Let $f$ $f$ be a differentiable function such that $2(x+2)^2f(x) - 3(x+2)^2 = 10int_0^x(t+2)f(t)dt$ for $x geq 0$ $x \geq 0$. Then $f(2)$ $f(2)$ is equal to ________.

Numerical Answer Type:

Enter a numerical value Answer: 19 +4 marks

Solution & Explanation

### Related Formula The Leibniz Integral Rule template allows direct differentiation of an integral with variable limits:

fracddxleft( int_0^x g(t) dt right) = g(x)

### Core Logic Differentiate both sides of the given functional equation with respect to

x

using the product rule:

fracddxleft[ 2(x+2)^2 f(x) - 3(x+2)^2 right] = fracddxleft[ 10int_0^x(t+2)f(t)dt right]

4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)

Since

x geq 0

, the factor

(x+2)

is strictly non-zero. Divide the entire equation by

2(x+2)

2f(x) + (x+2)f'(x) - 3 = 5f(x)

(x+2)f'(x) - 3f(x) = 3

### Step 1: Solve the First-Order Differential Equation Rearrange the expression into standard linear differential equation form where

y = f(x)

fracdydx - frac3x+2y = frac3x+2

Compute the Integrating Factor (I.F.):

textI.F. = e^int -frac3x+2 dx = e^-3ln(x+2) = (x+2)^-3

Multiply through by the I.F. and integrate:

y cdot (x+2)^-3 = int frac3x+2 cdot (x+2)^-3 dx = int 3(x+2)^-4 dx

fracf(x)(x+2)^3 = 3 cdot frac(x+2)^-3-3 + C = -(x+2)^-3 + C

f(x) = -1 + C(x+2)^3

### Step 2: Apply the Boundary Condition Find the boundary condition by substituting

x = 0

into the original integral equation equation:

2(0+2)^2 f(0) - 3(0+2)^2 = 10 int_0^0 (t+2)f(t) dt

8f(0) - 12 = 0 implies f(0) = frac128 = frac32

Substitute

x = 0

into our general solution formula:

f(0) = -1 + C(0+2)^3 implies frac32 = -1 + 8C

frac52 = 8C implies C = frac516

Thus, the explicit function is:

f(x) = -1 + frac516(x+2)^3

### Step 3: Evaluate at target point x = 2 Substitute

x = 2

into the final function equation:

f(2) = -1 + frac516(2+2)^3 = -1 + frac516(64)

f(2) = -1 + 5(4) = -1 + 20 = 19

### Pattern Recognition When an equation contains a variable integral limit

int_0^x

, differentiating both sides using the Leibniz rule converts it into a standard differential equation. The initial value is found by setting

x = 0

directly in the original expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Reference Study Guides

Definite Integrals

More Definite Integrals Previous-Year Questions

Q60 2025 Properties of Definite Integrals

The integral

int_0^pi frac(x + 3)sin x1 + 3cos^2x dx

is equal to:

A. $fracpisqrt3 (pi + 1)$ $\frac{\pi}{\sqrt{3}} (\pi + 1)$
B. $fracpisqrt3 (pi + 2)$ $\frac{\pi}{\sqrt{3}} (\pi + 2)$
C. $fracpi3sqrt3 (pi + 6)$ $\frac{\pi}{3\sqrt{3}} (\pi + 6)$
D. $fracpi2sqrt3 (pi + 4)$ $\frac{\pi}{2\sqrt{3}} (\pi + 4)$

Solution

### Related Formula King's Property of Definite Integrals:

int_a^b f(x) dx = int_a^b f(a + b - x) dx

### Core Logic Let the given integral be:

I = int_0^pi frac(x + 3)sin x1 + 3cos^2x dx quad dots (1)

Applying King's property (

x to pi - x

I = int_0^pi frac(pi - x + 3)sin(pi - x)1 + 3cos^2(pi - x) dx

I = int_0^pi frac(pi - x + 3)sin x1 + 3cos^2x dx quad dots (2)

### Step 1: Eliminate the x Variable Adding equations (1) and (2):

2I = int_0^pi frac[(x + 3) + (pi - x + 3)]sin x1 + 3cos^2x dx

2I = (pi + 6)int_0^pi fracsin x1 + 3cos^2x dx

Using the symmetric property

int_0^2a f(x)dx = 2int_0^a f(x)dx

f(2a-x)=f(x)

2I = 2(pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx

I = (pi + 6)int_0^pi/2 fracsin x1 + 3cos^2x dx

### Step 2: Solve Using Substitution Let

t = sqrt3cos x

. Then

dt = -sqrt3sin x dx implies sin x dx = -fracdtsqrt3

. Change in integration boundaries: - When

x = 0 implies t = sqrt3

- When

x = pi/2 implies t = 0

Substituting into the integral:

I = (pi + 6) int_sqrt3^0 frac-dt/sqrt31 + t^2 = fracpi + 6sqrt3 int_0^sqrt3 fracdt1 + t^2

I = fracpi + 6sqrt3 left[ tan^-1t right]_0^sqrt3 = fracpi + 6sqrt3 left( tan^-1sqrt3 - 0 right)

I = fracpi + 6sqrt3 cdot fracpi3 = fracpi3sqrt3(pi + 6)

### Pattern Recognition Whenever you encounter a linear

x

factor multiplying trigonometric components in a definite integral with symmetric limits like

0

pi

, executing King's property first is almost guaranteed to cleanly wipe out that variable element. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

Q58 2025 Properties of Definite Integrals

Let

f(x)

be a a positive function and

I_1 = int_-frac12^1 2xf(2x(1 - 2x)) \, dx

and

I_2 = int_-1^2 f(x(1 - x)) \, dx

. Then the value of

fracI_2I_1

is equal to

A. $9$ $9$
B. $6$ $6$
C. $12$ $12$
D. $4$ $4$

Solution

### Related Formula

int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx

### Core Logic Perform variable substitution to match the arguments and limit bounds across both separate integral functions before invoking King's property. ### Step 1: Perform Base Transformation Substitution In

I_1

, let

2x = t implies 2dx = dt

. Limits mapping:

x = -1/2 implies t = -1

;

x = 1 implies t = 2

I_1 = frac12 int_-1^2 t f(t(1-t)) \, dt implies 2I_1 = int_-1^2 t f(t(1-t)) \, dt

### Step 2: Invoke Integral Mirror Properties Apply the identity using parameters

(a+b-t) = (1-t)

2I_1 = int_-1^2 (1-t) f((1-t)(1-(1-t))) \, dt

2I_1 = int_-1^2 f(t(1-t)) \, dt - int_-1^2 t f(t(1-t)) \, dt

### Step 3: Final Matrix Matching Evaluation Notice component blocks align exactly with

I_2

definition values:

2I_1 = I_2 - 2I_1 implies 4I_1 = I_2

fracI_2I_1 = 4

### Pattern Recognition Symmetric transformations highlighting factor expressions like

x(1-x)

coupled with an external linear multiplier term

x

naturally simplify to half-weight area forms using reflection rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

Q61 2025 Integration of Absolute Value Functions

The integral

int_-1^frac32left(left|pi^2mathrmxsin (pi mathrmx)right|right) dx

is equal to :

A. $3 + 2pi$ $3 + 2\pi$
B. $4 + pi$ $4 + \pi$
C. $1 + 3pi$ $1 + 3\pi$
D. $2 + 3pi$ $2 + 3\pi$

Solution

### Related Formula

int x sin(pi x) \, dx = -fracxpicos(pi x) + fracsin(pi x)pi^2

### Core Logic Track sign configurations across the target integration segments to drop absolute modulus walls effectively at clean quadrant intervals. ### Step 1: Break Apart the Modulus Domain For

x in [-1, 1]

, product value elements

x sin(pi x) ge 0

. For

x in [1, 3/2]

, values drop below zero:

I = pi^2 left\ int_-1^1 x sin(pi x) \, dx - int_1^3/2 x sin(pi x) \, dx right\

### Step 2: Integrate the Even Function Block Since

x sin(pi x)

is symmetric and even:

int_-1^1 x sin(pi x) \, dx = 2 int_0^1 x sin(pi x) \, dx = 2 left[ -fracxpicos(pi x) + fracsin(pi x)pi^2 right]_0^1 = frac2pi

### Step 3: Subtract the Inverse Segment Evaluating boundary limits across the secondary phase track:

int_1^3/2 x sin(pi x) \, dx = left[ -fracxpicos(pi x) + fracsin(pi x)pi^2 right]_1^3/2 = left( 0 - frac1pi^2 right) - left( frac1pi right) = -frac1pi^2 - frac1pi

I = pi^2 left\ frac2pi - left(-frac1pi^2 - frac1piright) right\ = pi^2 left( frac3pi + frac1pi^2 right) = 3pi + 1

### Pattern Recognition Products of two odd tracking metrics (like linear variable

x

matched with sinusoidal waves) yield overall even systems, enabling rapid evaluation over center-aligned domains. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

Q63 2025 Properties of Definite Integrals (King's Property)

int_-fracpi2^fracpi2frac96x^2cos^2x(1 + e^x) dx = pi (alpha pi^2 +beta),alpha ,beta in mathbbZ,

then

(alpha + beta)^2

equals: (1) 144 (2) 196 (3) 100 (4) 64

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

Solution

### Related Formula King's property for definite integration:

int_a^b f(x) dx = int_a^b f(a+b-x) dx

### Core Logic Apply the identity

x to -x

to the integral:

I = int_-fracpi2^fracpi2 frac96x^2 cos^2 x1 + e^x dx = int_-fracpi2^fracpi2 frac96x^2 cos^2 x1 + e^-x dx

### Step 1: Adding both integral variations Adding the equations eliminates the exponential denominator term (

1+e^x

2I = int_-fracpi2^fracpi2 96x^2 cos^2 x cdot left[frac11+e^x + frace^x1+e^xright] dx

I = 48 int_0^fracpi2 x^2 (1 + cos 2x) dx

### Step 2: Evaluating the integrated components Integrating by parts gives:

I = pi (2pi^2 - 12)

Matching coefficients with the template:

alpha = 2

and \beta = -12.

(alpha + beta)^2 = (2 - 12)^2 = (-10)^2 = 100

### Pattern Recognition Exponential denominators like

1+e^x

in symmetric integral intervals are prime candidates for simplification using King's property. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Definite Integrals

Q68 2025 Properties of Definite Integrals

Let

f(x) + 2fleft(frac1xright) = x^2 + 5

and

2 mathrm g (mathrm x) - 3 mathrm g left(frac 12right) = mathrm x, mathrm x > 0

. If

alpha = int_ 1 ^ 2 f (x) d x

, and

beta = int_ 1 ^ 2 g (x) d x

, then the value of

9alpha + beta

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) + 2fleft(frac1xright) = x^2 + 5

Replace

x

with

frac1x

fleft(frac1xright) + 2f(x) = frac1x^2 + 5

Multiplying this new equation by

2

and subtracting the original Equation 1 eliminates the

fleft(frac1xright)

term:

4f(x) + 2fleft(frac1xright) - left(f(x) + 2fleft(frac1xright)right) = 2left(frac1x^2 + 5right) - (x^2 + 5)

3f(x) = frac2x^2 - x^2 + 5 implies f(x) = frac23x^2 - fracx^23 + frac53

### Step 1: Finding alpha Integrate

f(x)

from

1

2

alpha = int_1^2 left( frac23x^2 - fracx^23 + frac53 right) dx = left[ -frac23x - fracx^39 + frac5x3 right]_1^2

alpha = left( -frac13 - frac89 + frac103 right) - left( -frac23 - frac19 + frac53 right) = frac199 - frac89 = frac119

Thus,

9alpha = 11

. ### Step 2: Solving for g(x) and finding beta We are given

2g(x) - 3gleft(frac12right) = x

. Substitute

x = frac12

2gleft(frac12right) - 3gleft(frac12right) = frac12 implies -gleft(frac12right) = frac12 implies gleft(frac12right) = -frac12

Substitute this constant value back into the original equation:

2g(x) - 3left(-frac12right) = x implies 2g(x) + frac32 = x implies g(x) = fracx2 - frac34

Now find

beta

beta = int_1^2 left( fracx2 - frac34 right) dx = left[ fracx^24 - frac3x4 right]_1^2 = left( 1 - frac32 right) - left( frac14 - frac34 right) = -frac12 - left(-frac12right) = 0

### Step 3: Calculating 9alpha + beta Combining our values:

9alpha + beta = 11 + 0 = 11

### Pattern Recognition Functional equations involving

x to frac1x

are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Functional Equations

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Let $f$ $f$ be a differentiable function such that $2(x+2)^2f(x) - 3(x+2)^2 = 10int_0^x(t+2)f(t)dt$ for $x geq 0$ $x \geq 0$. Then $f(2)$ $f(2)$ is equal to ________.

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Let f$f$ be a differentiable function such that 2(x+2)^2f(x) - 3(x+2)^2 = 10int_0^x(t+2)f(t)dt$2(x+2)^{2}f(x) - 3(x+2)^{2} = 10\int_{0}^{x}(t+2)f(t)dt$ for x geq 0$x \geq 0$. Then f(2)$f(2)$ is equal to ________.

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Let $f$ $f$ be a differentiable function such that $2(x+2)^2f(x) - 3(x+2)^2 = 10int_0^x(t+2)f(t)dt$ for $x geq 0$ $x \geq 0$. Then $f(2)$ $f(2)$ is equal to ________.