JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f$ be a real valued continuous function on $[0, 1]$ and $f (x) = x + \int_{0}^{1} (x - t) f (t) d t$ . Then which of the following points $(x, y)$ lies on the curve $y = f (x)$ ?

Q: Why can we treat the integral as a constant? Isn't it a function of $x$?

That is a brilliant question! Look closely at the integral $\int_{0}^{1} (x - t) f(t) dt$. The variable of integration is $t$, and the limits are $0$ and $1$. Once you integrate with respect to $t$ and apply the limits, the variable $t$ disappears. The result is a number that depends only on the function $f(t)$ over the interval $[0, 1]$. Since $f(t)$ is fixed, the integral is just a constant. You are not integrating with respect to $x$, so $x$ behaves like a constant coefficient during the integration process.

Q: Why did we get a system of two equations?

Because our function $f(x)$ ended up having two unknown constants, $A$ and $B$. When we expanded the integral, we found that $f(x)$ depends on both $\int f(t) dt$ and $\int t f(t) dt$. Since we have two unknowns, we need two independent equations to solve for them. By substituting our expression for $f(x)$ back into the definitions of $A$ and $B$, we generated exactly those two equations. It is like solving a puzzle where every piece fits perfectly.

Select Answer:

Visualized Solution

$f (x) = x + \int_{0}^{1} (x - t) f (t) d t$

Given equation: $f (x) = x + \int_{0}^{1} (x - t) f (t) d t$
Domain: $x \in [0, 1]$
Goal: Find $f (x)$ and check which point lies on $y = f (x)$ .

$f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t f (t) d t$

Expand the integrand: $(x - t) f (t) = x f (t) - t f (t)$
Distribute the integral over the terms.
Since $t$ is the variable of integration, $x$ can be pulled out of the first integral.

$f (x) = (1 + A) x - B$

Let $A = \int_{0}^{1} f (t) d t$
Let $B = \int_{0}^{1} t f (t) d t$
Substitute $A$ and $B$ back: $f (x) = x + A x - B$
Simplify: $f (x) = (1 + A) x - B$

$A = \int_{0}^{1} ((1 + A) t - B) d t$

We defined $A = \int_{0}^{1} f (t) d t$ .
Replace $f (t)$ with $(1 + A) t - B$ .
$A = \int_{0}^{1} ((1 + A) t - B) d t$

$A + 2 B = 1$

Integrate: $A = [(1 + A) \frac{t ^{2}}{2} - B t]_{0}^{1}$
Substitute limits: $A = \frac{1 + A}{2} - B$
Multiply by $2$ : $2 A = 1 + A - 2 B$
Rearrange: $A + 2 B = 1$ (Equation 1)

$B = \int_{0}^{1} t ((1 + A) t - B) d t$

We defined $B = \int_{0}^{1} t f (t) d t$ .
Replace $f (t)$ with $(1 + A) t - B$ .
$B = \int_{0}^{1} ((1 + A) t^{2} - B t) d t$

$2 A - 9 B = - 2$

Integrate: $B = [(1 + A) \frac{t ^{3}}{3} - B \frac{t ^{2}}{2}]_{0}^{1}$
Substitute limits: $B = \frac{1 + A}{3} - \frac{B}{2}$
Multiply by $6$ : $6 B = 2 (1 + A) - 3 B$
Rearrange: $2 A - 9 B = - 2$ (Equation 2)

$A = \frac{5}{13}, B = \frac{4}{13}$

From Eq 1: $A = 1 - 2 B$
Substitute into Eq 2: $2 (1 - 2 B) - 9 B = - 2$
$2 - 4 B - 9 B = - 2 ⟹ - 13 B = - 4 ⟹ B = \frac{4}{13}$
Find $A$ : $A = 1 - 2 (\frac{4}{13}) = \frac{5}{13}$

$f (x) = \frac{18 x - 4}{13}$

Recall: $f (x) = (1 + A) x - B$
Substitute $A = \frac{5}{13}$ and $B = \frac{4}{13}$ :
$f (x) = (1 + \frac{5}{13}) x - \frac{4}{13}$
$f (x) = \frac{18}{13} x - \frac{4}{13} = \frac{18 x - 4}{13}$

$f (6) = \frac{18 ( 6 ) - 4}{13} = 8$

We need to check which point $(x, y)$ lies on $y = \frac{18 x - 4}{13}$ .
Test $(6, 8)$ : Substitute $x = 6$ .
$f (6) = \frac{108 - 4}{13} = \frac{104}{13} = 8$
The point $(6, 8)$ satisfies the equation.

Result: $(6, 8)$ lies on $f (x)$

Definite integrals with constant limits act as constant coefficients.
The function is a straight line: $13 y = 18 x - 4$ .
Other points $(2, 4), (1, 2), (4, 17)$ do not satisfy the equation.
Correct Option: $(6, 8)$

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The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

Imagine you are standing before a complex-looking equation:

f (x) = x + \int_{0}^{1} (x - t) f (t) d t

At first glance, it feels intimidating. It is an integral equation, where the function

f (x)

is defined by an integral that itself contains

f (t)

However, in the world of JEE Advanced, these problems are often elegant puzzles waiting to be unraveled.

The Power of Constants

The key to this problem lies in the integral

\int_{0}^{1} (x - t) f (t) d t

. Notice the limits of integration are

0

and

1

. These are constants!

When you integrate with respect to

t

and plug in the limits, the variable

t

vanishes. The result is just a number.

Let us expand the integrand:

(x - t) f (t) = x f (t) - t f (t)

. Now, we can distribute the integral:

f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t f (t) d t

Let us define two constants:

A = \int_{0}^{1} f (t) d t

and

B = \int_{0}^{1} t f (t) d t

Suddenly, the equation simplifies to

f (x) = x + A x - B

, or more cleanly:

f (x) = (1 + A) x - B

This reveals that

f (x)

is a linear function.

Solving the System

Now that we know

f (x)

is a line, we can find

A

and

B

. We substitute

f (t) = (1 + A) t - B

back into our definitions.

For

A

, we have:

A = \int_{0}^{1} ((1 + A) t - B) d t

Integrating this, we get:

A = [(1 + A) \frac{t ^{2}}{2} - B t]_{0}^{1} = \frac{1 + A}{2} - B

Multiplying by

2

gives

2 A = 1 + A - 2 B

, which simplifies to the first linear equation:

A + 2 B = 1

Similarly, for

B

, we have:

B = \int_{0}^{1} t ((1 + A) t - B) d t = \int_{0}^{1} ((1 + A) t^{2} - B t) d t

Integrating this, we get:

B = [(1 + A) \frac{t ^{3}}{3} - B \frac{t ^{2}}{2}]_{0}^{1} = \frac{1 + A}{3} - \frac{B}{2}

Multiplying by

6

gives

6 B = 2 (1 + A) - 3 B

, which simplifies to the second linear equation:

2 A - 9 B = - 2

We now have a system of two linear equations:

A + 2 B = 1

and

2 A - 9 B = - 2

. Solving this system, we find:

A = \frac{5}{13}, B = \frac{4}{13}

Final Calculation

With

A

and

B

in hand, our function becomes:

f (x) = (1 + \frac{5}{13}) x - \frac{4}{13} = \frac{18 x - 4}{13}

Now, we test the point

(6, 8)

. Substituting

x = 6

f (6) = \frac{18 ( 6 ) - 4}{13} = \frac{108 - 4}{13} = \frac{104}{13} = 8

The point

(6, 8)

lies perfectly on the curve. This journey shows that even the most daunting integral equations can be tamed by recognizing the constants hidden within them.

Common Traps & FAQs

Why can we treat the integral as a constant? Isn't it a function of

x

That is a brilliant question! Look closely at the integral

\int_{0}^{1} (x - t) f (t) d t

. The variable of integration is

t

, and the limits are

0

and

1

. Once you integrate with respect to

t

and apply the limits, the variable

t

disappears. The result is a number that depends only on the function

f (t)

over the interval

[0, 1]

. Since

f (t)

is fixed, the integral is just a constant. You are not integrating with respect to

x

, so

x

behaves like a constant coefficient during the integration process.

Can I just differentiate both sides with respect to

x

Yes, you absolutely can! Using the Leibniz Integral Rule, differentiating

f (x) = x + \int_{0}^{1} (x - t) f (t) d t

gives

f^{'} (x) = 1 + \int_{0}^{1} f (t) d t

. This means

f^{'} (x)

is a constant, which implies

f (x)

must be a linear function of the form

f (x) = m x + c

. This is a powerful alternative approach! However, the substitution method we used is often more direct for this specific type of problem. Both paths lead to the same beautiful result.

Why did we get a system of two equations?

Because our function

f (x)

ended up having two unknown constants,

A

and

B

. When we expanded the integral, we found that

f (x)

depends on both

\int f (t) d t

and

\int t f (t) d t

. Since we have two unknowns, we need two independent equations to solve for them. By substituting our expression for

f (x)

back into the definitions of

A

and

B

, we generated exactly those two equations. It is like solving a puzzle where every piece fits perfectly.

Animated Solution for Mathematics - Definite Integration: Let $f$ be a real valued continuous function on $[0, 1]$ and $f (x) = x + \int_{0}^{1} (x - t) f (t) d t$ . Then which of the following points $(x, y)$ lies on the curve $y = f (x)$ ?

Select Answer:

Visualized Solution

$f (x) = x + \int_{0}^{1} (x - t) f (t) d t$

$f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t f (t) d t$

$f (x) = (1 + A) x - B$

$A = \int_{0}^{1} ((1 + A) t - B) d t$

$A + 2 B = 1$

$B = \int_{0}^{1} t ((1 + A) t - B) d t$

$2 A - 9 B = - 2$

$A = \frac{5}{13}, B = \frac{4}{13}$

$f (x) = \frac{18 x - 4}{13}$

$f (6) = \frac{18 ( 6 ) - 4}{13} = 8$

Result: $(6, 8)$ lies on $f (x)$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Power of Constants

Solving the System

Final Calculation

Similar Questions

If $f$ is a continuous function with $\int_{0}^{x} f (t) d t \to \infty$ as $∣ x ∣ \to \infty$ , then show that every line $y = m x$ intersects the curve $y^{2} + \int_{0}^{x} f (t) d t = 2$

Let $f : [0, \infty) \to [0, \infty)$ be defined as $f (x) = \int_{0}^{x} [y] d y$ where $[x]$ is the greatest integer less than or equal to $x$ . Which of the following is true?

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 $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f (t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for all $t \in (1, 3]$ . The largest possible interval in which $g (3)$ lies is:

If $y = f (x)$ makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then $\int_{0}^{2} x f^{'} (x) d x$ is

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y$ , then $e + f (0)$ is equal to .........

Let $f : R \to R$ be a twice differentiable function such that $f (2) = 1.$ If $F (x) = x f (x)$ for all $x \in R$ , $\int_{0}^{2} x F^{'} (x) d x = 6$ and $\int_{0}^{2} x^{2} F^{''} (x) d x = 40$ then $F^{'} (2) + \int_{0}^{2} F (x) d x$ is equal to:

Let $f : R \to R$ be a differentiable function such that its derivative $f^{'}$ is continuous and $f (π) = - 6$ . If $F : [0, π] \to R$ is defined by $F (x) = \int_{0}^{x} f (t) d t$ , and if $\int_{0}^{π} (f^{'} (x) + F (x)) cos x d x = 2$ , then the value of $f (0)$ is ____.

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=x+∫01​(x−t)f(t)dt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=x+x∫01​f(t)dt−∫01​tf(t)dt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=(1+A)x−B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=∫01​((1+A)t−B)dt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+2B=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=∫01​t((1+A)t−B)dt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2A−9B=−2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=135​,B=134​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=1318x−4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(6)=1318(6)−4​=8

Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (6,8) lies on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Power of Constants

Solving the System

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f is a continuous function with ∫0x​f(t)dt→∞ as ∣x∣→∞, then show that every line y=mx intersects the curve y2+∫0x​f(t)dt=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[0,∞)→[0,∞) be defined as f(x)=∫0x​[y]dy where [x] is the greatest integer less than or equal to x. Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=f(x) makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then ∫02​xf′(x)dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) satisfies the relation f(x)=ex+∫01​(y+xex)f(y)dy, then e+f(0) is equal to .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0x​f(t)dt, where f is such that 21​≤f(t)≤1 for t∈[0,1] and 0≤f(t)≤21​ for t∈[1,2]. Then g(2) satisfies the inequality

$f (x) = x + \int_{0}^{1} (x - t) f (t) d t$

$f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t f (t) d t$

$f (x) = (1 + A) x - B$

$A = \int_{0}^{1} ((1 + A) t - B) d t$

$A + 2 B = 1$

$B = \int_{0}^{1} t ((1 + A) t - B) d t$

$2 A - 9 B = - 2$

$A = \frac{5}{13}, B = \frac{4}{13}$

$f (x) = \frac{18 x - 4}{13}$

$f (6) = \frac{18 ( 6 ) - 4}{13} = 8$

Result: $(6, 8)$ lies on $f (x)$

If $f$ is a continuous function with $\int_{0}^{x} f (t) d t \to \infty$ as $∣ x ∣ \to \infty$ , then show that every line $y = m x$ intersects the curve $y^{2} + \int_{0}^{x} f (t) d t = 2$

Let $f : [0, \infty) \to [0, \infty)$ be defined as $f (x) = \int_{0}^{x} [y] d y$ where $[x]$ is the greatest integer less than or equal to $x$ . Which of the following is true?

If $y = f (x)$ makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then $\int_{0}^{2} x f^{'} (x) d x$ is

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y$ , then $e + f (0)$ is equal to .........

Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality