JEE Main 2024 (08 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: The value of $k \in N$ for which the integral $I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x, n \in N$ , satisfies $147 I_{20} = 148 I_{21}$ is

Q: Why does the boundary term $[x(1-x^k)^n]_0^1$ vanish? I always get confused with limits.

Great question! Let's evaluate it: at the upper limit $x=1$, we get $1(1-1^k)^n = 1(0)^n = 0$. At the lower limit $x=0$, we get $0(1-0^k)^n = 0(1)^n = 0$. Since both evaluate to zero, the entire boundary term disappears. Always check your limits carefully—they are often designed to simplify the expression!

Q: Why can't we just use binomial expansion on $(1-x^k)^n$ and integrate term by term?

You absolutely could! However, for large $n$, that would result in a massive sum of terms, which is not efficient for finding a relationship between $I_{20}$ and $I_{21}$. The reduction formula approach is the 'JEE way' because it's elegant, faster, and works for any $n$.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): When you see a parameterized integral like

I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x

, your first instinct should be to look for a recurrence relation. The power

n

is a clear signal that Integration by Parts (IBP) will help you relate

I_{n}

I_{n - 1}

Hint 2

Hint 2 (The Arsenal): Use Integration by Parts with

u = (1 - x^{k})^{n}

and

d v = d x

. Remember to handle the chain rule carefully when differentiating

u

. The boundary terms will vanish, leaving you with an integral that looks slightly different from

I_{n}

Hint 3

Hint 3 (The Execution): To transform the resulting integral back into the form of

I_{n}

and

I_{n - 1}

, use the algebraic trick of writing

x^{k}

1 - (1 - x^{k})

. This will allow you to split the integral into two parts, creating the recurrence relation:

I_{n} = nk (I_{n - 1} - I_{n})

Visualized Solution

Understanding the Integral $I_{n}$

Given integral: $I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x$
Given condition: $147 I_{20} = 148 I_{21}$
Objective: Find the value of $k \in N$

Choosing the Integration Tool

To relate $I_{n}$ and $I_{n - 1}$ , we use Integration by Parts (IBP).
Let $u = (1 - x^{k})^{n}$ and $d v = 1 \cdot d x$

Calculating Differentials for IBP

Differentiating $u$ : $d u = n (1 - x^{k})^{n - 1} \cdot (- k x^{k - 1}) d x$
Integrating $d v$ : $v = x$

Applying the IBP Formula

$I_{n} = [x (1 - x^{k})^{n}]_{0}^{1} - \int_{0}^{1} x \cdot n (1 - x^{k})^{n - 1} (- k x^{k - 1}) d x$
Simplifying: $I_{n} = [x (1 - x^{k})^{n}]_{0}^{1} + nk \int_{0}^{1} x^{k} (1 - x^{k})^{n - 1} d x$

Evaluating Boundary Terms

At $x = 1$ : $1 (1 - 1^{k})^{n} = 0$
At $x = 0$ : $0 (1 - 0^{k})^{n} = 0$
Result: $I_{n} = nk \int_{0}^{1} x^{k} (1 - x^{k})^{n - 1} d x$

The Algebraic Manipulation Trick

Rewrite $x^{k}$ as $1 - (1 - x^{k})$ to relate back to $I_{n}$ terms.
$I_{n} = nk \int_{0}^{1} [1 - (1 - x^{k})] (1 - x^{k})^{n - 1} d x$

Splitting the Integral

$I_{n} = nk [\int_{0}^{1} (1 - x^{k})^{n - 1} d x - \int_{0}^{1} (1 - x^{k})^{n} d x]$
Substitute definitions: $I_{n} = nk [I_{n - 1} - I_{n}]$

Rearranging for the Ratio

$I_{n} = nk I_{n - 1} - nk I_{n}$
$(1 + nk) I_{n} = nk I_{n - 1}$
The reduction ratio: $\frac{I _{n - 1}}{I _{n}} = \frac{nk + 1}{nk}$

Substituting $n = 21$

Substitute $n = 21$ :
$\frac{I _{20}}{I _{21}} = \frac{21 k + 1}{21 k}$

Using the Given Condition

Given: $147 I_{20} = 148 I_{21}$
Rearranging: $\frac{I _{20}}{I _{21}} = \frac{148}{147}$

Equating the Ratios

Equating: $\frac{21 k + 1}{21 k} = \frac{148}{147}$
Simplifying: $1 + \frac{1}{21 k} = 1 + \frac{1}{147}$

Solving for $k$

$\frac{1}{21 k} = \frac{1}{147}$
$21 k = 147$
$k = \frac{147}{21} = 7$

Final Conclusion

Final Answer: $k = 7$
The correct option is (4).
Key Takeaway: Reduction formulas are powerful tools for handling parameterized integrals like $I_{n}$ .

00:00 / 00:00

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

The integral

I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x

appears complex due to the variable power

n

. In JEE Advanced mathematics, such problems are best approached by establishing a recurrence relation that connects

I_{n}

I_{n - 1}

The Tool of Choice

Integration by Parts

To derive the reduction formula, we employ Integration by Parts (IBP). We set:

u = (1 - x^{k})^{n}

d v = d x

Differentiating

u

using the chain rule gives:

d u = n (1 - x^{k})^{n - 1} \cdot (- k x^{k - 1}) d x

Integrating

d v

gives

v = x

. Applying the IBP formula

I_{n} = uv - \int v d u

, we obtain:

I_{n} = [x (1 - x^{k})^{n}]_{0}^{1} - \int_{0}^{1} x \cdot n (1 - x^{k})^{n - 1} (- k x^{k - 1}) d x

The boundary term evaluates to zero at both limits. This simplifies the expression to:

I_{n} = nk \int_{0}^{1} x^{k} (1 - x^{k})^{n - 1} d x

The Algebraic Pivot

To relate the integral back to

I_{n - 1}

, we manipulate the term

x^{k}

using the identity

x^{k} = 1 - (1 - x^{k})

. Substituting this into our expression yields:

I_{n} = nk \int_{0}^{1} [1 - (1 - x^{k})] (1 - x^{k})^{n - 1} d x

Distributing the terms, we split the integral into two parts:

I_{n} = nk \int_{0}^{1} (1 - x^{k})^{n - 1} d x - nk \int_{0}^{1} (1 - x^{k})^{n} d x

Recognizing the definitions of our integrals, we see that the first term is

I_{n - 1}

and the second is

I_{n}

. Thus, we arrive at the recurrence relation:

I_{n} = nk (I_{n - 1} - I_{n})

Closing the Loop

We rearrange the recurrence relation to isolate the ratio

\frac{I _{n - 1}}{I _{n}}

(1 + nk) I_{n} = nk I_{n - 1}

\frac{I _{n - 1}}{I _{n}} = \frac{1 + nk}{nk} = 1 + \frac{1}{nk}

The problem provides the condition

147 I_{20} = 148 I_{21}

, which implies:

\frac{I _{20}}{I _{21}} = \frac{148}{147} = 1 + \frac{1}{147}

Setting

n = 21

in our derived ratio formula, we equate:

1 + \frac{1}{21 k} = 1 + \frac{1}{147}

Solving for

k

, we find

21 k = 147

. This yields the final result: $k = 7$

Common Traps & FAQs

Why does the boundary term

[x (1 - x^{k})^{n}]_{0}^{1}

vanish? I always get confused with limits.

Great question! Let's evaluate it: at the upper limit

x = 1

, we get

1 (1 - 1^{k})^{n} = 1 (0)^{n} = 0

. At the lower limit

x = 0

, we get

0 (1 - 0^{k})^{n} = 0 (1)^{n} = 0

. Since both evaluate to zero, the entire boundary term disappears. Always check your limits carefully—they are often designed to simplify the expression!

Why can't we just use binomial expansion on

(1 - x^{k})^{n}

and integrate term by term?

You absolutely could! However, for large

n

, that would result in a massive sum of terms, which is not efficient for finding a relationship between

I_{20}

and

I_{21}

. The reduction formula approach is the 'JEE way' because it's elegant, faster, and works for any

n

How did we know to rewrite

x^{k}

1 - (1 - x^{k})

This is a classic 'manipulation trick' in calculus. We look at the integral we have, which contains

x^{k} (1 - x^{k})^{n - 1}

, and we compare it to our target

I_{n}

and

I_{n - 1}

. We need to create a

(1 - x^{k})^{n}

term to get

I_{n}

and a

(1 - x^{k})^{n - 1}

term to get

I_{n - 1}

. Adding and subtracting 1 is the standard way to force that structure.

Animated Solution for Mathematics - Definite Integration: The value of $k \in N$ for which the integral $I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x, n \in N$ , satisfies $147 I_{20} = 148 I_{21}$ is

Select Answer:

Visualized Solution

Understanding the Integral $I_{n}$

Choosing the Integration Tool

Calculating Differentials for IBP

Applying the IBP Formula

Evaluating Boundary Terms

The Algebraic Manipulation Trick

Splitting the Integral

Rearranging for the Ratio

Substituting $n = 21$

Using the Given Condition

Equating the Ratios

Solving for $k$

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Tool of Choice

The Algebraic Pivot

Closing the Loop

Similar Questions

Determine a positive integer $n \leq 5$ , such that $\int_{0}^{1} e^{x} (x - 1)^{n} d x = 16 - 6 e$

Let $β (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x, m, n > 0$ . If $\int_{0}^{1} (1 - x^{10})^{20} d x = a \times β (b, c)$ , then $100 (a + b + c)$ equals

For $m, n > 0$ , let $α (m, n) = \int_{0}^{2} t^{m} (1 + 3 t)^{n} d t$ . If $11 α (10, 6) + 18 α (11, 5) = p (14)^{6}$ , then $p$ is equal to

The value of $5050 \frac{\int _{0}^{1} ( 1 - x ^{50} ) ^{100} d x}{\int _{0}^{1} ( 1 - x ^{50} ) ^{101} d x}$ is.

$I_{n} = \int_{0}^{π /4} tan^{n} x d x$ then $lim_{n \to \infty} n [I_{n} + I_{n + 2}]$ equals

$lim_{x \to 0} \frac{48}{x ^{4}} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t$ is equal to ______.

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a > 0$ , $\int_{0}^{1} f (λ x) d λ = a f (x)$ , $f (1) = 1$ and $f (16) = \frac{1}{8}$ , then $16 - f^{'} (\frac{1}{16})$ is equal to

Let $I_{n} (x) = \int_{0}^{x} \frac{d t}{( t ^{2} + 5 ) ^{n}}, n = 1, 2, 3, \dots$ . Then

Let f be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ then value of $\sum_{r = 1}^{15} f (r^{3})$ is:

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of k∈N for which the integral In​=∫01​(1−xk)ndx,n∈N, satisfies 147I20​=148I21​ is

Select Answer:

Visualized Solution

Understanding the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In​

Choosing the Integration Tool

Calculating Differentials for IBP

Applying the IBP Formula

Evaluating Boundary Terms

The Algebraic Manipulation Trick

Splitting the Integral

Rearranging for the Ratio

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=21

Using the Given Condition

Equating the Ratios

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

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Tool of Choice

The Algebraic Pivot

Closing the Loop

Similar Questions

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Animated Solution for Mathematics - Definite Integration: The value of $k \in N$ for which the integral $I_{n} = \int_{0}^{1} (1 - x^{k})^{n} d x, n \in N$ , satisfies $147 I_{20} = 148 I_{21}$ is

Understanding the Integral $I_{n}$

Substituting $n = 21$

Solving for $k$

Determine a positive integer $n \leq 5$ , such that $\int_{0}^{1} e^{x} (x - 1)^{n} d x = 16 - 6 e$

Let $β (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x, m, n > 0$ . If $\int_{0}^{1} (1 - x^{10})^{20} d x = a \times β (b, c)$ , then $100 (a + b + c)$ equals

For $m, n > 0$ , let $α (m, n) = \int_{0}^{2} t^{m} (1 + 3 t)^{n} d t$ . If $11 α (10, 6) + 18 α (11, 5) = p (14)^{6}$ , then $p$ is equal to

The value of $5050 \frac{\int _{0}^{1} ( 1 - x ^{50} ) ^{100} d x}{\int _{0}^{1} ( 1 - x ^{50} ) ^{101} d x}$ is.

$I_{n} = \int_{0}^{π /4} tan^{n} x d x$ then $lim_{n \to \infty} n [I_{n} + I_{n + 2}]$ equals

$lim_{x \to 0} \frac{48}{x ^{4}} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t$ is equal to ______.

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a > 0$ , $\int_{0}^{1} f (λ x) d λ = a f (x)$ , $f (1) = 1$ and $f (16) = \frac{1}{8}$ , then $16 - f^{'} (\frac{1}{16})$ is equal to

Let $I_{n} (x) = \int_{0}^{x} \frac{d t}{( t ^{2} + 5 ) ^{n}}, n = 1, 2, 3, \dots$ . Then

Let f be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ then value of $\sum_{r = 1}^{15} f (r^{3})$ is: