JEE Main 2021 (February) (24 February Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

Q: Why does the constant of integration result in $f(x) + f(2-x) = C$?

When you integrate $f'(x) = f'(2-x)$, the left side becomes $f(x) + C_1$. The right side becomes $-f(2-x) + C_2$. When you move the terms to one side, the constants $C_1$ and $C_2$ merge into a single constant $C$. This relation $f(x) + f(2-x) = C$ is a powerful invariant. It tells us that the sum of the function values at symmetric points is constant throughout the interval.

Q: How do I know when to use King's Property?

King's Property, $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, is your best friend whenever you see a definite integral with limits $a$ and $b$, especially when the integrand contains terms like $(a+b-x)$. If you see symmetry in the limits and the function, or if the integrand looks difficult to integrate directly, King's Property is almost certainly the intended path to simplify the expression.

Select Answer:

Progressive Hints

Hint 1

The Spark: Do not get overwhelmed trying to find the explicit form of

f (x)

. The problem gives you a symmetry condition in the derivative:

f^{'} (x) = f^{'} (2 - x)

. This implies that the slope at any point

x

is identical to the slope at the reflected point

2 - x

. This is a massive hint that the function itself possesses a hidden symmetry. Focus on the relationship between

f (x)

and

f (2 - x)

rather than the function's specific algebraic form.

Hint 2

The Arsenal: You need to bridge the gap between the derivative and the function. Integrate both sides of the given equation

f^{'} (x) = f^{'} (2 - x)

. Remember the chain rule when integrating

f^{'} (2 - x)

—it introduces a negative sign because the derivative of the inner function

(2 - x)

- 1

. This will lead you to a constant relation:

f (x) + f (2 - x) = C

Hint 3

The Execution: Once you establish that

f (x) + f (2 - x) = C

, use the boundary conditions

f (0) = 1

and

f (2) = e^{2}

to find the value of

C

. Finally, define your integral

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

and apply King's Property:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

. By adding the two versions of the integral together, the

f (x)

and

f (2 - x)

terms will combine into the constant

C

, making the integration trivial.

Visualized Solution

Problem Setup for $f (x)$

Given function $f (x)$ defined on $[0, 2]$ .
Boundary conditions: $f (0) = 1$ and $f (2) = e^{2}$ .
We need to find the area under the curve: $I = \int_{0}^{2} f (x) d x$ .

Analyzing $f^{'} (x) = f^{'} (2 - x)$

We are given a special relation for the derivative:
$f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ .
This tells us about the symmetry of the slopes.

Integrating the Equation

To find information about $f (x)$ , we integrate both sides with respect to $x$ .
$\int f^{'} (x) d x = \int f^{'} (2 - x) d x$

Applying the Chain Rule

Left side: $\int f^{'} (x) d x = f (x)$
Right side: $\int f^{'} (2 - x) d x = - f (2 - x)$ (due to chain rule on $- x$ )
Result: $f (x) = - f (2 - x) + C$

Finding the Constant $C$

We have the relation: $f (x) + f (2 - x) = C$
We need to find the value of $C$ .
We will use the given boundary conditions: $f (0) = 1$ and $f (2) = e^{2}$ .

Substituting $x = 0$

Substitute $x = 0$ into the relation:
$f (0) + f (2 - 0) = C$
$f (0) + f (2) = C$

Calculating $C = 1 + e^{2}$

Substitute the known values: $f (0) = 1$ and $f (2) = e^{2}$ .
$1 + e^{2} = C$
Therefore, the functional relation is: $f (x) + f (2 - x) = 1 + e^{2}$

Setting Up $I = \int_{0}^{2} f (x) d x$

Let the required integral be $I$ .
$I = \int_{0}^{2} f (x) d x$
We need a property of definite integrals to utilize our functional relation.

Applying King's Property

Recall King's Property: $\int_{a}^{b} g (x) d x = \int_{a}^{b} g (a + b - x) d x$
Here, $a = 0$ and $b = 2$ .
Applying this to our integral: $I = \int_{0}^{2} f (2 - x) d x$

Adding the Two Integrals

We now have two expressions for $I$ :
$I = \int_{0}^{2} f (x) d x$
$I = \int_{0}^{2} f (2 - x) d x$
Adding them vertically: $2 I = \int_{0}^{2} f (x) d x + \int_{0}^{2} f (2 - x) d x$

Combining the Integrands

Since the limits of integration are the same, we can combine the integrands.
$2 I = \int_{0}^{2} [f (x) + f (2 - x)] d x$
Notice the term inside the bracket!

Substituting $f (x) + f (2 - x)$

From earlier: $f (x) + f (2 - x) = 1 + e^{2}$
Substitute this constant value into the integral:
$2 I = \int_{0}^{2} (1 + e^{2}) d x$

Evaluating the Constant Integral

Since $(1 + e^{2})$ is a constant, pull it out of the integral.
$2 I = (1 + e^{2}) \int_{0}^{2} 1 d x$
$2 I = (1 + e^{2}) [x]_{0}^{2}$

Final Calculation for $I$

Apply the limits: $[x]_{0}^{2} = (2 - 0) = 2$
$2 I = (1 + e^{2}) \cdot 2$
Divide both sides by $2$ :
$I = 1 + e^{2}$

Conclusion: $I = 1 + e^{2}$

The value of the integral $\int_{0}^{2} f (x) d x$ is $1 + e^{2}$ .
Comparing with the given options:
(A) $1 + e^{2}$
(B) $1 - e^{2}$
(C) $2 (1 - e^{2})$
(D) $2 (1 + e^{2})$
Correct Option is (A).

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

The condition

f^{'} (x) = f^{'} (2 - x)

is not merely an equation; it is a profound statement of symmetry. It indicates that the rate of change of the function at any point

x

is perfectly mirrored at the point

2 - x

To recover the function

f (x)

from its derivative, we must perform the inverse operation: integration.

The Master Equation

When we integrate both sides of the given condition, we must be careful with the chain rule. Integrating the left side yields

f (x)

, while integrating the right side

\int f^{'} (2 - x) d x

results in

- f (2 - x) + C

This leads us to the elegant relation:

f (x) + f (2 - x) = C

To determine the constant

C

, we utilize the provided boundary conditions

f (0) = 1

and

f (2) = e^{2}

. Substituting

x = 0

into our relation gives:

f (0) + f (2) = C

1 + e^{2} = C

Thus, our functional relation is fully defined as:

f (x) + f (2 - x) = 1 + e^{2}

The King's Property Finale

We now evaluate the integral

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

. We deploy the King's Property, which states that

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

Applying this with

a = 0

and

b = 2

, we obtain:

I = \int_{0}^{2} f (2 - x) d x

By adding the two expressions for

I

, we get:

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

Substituting our known constant sum into the integrand:

2 I = \int_{0}^{2} (1 + e^{2}) d x

Since

(1 + e^{2})

is a constant, we extract it from the integral:

2 I = (1 + e^{2}) \int_{0}^{2} 1 d x

2 I = (1 + e^{2}) \cdot [x]_{0}^{2}

2 I = (1 + e^{2}) \cdot 2

Dividing both sides by

2

, we arrive at the final result:

$I = 1 + e^{2}$

Common Traps & FAQs

Why can't we just solve for

f (x)

directly?

In many JEE Advanced problems, you are not expected to find the explicit function

f (x)

. The problem provides just enough information to determine the *integral* of the function, not the function itself. Trying to solve for

f (x)

is a trap that will waste your time. Always look at what the question asks for—if it asks for an integral, look for properties of integrals like King's Property or symmetry, rather than trying to reconstruct the function.

Why does the constant of integration result in

f (x) + f (2 - x) = C

When you integrate

f^{'} (x) = f^{'} (2 - x)

, the left side becomes

f (x) + C_{1}

. The right side becomes

- f (2 - x) + C_{2}

. When you move the terms to one side, the constants

C_{1}

and

C_{2}

merge into a single constant

C

. This relation

f (x) + f (2 - x) = C

is a powerful invariant. It tells us that the sum of the function values at symmetric points is constant throughout the interval.

How do I know when to use King's Property?

King's Property,

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

, is your best friend whenever you see a definite integral with limits

a

and

b

, especially when the integrand contains terms like

(a + b - x)

. If you see symmetry in the limits and the function, or if the integrand looks difficult to integrate directly, King's Property is almost certainly the intended path to simplify the expression.

Animated Solution for Mathematics - Definite Integration: Let $f (x)$ be a differentiable function defined on $[0, 2]$ such that $f^{'} (x) = f^{'} (2 - x)$ for all $x \in (0, 2)$ , $f (0) = 1$ and $f (2) = e^{2}$ . Then the value of $\int_{0}^{2} f (x) d x$ is:

Select Answer:

Visualized Solution

Problem Setup for $f (x)$

Analyzing $f^{'} (x) = f^{'} (2 - x)$

Integrating the Equation

Applying the Chain Rule

Finding the Constant $C$

Substituting $x = 0$

Calculating $C = 1 + e^{2}$

Setting Up $I = \int_{0}^{2} f (x) d x$

Applying King's Property

Adding the Two Integrals

Combining the Integrands

Substituting $f (x) + f (2 - x)$

Evaluating the Constant Integral

Final Calculation for $I$

Conclusion: $I = 1 + e^{2}$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Master Equation

The King's Property Finale

Similar Questions

Let the function $f : [0, 2] \to R$ be defined as $f (x) = {e^{m i n {x^{2}, x - [x]}}, e^{[x - l n x]}, x \in [0, 1) x \in [1, 2]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is

$\int_{0}^{2} [x^{2}] d x$ is

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Select Answer:

Visualized Solution

Problem Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)=f′(2−x)

Integrating the Equation

Applying the Chain Rule

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C=1+e2

Setting Up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=∫02​f(x)dx

Applying King's Property

Adding the Two Integrals

Combining the Integrands

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)+f(2−x)

Evaluating the Constant Integral

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Conclusion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=1+e2

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Master Equation

The King's Property Finale

Similar Questions

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Problem Setup for $f (x)$

Analyzing $f^{'} (x) = f^{'} (2 - x)$

Finding the Constant $C$

Substituting $x = 0$

Calculating $C = 1 + e^{2}$

Setting Up $I = \int_{0}^{2} f (x) d x$

Substituting $f (x) + f (2 - x)$

Final Calculation for $I$

Conclusion: $I = 1 + e^{2}$

$\int_{0}^{2} [x^{2}] d x$ is

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

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 .........

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

Let $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0$ . Then $18 \int_{1}^{2} f (x) d x$ is equal to

The value of the integral $\int_{e^{- 1}}^{e^{2}} ∣ \frac{l o g _{e} x}{x} ∣ d x$ is

The integral $\int_{1}^{e} {(\frac{x}{e})^{2 x} - (\frac{e}{x})^{x}} lo g_{e} x d x$ is equal to :