JEE Main 2021 (27 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let the domain of the function $f (x) = lo g_{4} (lo g_{5} (lo g_{3} (18 x - x^{2} - 77)))$ be $(a, b)$ . Then the value of the integral $\int_{a}^{b} \frac{s i n ^{3} x}{( s i n ^{3} x + s i n ^{3} ( a + b - x ))} d x$ is equal to

Q: Why did the inequality sign flip when we multiplied by $-1$?

It is a fundamental rule of inequalities! When you multiply or divide by a negative number, the direction of the inequality reverses. Think of it as reflecting the number line: if $5 > 2$, then $-5 < -2$. We did this to make the quadratic easier to factor.

Q: How do I know the domain is $(8, 10)$ and not something else?

We solved the quadratic inequality $x^2 - 18x + 80 < 0$. The roots are $8$ and $10$. Since the parabola opens upward, it is negative *between* the roots. Thus, the domain is strictly the interval $(8, 10)$.

Q: Is there any other way to solve this integral?

Technically, yes, but it would be a mathematical nightmare. Without King's Property, you would be stuck with a non-elementary integral that cannot be solved using standard techniques. In JEE, if you see this specific structure, it is a clear signal to use the symmetry property.

Enter Numerical Value:

Visualized Solution

The Nested Logarithm and Integral

Function: $f (x) = lo g_{4} (lo g_{5} (lo g_{3} (18 x - x^{2} - 77)))$
Domain is given as $(a, b)$ .
Goal: Find $a, b$ and evaluate $I = \int_{a}^{b} \frac{s i n ^{3} x}{s i n ^{3} x + s i n ^{3} ( a + b - x )} d x$ .

Unwrapping the Logarithms

$lo g_{4} (X)$ requires $X > 0$ .
$lo g_{5} (lo g_{3} (18 x - x^{2} - 77)) > 0$
$lo g_{3} (18 x - x^{2} - 77) > 5^{0} = 1$
$18 x - x^{2} - 77 > 3^{1} = 3$

Forming the Quadratic Inequality

$18 x - x^{2} - 77 > 3$
$- x^{2} + 18 x - 80 > 0$
Multiply by $- 1$ : $x^{2} - 18 x + 80 < 0$

Factoring the Quadratic

Factorizing $x^{2} - 18 x + 80 < 0$
Roots are $8$ and $10$ .
$(x - 8) (x - 10) < 0$

The Domain Interval $(a, b)$

Inequality holds for $x \in (8, 10)$ .
Comparing with $(a, b)$ , we get $a = 8$ and $b = 10$ .

Substituting Limits into the Integral

$I = \int_{a}^{b} \frac{s i n ^{3} x}{s i n ^{3} x + s i n ^{3} ( a + b - x )} d x$
Substitute $a = 8, b = 10$ :
$I = \int_{8}^{10} \frac{s i n ^{3} x}{s i n ^{3} x + s i n ^{3} ( 18 - x )} d x$

Applying King's Property

King's Property: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
Replace $x$ with $18 - x$ :
$I = \int_{8}^{10} \frac{s i n ^{3} ( 18 - x )}{s i n ^{3} ( 18 - x ) + s i n ^{3} ( 18 - ( 18 - x ))} d x$
$I = \int_{8}^{10} \frac{s i n ^{3} ( 18 - x )}{s i n ^{3} ( 18 - x ) + s i n ^{3} x} d x$

Summing the Two Equations

Add the original and new integrals:
$2 I = \int_{8}^{10} \frac{s i n ^{3} x + s i n ^{3} ( 18 - x )}{s i n ^{3} x + s i n ^{3} ( 18 - x )} d x$
$2 I = \int_{8}^{10} 1 d x$

Computing the Final Answer

$2 I = [x]_{8}^{10}$
$2 I = 10 - 8 = 2$
$I = 1$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

The function

f (x) = lo g_{4} (lo g_{5} (lo g_{3} (18 x - x^{2} - 77)))

presents a nested structure. To determine the domain

(a, b)

, we must ensure each logarithmic argument is strictly positive and satisfies the constraints imposed by the outer layers.

The outermost logarithm requires its argument to be positive. Moving inward, the condition

lo g_{5} (Y) > 0

implies

Y > 5^{0} = 1

. Finally, the innermost condition

lo g_{3} (Z) > 1

implies

Z > 3^{1} = 3

This leads us to the core inequality:

18 x - x^{2} - 77 > 3

Solving the Quadratic Inequality

We simplify the expression by bringing the constant to the left side:

- x^{2} + 18 x - 80 > 0

Multiplying by

- 1

reverses the inequality sign, yielding:

x^{2} - 18 x + 80 < 0

Factoring the quadratic, we obtain:

(x - 8) (x - 10) < 0

Since the parabola opens upward, the expression is negative between the roots

x = 8

and

x = 10

. Thus, the domain is

x \in (8, 10)

, which identifies our limits of integration as

a = 8

and

b = 10

The Calculus Bridge

With the limits established, we evaluate the integral:

I = \int_{8}^{10} \frac{sin ^{3} x}{sin ^{3} x + sin ^{3} ( 18 - x )} d x

This problem utilizes the King's Property, which states that

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

. Here,

a + b = 18

If we substitute

x

with

18 - x

, the numerator

sin^{3} x

transforms into

sin^{3} (18 - x)

, while the denominator remains invariant.

The Elegant Cancellation

By adding the original integral

I

to the transformed integral

I

, we obtain:

2 I = \int_{8}^{10} \frac{sin ^{3} x + sin ^{3} ( 18 - x )}{sin ^{3} x + sin ^{3} ( 18 - x )} d x

The integrand simplifies to

1

. Consequently, we evaluate:

2 I = \int_{8}^{10} 1 d x = [x]_{8}^{10} = 10 - 8 = 2

Solving for

I

, we find the final result: $I = 1$

Common Traps & FAQs

Why did the inequality sign flip when we multiplied by

- 1

It is a fundamental rule of inequalities! When you multiply or divide by a negative number, the direction of the inequality reverses. Think of it as reflecting the number line: if

5 > 2

, then

- 5 < - 2

. We did this to make the quadratic easier to factor.

How do I know the domain is

(8, 10)

and not something else?

We solved the quadratic inequality

x^{2} - 18 x + 80 < 0

. The roots are

8

and

10

. Since the parabola opens upward, it is negative *between* the roots. Thus, the domain is strictly the interval

(8, 10)

Is there any other way to solve this integral?

Technically, yes, but it would be a mathematical nightmare. Without King's Property, you would be stuck with a non-elementary integral that cannot be solved using standard techniques. In JEE, if you see this specific structure, it is a clear signal to use the symmetry property.

Animated Solution for Mathematics - Definite Integration: Let the domain of the function $f (x) = lo g_{4} (lo g_{5} (lo g_{3} (18 x - x^{2} - 77)))$ be $(a, b)$ . Then the value of the integral $\int_{a}^{b} \frac{s i n ^{3} x}{( s i n ^{3} x + s i n ^{3} ( a + b - x ))} d x$ is equal to

Enter Numerical Value:

Visualized Solution

The Nested Logarithm and Integral

Unwrapping the Logarithms

Forming the Quadratic Inequality

Factoring the Quadratic

The Domain Interval $(a, b)$

Substituting Limits into the Integral

Applying King's Property

Summing the Two Equations

Computing the Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Solving the Quadratic Inequality

The Calculus Bridge

The Elegant Cancellation

Similar Questions

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Enter Numerical Value:

Visualized Solution

The Nested Logarithm and Integral

Unwrapping the Logarithms

Forming the Quadratic Inequality

Factoring the Quadratic

The Domain Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b)

Substituting Limits into the Integral

Applying King's Property

Summing the Two Equations

Computing the Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Solving the Quadratic Inequality

The Calculus Bridge

The Elegant Cancellation

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