JEE Advanced 1996

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ . Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

Q: Why is $\tan^n x < \tan^{n-2} x$ for $x \in (0, \pi/4)$?

In the interval $(0, \pi/4)$, the value of $\tan x$ is a fraction strictly between $0$ and $1$. When you raise a fraction between $0$ and $1$ to a higher power, it becomes smaller. Thus, $\tan^n x$ is smaller than $\tan^{n-2} x$.

Q: Why did we replace $n$ with $n+2$ to find the lower bound?

We needed a relationship involving $A_n$ and $A_{n+2}$ to create an inequality $A_{n+2} + A_n = \frac{1}{n+1}$. Since $A_{n+2} < A_n$, we can bound $2A_n$ by $A_{n+2} + A_n$, which helps us isolate $A_n$.

Q: Is there a simpler way to solve this without recurrence?

While you could try integration by parts, recurrence relations are the standard, most elegant way to handle integrals of trigonometric powers. It turns a complex calculus problem into a simple algebraic one, which is exactly what JEE examiners love to see.

Visualized Solution

Defining the Area $A_{n}$

The area $A_{n}$ is bounded by $y = tan^{n} x$ , $x = 0$ , $y = 0$ , and $x = \frac{π}{4}$ .
Mathematically, this is expressed as the definite integral:
$A_{n} = \int_{0}^{\frac{π}{4}} tan^{n} x d x$

Setting up $A_{n} + A_{n - 2}$

Consider the sum of two consecutive terms with a gap of two:
$A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} tan^{n} x d x + \int_{0}^{\frac{π}{4}} tan^{n - 2} x d x$

Combining the Integrals

Combine the integrals since the limits are identical:
$A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} (tan^{n} x + tan^{n - 2} x) d x$
Factor out $tan^{n - 2} x$ :
$A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} tan^{n - 2} x (tan^{2} x + 1) d x$

Applying Trigonometric Identity

Use the fundamental identity: $1 + tan^{2} x = sec^{2} x$ .
Substitute this into the integral:
$A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} tan^{n - 2} x sec^{2} x d x$

Integration by Substitution

Let $t = tan x$ .
Differentiating both sides gives:
$d t = sec^{2} x d x$

Changing the Limits of Integration

When $x = 0$ , $t = tan (0) = 0$ .
When $x = \frac{π}{4}$ , $t = tan (\frac{π}{4}) = 1$ .
The integral transforms to:
$A_{n} + A_{n - 2} = \int_{0}^{1} t^{n - 2} d t$

Evaluating the Integral

Apply the power rule for integration: $\int t^{m} d t = \frac{t ^{m + 1}}{m + 1}$ .
$A_{n} + A_{n - 2} = [\frac{t ^{n - 1}}{n - 1}]_{0}^{1}$
$A_{n} + A_{n - 2} = \frac{1 ^{n - 1}}{n - 1} - \frac{0 ^{n - 1}}{n - 1} = \frac{1}{n - 1}$

Analyzing Monotonicity of the Curves

For $x \in (0, \frac{π}{4})$ , $0 < tan x < 1$ .
For any $0 < u < 1$ , as the exponent increases, the value decreases.
Thus, $tan^{n} x < tan^{n - 2} x$ for $x \in (0, \frac{π}{4})$ .
Integrating both sides gives: $A_{n} < A_{n - 2}$ .

Deducing the Upper Bound

We have $A_{n} < A_{n - 2}$ .
Add $A_{n}$ to both sides: $2 A_{n} < A_{n} + A_{n - 2}$ .
Substitute the recurrence result:
$2 A_{n} < \frac{1}{n - 1}$
Divide by $2$ :
$A_{n} < \frac{1}{2 n - 2}$

Deducing the Lower Bound

Replace $n$ with $n + 2$ in the recurrence formula:
$A_{n + 2} + A_{n} = \frac{1}{( n + 2 ) - 1} = \frac{1}{n + 1}$
Since $A_{n + 2} < A_{n}$ , we have:
$A_{n + 2} + A_{n} < 2 A_{n}$
Substitute the value: $\frac{1}{n + 1} < 2 A_{n}$
Divide by $2$ : $A_{n} > \frac{1}{2 n + 2}$

Final Conclusion

Final Result:
1. $A_{n} + A_{n - 2} = \frac{1}{n - 1}$
2. $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$
Key Takeaway: Recurrence relations are excellent tools for analyzing integrals that depend on an integer parameter $n$ .

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The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

The area

A_{n}

bounded by the curve

y = (tan x)^{n}

and the lines

x = 0

y = 0

, and

x = \frac{π}{4}

is defined by the integral:

A_{n} = \int_{0}^{\frac{π}{4}} tan^{n} x d x

The Algebraic Spark

To find the relationship between

A_{n}

and

A_{n - 2}

, we consider their sum. Since the limits of integration are identical, we combine the integrals:

A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} (tan^{n} x + tan^{n - 2} x) d x

Factoring out

tan^{n - 2} x

from the integrand, we obtain:

A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} tan^{n - 2} x (tan^{2} x + 1) d x

Using the fundamental trigonometric identity

1 + tan^{2} x = sec^{2} x

, the expression simplifies to:

A_{n} + A_{n - 2} = \int_{0}^{\frac{π}{4}} tan^{n - 2} x sec^{2} x d x

The Substitution

We recognize that

sec^{2} x

is the derivative of

tan x

. We perform the substitution

t = tan x

, which implies

d t = sec^{2} x d x

Updating the limits of integration: when

x = 0

t = 0

; when

x = \frac{π}{4}

t = 1

. The integral becomes:

A_{n} + A_{n - 2} = \int_{0}^{1} t^{n - 2} d t

Applying the power rule for integration, we find:

A_{n} + A_{n - 2} = [\frac{t ^{n - 1}}{n - 1}]_{0}^{1} = \frac{1}{n - 1}

The Art of Bounding

In the interval

(0, \frac{π}{4})

, the function

tan x

satisfies

0 < tan x < 1

. Consequently, for increasing powers,

tan^{n} x < tan^{n - 2} x

, which implies

A_{n} < A_{n - 2}

Adding

A_{n}

to both sides of the inequality

A_{n} < A_{n - 2}

yields

2 A_{n} < A_{n} + A_{n - 2}

. Substituting our recurrence result:

2 A_{n} < \frac{1}{n - 1} \Rightarrow A_{n} < \frac{1}{2 n - 2}

To find the lower bound, we use the recurrence relation

A_{n + 2} + A_{n} = \frac{1}{n + 1}

. Since

A_{n + 2} < A_{n}

, it follows that

2 A_{n + 2} < A_{n + 2} + A_{n} = \frac{1}{n + 1}

By shifting the index, we conclude:

\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}

You have successfully mastered this classic JEE technique.

Common Traps & FAQs

Why is

tan^{n} x < tan^{n - 2} x

for

x \in (0, π /4)

In the interval

(0, π /4)

, the value of

tan x

is a fraction strictly between

0

and

1

. When you raise a fraction between

0

and

1

to a higher power, it becomes smaller. Thus,

tan^{n} x

is smaller than

tan^{n - 2} x

Why did we replace

n

with

n + 2

to find the lower bound?

We needed a relationship involving

A_{n}

and

A_{n + 2}

to create an inequality

A_{n + 2} + A_{n} = \frac{1}{n + 1}

. Since

A_{n + 2} < A_{n}

, we can bound

2 A_{n}

A_{n + 2} + A_{n}

, which helps us isolate

A_{n}

Is there a simpler way to solve this without recurrence?

While you could try integration by parts, recurrence relations are the standard, most elegant way to handle integrals of trigonometric powers. It turns a complex calculus problem into a simple algebraic one, which is exactly what JEE examiners love to see.

Animated Solution for Mathematics - Definite Integration: Let $A_{n}$ be the area bounded by the curve $y = (tan x)^{n}$ and the lines $x = 0, y = 0$ and $x = \frac{π}{4}$ . Prove that for $n > 2, A_{n} + A_{n - 2} = \frac{1}{n - 1}$ and deduce $\frac{1}{2 n + 2} < A_{n} < \frac{1}{2 n - 2}$ .

Visualized Solution

Defining the Area $A_{n}$

Setting up $A_{n} + A_{n - 2}$

Combining the Integrals

Applying Trigonometric Identity

Integration by Substitution

Changing the Limits of Integration

Evaluating the Integral

Analyzing Monotonicity of the Curves

Deducing the Upper Bound

Deducing the Lower Bound

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Algebraic Spark

The Substitution

The Art of Bounding

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Visualized Solution

Defining the Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An​

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An​+An−2​

Combining the Integrals

Applying Trigonometric Identity

Integration by Substitution

Changing the Limits of Integration

Evaluating the Integral

Analyzing Monotonicity of the Curves

Deducing the Upper Bound

Deducing the Lower Bound

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Algebraic Spark

The Substitution

The Art of Bounding

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Setting up $A_{n} + A_{n - 2}$

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$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (t^{\frac{1}{3}}) d t)$ is equal to

Let $a + b = 4$ , where $a < 2$ , and let $g (x)$ be a differentiable function. If $d g / d x > 0$ for all $x$ , prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a)$ increases.

$lim_{x \to \frac{π}{2}} \frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} )) d t}{( x - \frac{π}{2} ) ^{2}}$ is equal to

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

The intercepts on $x$ -axis made by tangents to the curve, $y = \int_{0}^{x} ∣ t ∣ d t, x \in R$ , which are parallel to the line $y = 2 x$ , are equal to

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Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then