ + x⁷. Let y = x³ x = y1/3. Substituting this into the expression for g:

g(y) = (y1/3)⁶ + (y1/3)⁷ = y² + y7/3

Thus, replacing y back with x:

g(x) = x² + x7/3
Step 1: Differentiate g(x) to find f(x)
g'(x) = 2x + (7)/(3)x4/3

Now find f(x):

f(x) = (g'(x))/(x) = 2x + (7)/(3)x4/3x = 2 + (7)/(3)x1/3
Step 2: Evaluate the Summation

We need to find Σr=1¹⁵ f(r³):

f(r³) = 2 + (7)/(3)(r³)1/3 = 2 + (7)/(3)r

Now, compute the summation from r=1 to 15:

Σr=1¹⁵ f(r³) = Σr=1¹⁵ ( 2 + (7)/(3)r ) = Σr=1¹⁵ 2 + (7)/(3)Σr=1¹⁵ r (2 × 15) + (7)/(3) × (15 × 16)/(2) 30 + (7)/(3) × 120 = 30 + 7 × 40 = 30 + 280 = 310
Pattern Recognition

Converting g(x³) directly into a function of variable y=x³ prevents multi-layer chain rule complications when applying differentiation immediately.

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Definite Integration

Q66 jee_main_2025_29_jan_morning Integration by Substitution
The integral 80∫₀(π)/(4)(( θ + θ)/(9 + 16 2θ))dθ is equal to:
  • A. 3 ₑ4
  • B. 6 ₑ₄
  • C. 4 ₑ₃
  • D. 2 ₑ3

Solution

Related Formula
∫ (dt)/(a² - b² t²) = (1)/(2a) ln | (a+bt)/(a-bt) | 2θ = 1 - ( θ - θ)²
Core Logic

Let θ - θ = t. Then ( θ + θ)dθ = dt. Transform limits: When θ = 0 t = 0 - 1 = -1 When θ = (π)/(4) t = 1√(2) - 1√(2) = 0

Step 1: Perform the algebraic substitution

Express the denominator base:

9 + 16 2θ = 9 + 16[1 - t²] = 25 - 16t²

The integral transforms to:

I = 80 ∫₋₁⁰ (dt)/(25 - 16t²) = (80)/(16) ∫₋₁⁰ (dt)/(((5)/(4))² - t²)
Step 2: Execute Integral Calculation
I = 5 [ (1)/(2((5)/(4))) ln | ((5)/(4) + t)/((5)/(4) - t) | ]₋₁⁰ I = 2 [ ln(1) - ln( (1/4)/(9/4) ) ] = 2 [ 0 - ln((1)/(9)) ] = 2ln(9) = 4ln(3)
Pattern Recognition

Whenever ( θ + θ) sits inside the numerator, instantly choose t = θ - θ as your core linear substitution engine to clean denominators.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q71 jee_main_2025_29_jan_morning Functional Equations with Integrals
Let f: (0, ∞) → R be a twice differentiable function. If for some a ≠ 0 , ∫₀¹ f(λ x) dλ = a f(x) , f(1) = 1 and f(16) = (1)/(8) , then 16 - f'((1)/(16)) is equal to
Numerical Answer. Answer: 112

Solution

Related Formula
Leibniz Integral Rule for differentiation: (d)/(dx)∫₀x f(t) dt = f(x)
Core Logic

Perform variable substitution inside the integral: let λ x = t dλ = (1)/(x) dt. When λ = 0 t = 0; when λ = 1 t = x. The equation transforms to:

(1)/(x) ∫₀x f(t) dt = a f(x) ∫₀x f(t) dt = a x f(x)
Step 1: Differentiate with respect to x

Using Leibniz rule and product rule:

f(x) = a [x f'(x) + f(x)] (1 - a)f(x) = a x f'(x) (f'(x))/(f(x)) = (1-a)/(a) (1)/(x)

Integrating both sides yields:

ln f(x) = ((1-a)/(a))ln x + c f(x) = C x(1-a)/(a)
Step 2: Calculate Constants using boundaries

Given f(1) = 1 C = 1. Given f(16) = (1)/(8) (1)/(8) = (16)(1-a)/(a) 2⁻³ = (2⁴)(1-a)/(a)

-3 = (4(1-a))/(a) -3a = 4 - 4a a = 4

Therefore, power exponent = (1-4)/(4) = -(3)/(4) f(x) = x-(3)/(4).

Step 3: Evaluate target derivative value

Find the derivative:

f'(x) = -(3)/(4) x-(7)/(4)

Substitute x = (1)/(16):

f'((1)/(16)) = -(3)/(4) (2⁻⁴)-(7)/(4) = -(3)/(4) (2⁷) = -(3)/(4) × 128 = -96

Final requested computation calculation:

16 - f'((1)/(16)) = 16 - (-96) = 112
Pattern Recognition

Scaling inputs inside functional definite integrals tracks closely to homogenous Euler equation properties. Converting integrations quickly to local power functions reduces processing parameters.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Q2 jee_main_2024_01_february_morning Properties of Definite Integrals
The value of the integral ∫₀(π)/(4) x dx ⁴(2x)+ ⁴(2x) equals:
  • A. √(2)π²8
  • B. √(2)π²16
  • C. √(2)π²32
  • D. √(2)π²64

Solution

Related Formula

King's Property of Definite Integrals:

∫ₐb f(x) dx = ∫ₐb f(a+b-x) dx
Core Logic

Let the given integral be I:

I = ∫₀(π)/(4) x dx ⁴(2x)+ ⁴(2x)

Substitute 2x = t 2dx = dt dx = (1)/(2)dt. When x = 0 t = 0, and when x = (π)/(4) t = (π)/(2).

I = (1)/(4) ∫₀(π)/(2) t dt ⁴t + ⁴t (1)
Step 1: Apply King's Property

Applying the property ∫₀a f(t) dt = ∫₀a f(a-t) dt:

I = (1)/(4) ∫₀(π)/(2) ((π)/(2) - t) dt ⁴((π)/(2)-t) + ⁴((π)/(2)-t) I = (1)/(4) ∫₀(π)/(2) ((π)/(2) - t) dt ⁴t + ⁴t (2)

Adding equations (1) and (2):

2I = (1)/(4) ∫₀(π)/(2) (π)/(2) dt ⁴t + ⁴t 2I = (π)/(8) ∫₀(π)/(2) dt ⁴t + ⁴t 2I = (π)/(8) ∫₀(π)/(2) ⁴t dt ⁴t + 1 2I = (π)/(8) ∫₀(π)/(2) (1 + ²t) ²t dt ⁴t + 1
Step 2: Substitution and Algebraic Limits

Let t = y ²t dt = dy. When t = 0 y = 0, and when t = (π)/(2) y = ∞.

2I = (π)/(8) ∫₀∞ (1 + y²) dy1 + y⁴ I = (π)/(16) ∫₀∞ 1 + 1y²y² + 1y² dy

Now put y - (1)/(y) = p (1 + 1y²) dy = dp. When y → 0^+ p → -∞, and when y → ∞ p → ∞. Also, y² + 1y² = p² + 2 = p² + (√(2))².

I = (π)/(16) ∫-∞∞ dpp² + (√(2))² I = π16√(2) [ ⁻¹( p√(2)) ]-∞∞

$I = π16√(2) ( (π)/(2) - (-(π)/(2)) ) = π²16√(2) = √(2)π²32

Pattern Recognition

Sees: Integrand containing x in the numerator and symmetric trigonometric functions in the denominator. Shortcut: The elimination of x using King's property is standard. For integrals containing (1+y²)/(1+y⁴), dividing by y² transforms the denominator into a perfect square form (y-(1)/(y))²+2, making substitution trivial.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 11 Mathematics: Trigonometric Identities

More Definite Integrals Questions — jee_main_2025_04_april_evening

Practice all Definite Integrals previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)