JEE Main 2024 (31 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be a function defined by $f (x) = \frac{4 ^{x}}{4 ^{x} + 2}$ and $M = \int_{f (a)}^{f (1 - a)} x sin^{4} (x (1 - x)) d x, N = \int_{f (a)}^{f (1 - a)} sin^{4} (x (1 - x)) d x; a \neq = \frac{1}{2}$ . If $α M = β N, α, β \in N$ , then the least value of $α^{2} + β^{2}$ is equal to _______.

Q: Why does $f(x) + f(1-x) = 1$?

Let's look at the algebra: $f(x) = \frac{4^x}{4^x+2}$. Then $f(1-x) = \frac{4^{1-x}}{4^{1-x}+2}$. Multiply the numerator and denominator of the second term by $4^x$, and you get $\frac{4}{4 + 2 \cdot 4^x}$, which simplifies to $\frac{2}{2 + 4^x}$. Adding $\frac{4^x}{4^x+2} + \frac{2}{4^x+2}$ gives $\frac{4^x+2}{4^x+2} = 1$. It is a perfect, elegant cancellation!

Enter Numerical Value:

Visualized Solution

Analyzing $f (x)$

Given function: $f (x) = \frac{4 ^{x}}{4 ^{x} + 2}$
We need to evaluate integrals with limits $f (a)$ and $f (1 - a)$ .

Property of $f (x)$

Let's evaluate $f (1 - x)$ .
$f (1 - x) = \frac{4 ^{1 - x}}{4 ^{1 - x} + 2} = \frac{2}{2 + 4 ^{x}}$
Summing them: $f (x) + f (1 - x) = 1$

Integral Limits

Let lower limit $p = f (a)$
Let upper limit $q = f (1 - a)$
From our property: $p + q = 1$

The Integral $M$

$M = \int_{p}^{q} x sin^{4} (x (1 - x)) d x$
The limits sum to $1$ ( $p + q = 1$ )

King's Rule

King's Rule: $\int_{p}^{q} g (x) d x = \int_{p}^{q} g (p + q - x) d x$
Since $p + q = 1$ , substitute $x \to 1 - x$

Applying the Substitution

$M = \int_{p}^{q} (1 - x) sin^{4} ((1 - x) (1 - (1 - x))) d x$
Simplify the argument: $(1 - x) (1 - (1 - x)) = (1 - x) x$

The Unchanged Sine Term

$M = \int_{p}^{q} (1 - x) sin^{4} (x (1 - x)) d x$
Notice: The sine term is completely unchanged!

Expanding the Expression

Expand: $(1 - x) sin^{4} (x (1 - x))$
$M = \int_{p}^{q} sin^{4} (x (1 - x)) d x - \int_{p}^{q} x sin^{4} (x (1 - x)) d x$

Identifying $M$ and $N$

Recognize: $\int_{p}^{q} sin^{4} (x (1 - x)) d x = N$
Recognize: $\int_{p}^{q} x sin^{4} (x (1 - x)) d x = M$
Substitute: $M = N - M$

The Ratio of $M$ and $N$

Rearrange: $M + M = N$
$2 M = N$
Given condition: $α M = β N$

Comparing Coefficients

Compare $2 M = 1 N$ with $α M = β N$
For least $α, β \in N$ :
$α = 2$
$β = 1$

Calculating $α^{2} + β^{2}$

We need the least value of $α^{2} + β^{2}$
Substitute: $2^{2} + 1^{2}$
Result: $4 + 1 = 5$
Final Answer: $5$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Symmetry

The function is defined as

f (x) = \frac{4 ^{x}}{4 ^{x} + 2}

. To understand its behavior, we evaluate

f (1 - x)

f (1 - x) = \frac{4 ^{1 - x}}{4 ^{1 - x} + 2}

By multiplying the numerator and denominator by

4^{x}

, we transform the expression:

f (1 - x) = \frac{4}{4 + 2 \cdot 4 ^{x}} = \frac{2}{2 + 4 ^{x}}

Now, we observe the sum of the function and its symmetric counterpart:

f (x) + f (1 - x) = \frac{4 ^{x}}{4 ^{x} + 2} + \frac{2}{4 ^{x} + 2} = \frac{4 ^{x} + 2}{4 ^{x} + 2} = 1

This confirms that the function is symmetric about the point

(\frac{1}{2}, \frac{1}{2})

Applying the King's Rule

We define the integral

M = \int_{p}^{q} x sin^{4} (x (1 - x)) d x

, where

p = f (a)

and

q = f (1 - a)

. Note that

p + q = f (a) + f (1 - a) = 1

The King's Rule states that

\int_{p}^{q} g (x) d x = \int_{p}^{q} g (p + q - x) d x

. Substituting

p + q = 1

and

x \to 1 - x

M = \int_{p}^{q} (1 - x) sin^{4} ((1 - x) (1 - (1 - x))) d x

Simplifying the argument inside the sine function, we get

(1 - x) (x)

, which is identical to

x (1 - x)

. Thus:

M = \int_{p}^{q} (1 - x) sin^{4} (x (1 - x)) d x

The Elegant Cancellation

We expand the integral

M

as follows:

M = \int_{p}^{q} sin^{4} (x (1 - x)) d x - \int_{p}^{q} x sin^{4} (x (1 - x)) d x

Let

N = \int_{p}^{q} sin^{4} (x (1 - x)) d x

. The equation simplifies to:

M = N - M \Rightarrow 2 M = N

Given the relation

α M = β N

, we substitute

N = 2 M

α M = β (2 M) \Rightarrow α = 2 β

For the least natural numbers, we choose

β = 1

and

α = 2

Final Calculation

We are asked to find the value of

α^{2} + β^{2}

α^{2} + β^{2} = 2^{2} + 1^{2} = 4 + 1 = 5

The final result is 5.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

In this problem, we are not dealing with energy conservation, but with the symmetry of functions! The 'negative root' confusion usually arises in quadratic equations, but here, we are using the King's Rule. The beauty of this problem lies in the fact that

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

, which makes the sum of the limits

p + q = 1

. There are no roots to reject; we are simply transforming the variable to reveal the hidden structure of the integral.

How do I know when to use the King's Rule?

Great question! The King's Rule is your best friend whenever you see a definite integral

\int_{p}^{q} g (x) d x

where the sum of the limits

p + q

is a constant, especially if the integrand

g (x)

looks complicated but has a symmetric structure. Whenever you see

x

and

1 - x

appearing together, it is a massive signal to test the King's Rule.

Why does

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

Let's look at the algebra:

f (x) = \frac{4 ^{x}}{4 ^{x} + 2}

. Then

f (1 - x) = \frac{4 ^{1 - x}}{4 ^{1 - x} + 2}

. Multiply the numerator and denominator of the second term by

4^{x}

, and you get

\frac{4}{4 + 2 \cdot 4 ^{x}}

, which simplifies to

\frac{2}{2 + 4 ^{x}}

. Adding

\frac{4 ^{x}}{4 ^{x} + 2} + \frac{2}{4 ^{x} + 2}

gives

\frac{4 ^{x} + 2}{4 ^{x} + 2} = 1

. It is a perfect, elegant cancellation!

Enter Numerical Value:

Visualized Solution

Analyzing $f (x)$

Property of $f (x)$

Integral Limits

The Integral $M$

King's Rule

Applying the Substitution

The Unchanged Sine Term

Expanding the Expression

Identifying $M$ and $N$

The Ratio of $M$ and $N$

Comparing Coefficients

Calculating $α^{2} + β^{2}$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Symmetry

Applying the King's Rule

The Elegant Cancellation

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

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

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

Integral Limits

The Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

King's Rule

Applying the Substitution

The Unchanged Sine Term

Expanding the Expression

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

The Ratio of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Comparing Coefficients

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Symmetry

Applying the King's Rule

The Elegant Cancellation

Final Calculation

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Analyzing $f (x)$

Property of $f (x)$

The Integral $M$

Identifying $M$ and $N$

The Ratio of $M$ and $N$

Calculating $α^{2} + β^{2}$

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

Let $f (θ) = sin θ + \int_{- π /2}^{π /2} (sin θ + t cos θ) f (t) d t$ . Then the value of $\int_{0}^{π /2} f (θ) d θ$ is ______.

Let $f (x) = x + \frac{a}{π ^{2} - 4} sin x + \frac{b}{π ^{2} - 4} cos x, x \in R$ be a function which satisfies $f (x) = x + \int_{0}^{π /2} sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to

Let $f (x) = \int_{0}^{x} g (t) lo g_{e} (\frac{1 - t}{1 + t}) d t$ , where $g$ is a continuous odd function. If $\int_{- π /2}^{π /2} (f (x) + \frac{x ^{2} c o s x}{1 + e ^{x}}) d x = (\frac{π}{α})^{2} - α$ , then $α$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x, x > 0$ . If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ 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

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