JEE Main 2022 (28 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be continuous function satisfying $f (x) + f (x + k) = n$ , for all $x \in R$ where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4 nk} f (x) d x$ and $I_{2} = \int_{- k}^{3 k} f (x) d x$ , then

Q: Does the starting point of the integral matter for periodic functions?

Great question! For a periodic function with period $T$, the integral $\int_{a}^{a+T} f(x) dx$ is the same for any $a$. Think of it as sliding a window of length $T$ along the wave; the area trapped inside remains constant.

Q: How did we know to split the integral at $k$?

The functional equation gives us a relationship between $f(x)$ and $f(x+k)$. Since our period is $2k$, splitting the integral at $k$ allows us to pair $f(x)$ and $f(x+k)$ together, which simplifies perfectly to the constant $n$.

Select Answer:

Visualized Solution

Given Functional Equation

Given: $f (x) + f (x + k) = n$
Where $k > 0$ and $n \in Z^{+}$
We need to evaluate $I_{1}$ and $I_{2}$ .

Finding the Period

Substitute $x \to x + k$ in the given equation.
$f (x + k) + f (x + 2 k) = n$
Original equation: $f (x) + f (x + k) = n$

Establishing Periodicity

Equate the two expressions:
$f (x) + f (x + k) = f (x + k) + f (x + 2 k)$
Cancel $f (x + k)$ from both sides.
$f (x) = f (x + 2 k)$
Conclusion: $f (x)$ is periodic with period $T = 2 k$ .

Integral Over One Period

Let's evaluate the integral over one full period: $\int_{0}^{2 k} f (x) d x$
Split the integral at $x = k$ :
$\int_{0}^{2 k} f (x) d x = \int_{0}^{k} f (x) d x + \int_{k}^{2 k} f (x) d x$

Applying Substitution

Consider the second integral: $\int_{k}^{2 k} f (x) d x$
Substitute $x = t + k ⟹ d x = d t$
Limits change: When $x = k, t = 0$ . When $x = 2 k, t = k$ .
$\int_{k}^{2 k} f (x) d x = \int_{0}^{k} f (t + k) d t$

Evaluating the Fundamental Integral

Combine the integrals: $\int_{0}^{2 k} f (x) d x = \int_{0}^{k} f (x) d x + \int_{0}^{k} f (x + k) d x$
$\int_{0}^{2 k} f (x) d x = \int_{0}^{k} [f (x) + f (x + k)] d x$
Substitute $f (x) + f (x + k) = n$ :
$\int_{0}^{k} n d x = [n x]_{0}^{k} = nk$

Calculating $I_{1}$

$I_{1} = \int_{0}^{4 nk} f (x) d x$
The period is $T = 2 k$ .
Number of periods in the interval $[0, 4 nk]$ is $m = \frac{4 nk}{2 k} = 2 n$ .

Evaluating $I_{1}$

Property of periodic functions: $\int_{0}^{m T} f (x) d x = m \int_{0}^{T} f (x) d x$
$I_{1} = 2 n \int_{0}^{2 k} f (x) d x$
Substitute $\int_{0}^{2 k} f (x) d x = nk$ :
$I_{1} = 2 n (nk) = 2 n^{2} k$

Calculating $I_{2}$

$I_{2} = \int_{- k}^{3 k} f (x) d x$
Length of the interval $= 3 k - (- k) = 4 k$ .
Since $4 k = 2 (2 k) = 2 T$ , the interval covers exactly 2 periods.

Evaluating $I_{2}$

Property: Integral over any interval of length $m T$ is $m \int_{0}^{T} f (x) d x$ .
$I_{2} = 2 \int_{0}^{2 k} f (x) d x$
Substitute $\int_{0}^{2 k} f (x) d x = nk$ :
$I_{2} = 2 (nk) = 2 nk$

Finding the Final Relation

We have $I_{1} = 2 n^{2} k$ and $I_{2} = 2 nk$ .
Let's test the given options. Consider $I_{1} + n I_{2}$ :
$I_{1} + n I_{2} = 2 n^{2} k + n (2 nk)$
$I_{1} + n I_{2} = 2 n^{2} k + 2 n^{2} k = 4 n^{2} k$
This matches the third option perfectly.

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The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

We are given the functional equation:

f (x) + f (x + k) = n

This equation implies that the function is periodic. To prove this, we consider the equation at two different points.

First, we have:

f (x) + f (x + k) = n

Next, replacing

x

with

x + k

, we obtain:

f (x + k) + f (x + 2 k) = n

Unmasking the Periodicity

Since both expressions equal

n

, we can equate them:

f (x) + f (x + k) = f (x + k) + f (x + 2 k)

By subtracting

f (x + k)

from both sides, we arrive at the fundamental property:

f (x) = f (x + 2 k)

This confirms that the function

f (x)

is periodic with a period of

T = 2 k

The Integral over the Period

To understand the behavior of the function, we evaluate the integral over one full period,

2 k

\int_{0}^{2 k} f (x) d x = \int_{0}^{k} f (x) d x + \int_{k}^{2 k} f (x) d x

For the second integral, we use the substitution

x = t + k

, which implies

d x = d t

. The limits change from

[k, 2 k]

[0, k]

\int_{k}^{2 k} f (x) d x = \int_{0}^{k} f (t + k) d t

Combining these, we get:

\int_{0}^{2 k} f (x) d x = \int_{0}^{k} [f (x) + f (x + k)] d x

Since

f (x) + f (x + k) = n

, the integral simplifies to:

\int_{0}^{2 k} f (x) d x = \int_{0}^{k} n d x = nk

Solving for $I_{1}$ and $I_{2}$

We now calculate

I_{1} = \int_{0}^{4 nk} f (x) d x

. The interval length is

4 nk

, which contains

\frac{4 nk}{2 k} = 2 n

periods.

I_{1} = 2 n \times (nk) = 2 n^{2} k

Next, we calculate

I_{2} = \int_{- k}^{3 k} f (x) d x

. The interval length is

3 k - (- k) = 4 k

, which is exactly two periods.

I_{2} = 2 \times (nk) = 2 nk

Final Calculation

We evaluate the expression

I_{1} + n I_{2}

I_{1} + n I_{2} = 2 n^{2} k + n (2 nk)

I_{1} + n I_{2} = 2 n^{2} k + 2 n^{2} k

The final result is $4 n^{2} k$ .

Common Traps & FAQs

Why does

f (x) + f (x + k) = n

imply periodicity? I thought it just meant the function is constant.

That is a common trap! It does not mean the function is constant. It means the function has a 'memory' of its values separated by

k

. By shifting the equation, we prove that the function repeats its values every

2 k

. It is a wave-like property, not a flat line.

Does the starting point of the integral matter for periodic functions?

Great question! For a periodic function with period

T

, the integral

\int_{a}^{a + T} f (x) d x

is the same for any

a

. Think of it as sliding a window of length

T

along the wave; the area trapped inside remains constant.

How did we know to split the integral at

k

The functional equation gives us a relationship between

f (x)

and

f (x + k)

. Since our period is

2 k

, splitting the integral at

k

allows us to pair

f (x)

and

f (x + k)

together, which simplifies perfectly to the constant

n

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be continuous function satisfying $f (x) + f (x + k) = n$ , for all $x \in R$ where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4 nk} f (x) d x$ and $I_{2} = \int_{- k}^{3 k} f (x) d x$ , then

Select Answer:

Visualized Solution

Given Functional Equation

Finding the Period

Establishing Periodicity

Integral Over One Period

Applying Substitution

Evaluating the Fundamental Integral

Calculating $I_{1}$

Evaluating $I_{1}$

Calculating $I_{2}$

Evaluating $I_{2}$

Finding the Final Relation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Unmasking the Periodicity

The Integral over the Period

Solving for $I_{1}$ and $I_{2}$

Final Calculation

Similar Questions

Let $f : R \to R$ be a continuous function such that $f (x) + f (x + 1) = 2$ , for all $x \in R$ . If $I_{1} = \int_{0}^{8} f (x) d x$ and $I_{2} = \int_{- 1}^{3} f (x) d x$ , then the value of $I_{1} + 2 I_{2}$ is equal to ____

Let $T > 0$ be a fixed real number. Suppose $f$ is a continuous function such that for all $x \in R$ , $f (x + T) = f (x)$ . If $I = \int_{0}^{T} f (x) d x$ then the value of $\int_{3}^{3 + 3 T} f (2 x) d x$ is

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If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

Let for $f (x) = 7 tan^{8} x + 7 tan^{6} x - 3 tan^{4} x - 3 tan^{2} x$ , $I_{1} = \int_{0}^{π /4} f (x) d x$ and $I_{2} = \int_{0}^{π /4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to:

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

If $f$ and $g$ are continuous function on $[0, a]$ satisfying $f (x) = f (a - x)$ and $g (x) + g (a - x) = 2$ , then show that $\int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (x) d x$

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Let $f (x)$ be a positive function and $I_{1} = \int_{1/2}^{1} 2 x f (2 x (1 - 2 x)) d x$ and $I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I _{2}}{I _{1}}$ is equal to ________

Select Answer:

Visualized Solution

Given Functional Equation

Finding the Period

Establishing Periodicity

Integral Over One Period

Applying Substitution

Evaluating the Fundamental Integral

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Finding the Final Relation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Unmasking the Periodicity

The Integral over the Period

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Final Calculation

Similar Questions

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Calculating $I_{1}$

Evaluating $I_{1}$

Calculating $I_{2}$

Evaluating $I_{2}$

Solving for $I_{1}$ and $I_{2}$

Let $T > 0$ be a fixed real number. Suppose $f$ is a continuous function such that for all $x \in R$ , $f (x + T) = f (x)$ . If $I = \int_{0}^{T} f (x) d x$ then the value of $\int_{3}^{3 + 3 T} f (2 x) d x$ is

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

Let for $f (x) = 7 tan^{8} x + 7 tan^{6} x - 3 tan^{4} x - 3 tan^{2} x$ , $I_{1} = \int_{0}^{π /4} f (x) d x$ and $I_{2} = \int_{0}^{π /4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to:

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

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