JEE Main 2023 (06 April Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}, x \in R - {- 1}, n \in N, n > 2$ . If $f_{n} (x) = (f \circ f \circ \dots upto n times) (x)$ , then $lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f_{n} (x)) d x$ is equal to

Q: Why did we set $1 + nx^n = t^n$ instead of just $u = 1 + nx^n$?

Great question! If you set $u = 1 + nx^n$, you would still have the $(u)^{1/n}$ in the denominator, which makes the integration messy. By setting it to $t^n$, the $1/n$ power cancels out immediately, turning a difficult radical integral into a simple polynomial integral. Always look for substitutions that simplify the exponents!

Q: How do I handle the limit $\lim_{n \to \infty} \frac{(1+n)^{1 - 1/n} - 1}{n^2 - n}$? It looks like an indeterminate form.

It does look tricky, but focus on the dominant terms. As $n$ grows, $1/n$ vanishes to zero. The term $(1+n)^{1-1/n}$ essentially becomes $(1+n)^1$. The $-1$ cancels the $1$, leaving you with $n$ in the numerator. The denominator is dominated by $n^2$. Thus, $n/n^2$ clearly heads to zero.

Enter Numerical Value:

Visualized Solution

Analyze the Function $f (x)$

Given function: $f (x) = \frac{x}{( 1 + x ^{n} ) ^{\frac{1}{n}}}$
We need to find the $n$ -th composition: $f_{n} (x) = (f \circ f \circ \dots upto n times) (x)$
Constraint: $x \in R - {- 1}, n \in N, n > 2$

Calculate the Second Composition $f_{2} (x)$

Substitute $f (x)$ into itself: $f_{2} (x) = f (f (x))$
$f_{2} (x) = \frac{f ( x )}{( 1 + ( f ( x ) ) ^{n} ) ^{\frac{1}{n}}}$
Plugging the expression: $f_{2} (x) = \frac{\frac{x}{( 1 + x ^{n} ) ^{\frac{1}{n}}}}{( 1 + ( \frac{x}{( 1 + x ^{n} ) ^{\frac{1}{n}}} ) ^{n} ) ^{\frac{1}{n}}}$

Simplify $f_{2} (x)$

Simplify the denominator's inner term: $(\frac{x}{( 1 + x ^{n} ) ^{\frac{1}{n}}})^{n} = \frac{x ^{n}}{1 + x ^{n}}$
Add $1$ : $1 + \frac{x ^{n}}{1 + x ^{n}} = \frac{1 + 2 x ^{n}}{1 + x ^{n}}$
Resulting expression: $f_{2} (x) = \frac{\frac{x}{( 1 + x ^{n} ) ^{\frac{1}{n}}}}{( \frac{1 + 2 x ^{n}}{1 + x ^{n}} ) ^{\frac{1}{n}}} = \frac{x}{( 1 + 2 x ^{n} ) ^{\frac{1}{n}}}$

Generalize to $f_{n} (x)$ via Induction

By observing the pattern:
$f_{1} (x) = \frac{x}{( 1 + 1 \cdot x ^{n} ) ^{\frac{1}{n}}}$
$f_{2} (x) = \frac{x}{( 1 + 2 \cdot x ^{n} ) ^{\frac{1}{n}}}$
General form: $f_{n} (x) = \frac{x}{( 1 + n \cdot x ^{n} ) ^{\frac{1}{n}}}$

Set up the Integral $I_{n}$

Substitute $f_{n} (x)$ into the integral: $I_{n} = \int_{0}^{1} x^{n - 2} \frac{x}{( 1 + n x ^{n} ) ^{\frac{1}{n}}} d x$
Combine the powers of $x$ : $x^{n - 2} \cdot x = x^{n - 1}$
Simplified Integral: $I_{n} = \int_{0}^{1} \frac{x ^{n - 1}}{( 1 + n x ^{n} ) ^{\frac{1}{n}}} d x$

Apply Substitution Method

Let the denominator's inner term be $t^{n}$ : $1 + n x^{n} = t^{n}$
Differentiating both sides: $n^{2} x^{n - 1} d x = n t^{n - 1} d t$
Simplify the differential: $x^{n - 1} d x = \frac{1}{n} t^{n - 1} d t$

Update the Limits of Integration

Original limits are for $x$ : Lower limit $x = 0$ , Upper limit $x = 1$
Find new limits for $t$ using $t = (1 + n x^{n})^{\frac{1}{n}}$
Lower limit: At $x = 0$ , $t = (1 + 0)^{\frac{1}{n}} = 1$
Upper limit: At $x = 1$ , $t = (1 + n (1)^{n})^{\frac{1}{n}} = (1 + n)^{\frac{1}{n}}$

Evaluate the Definite Integral

Substitute into the integral: $I_{n} = \int_{1}^{(1 + n)^{\frac{1}{n}}} \frac{\frac{1}{n} t ^{n - 1} d t}{( t ^{n} ) ^{\frac{1}{n}}}$
Simplify the denominator: $(t^{n})^{\frac{1}{n}} = t$
Simplified integrand: $I_{n} = \frac{1}{n} \int_{1}^{(1 + n)^{\frac{1}{n}}} \frac{t ^{n - 1}}{t} d t = \frac{1}{n} \int_{1}^{(1 + n)^{\frac{1}{n}}} t^{n - 2} d t$

Perform the Integration

Integrate $t^{n - 2}$ : $\int t^{n - 2} d t = \frac{t ^{n - 1}}{n - 1}$
Apply the limits: $I_{n} = \frac{1}{n} [\frac{t ^{n - 1}}{n - 1}]_{1}^{(1 + n)^{\frac{1}{n}}}$
Substitute upper and lower bounds: $I_{n} = \frac{1}{n ( n - 1 )} (((1 + n)^{\frac{1}{n}})^{n - 1} - 1^{n - 1})$

Simplify the Expression for $I_{n}$

Simplify the upper bound term: $((1 + n)^{\frac{1}{n}})^{n - 1} = (1 + n)^{\frac{n - 1}{n}} = (1 + n)^{1 - \frac{1}{n}}$
Final expression for $I_{n}$ : $I_{n} = \frac{( 1 + n ) ^{1 - \frac{1}{n}} - 1}{n ^{2} - n}$

Apply the Limit $n \to \infty$

We need to evaluate $lim_{n \to \infty} I_{n} = lim_{n \to \infty} \frac{( 1 + n ) ^{1 - \frac{1}{n}} - 1}{n ^{2} - n}$
As $n \to \infty$ , the exponent $1 - \frac{1}{n} \to 1 - 0 = 1$
The numerator behaves like $(1 + n)^{1} - 1 = n$
The expression simplifies to $lim_{n \to \infty} \frac{n}{n ^{2} - n}$

Conclusion and Final Answer

Divide numerator and denominator by $n^{2}$ : $lim_{n \to \infty} \frac{\frac{1}{n}}{1 - \frac{1}{n}}$
As $n \to \infty$ , $\frac{1}{n} \to 0$
Limit evaluates to $\frac{0}{1 - 0} = 0$
Final Answer: $0$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

We are given the function

f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}

. We aim to evaluate the limit of an integral involving the

n

-th composition of this function, denoted as

f_{n} (x) = (f \circ f \circ \dots \circ f) (x)

In JEE Advanced, complexity is often a mask for symmetry. Let us peel back the layers of this composition to find a general form.

The Pattern Hunt

We begin by calculating the second composition,

f_{2} (x) = f (f (x))

. Substituting

f (x)

into itself, we obtain:

f_{2} (x) = \frac{\frac{x}{( 1 + x ^{n} ) ^{1/ n}}}{( 1 + ( \frac{x}{( 1 + x ^{n} ) ^{1/ n}} ) ^{n} ) ^{1/ n}}

Look closely at the denominator. The inner term raised to the power of

n

simplifies to

\frac{x ^{n}}{1 + x ^{n}}

. Adding

1

to this yields

\frac{1 + 2 x ^{n}}{1 + x ^{n}}

When we take the

n

-th root and divide, the

(1 + x^{n})^{1/ n}

terms cancel out. This leaves us with:

f_{2} (x) = \frac{x}{( 1 + 2 x ^{n} ) ^{1/ n}}

The pattern is undeniable. For

f_{1} (x)

, the coefficient of

x^{n}

was

1

. For

f_{2} (x)

, it is

2

. By induction, the general form is:

f_{n} (x) = \frac{x}{( 1 + n x ^{n} ) ^{1/ n}}

The Integral Setup

Now, we apply this result to our integral

I_{n} = \int_{0}^{1} x^{n - 2} f_{n} (x) d x

. Substituting our generalized function, we get:

I_{n} = \int_{0}^{1} \frac{x ^{n - 1}}{( 1 + n x ^{n} ) ^{1/ n}} d x

The numerator

x^{n - 1}

is proportional to the derivative of

x^{n}

. To make the integration effortless, we use the substitution

1 + n x^{n} = t^{n}

Differentiating both sides, we find

n^{2} x^{n - 1} d x = n t^{n - 1} d t

, which simplifies to

x^{n - 1} d x = \frac{1}{n} t^{n - 1} d t

. We must also update our limits: when

x = 0

t = 1

; when

x = 1

t = (1 + n)^{1/ n}

The Grand Finale

Our integral transforms into:

I_{n} = \frac{1}{n} \int_{1}^{(1 + n)^{1/ n}} \frac{t ^{n - 1}}{t} d t = \frac{1}{n} \int_{1}^{(1 + n)^{1/ n}} t^{n - 2} d t

Integrating this polynomial, we get:

I_{n} = \frac{1}{n ( n - 1 )} [t^{n - 1}]_{1}^{(1 + n)^{1/ n}} = \frac{( 1 + n ) ^{\frac{n - 1}{n}} - 1}{n ^{2} - n}

Finally, we take the limit as

n \to \infty

. The exponent

\frac{n - 1}{n}

approaches

1

, so the numerator behaves like

(1 + n) - 1 = n

. The expression becomes:

n \to \infty lim \frac{n}{n ^{2} - n} = 0

We have journeyed from a terrifying composition to a simple result. The final answer is 0.

Common Traps & FAQs

Wait, why does the coefficient just become

n

? Is it really that simple?

It feels like magic, doesn't it? When you compose the function, you are essentially nesting the denominator. Each layer adds another

x^{n}

term to the sum inside the root. It is a beautiful example of how complex-looking recursive structures often collapse into simple, elegant arithmetic progressions.

Why did we set

1 + n x^{n} = t^{n}

instead of just

u = 1 + n x^{n}

Great question! If you set

u = 1 + n x^{n}

, you would still have the

(u)^{1/ n}

in the denominator, which makes the integration messy. By setting it to

t^{n}

, the

1/ n

power cancels out immediately, turning a difficult radical integral into a simple polynomial integral. Always look for substitutions that simplify the exponents!

How do I handle the limit

lim_{n \to \infty} \frac{( 1 + n ) ^{1 - 1/ n} - 1}{n ^{2} - n}

? It looks like an indeterminate form.

It does look tricky, but focus on the dominant terms. As

n

grows,

1/ n

vanishes to zero. The term

(1 + n)^{1 - 1/ n}

essentially becomes

(1 + n)^{1}

. The

- 1

cancels the

1

, leaving you with

n

in the numerator. The denominator is dominated by

n^{2}

. Thus,

n / n^{2}

clearly heads to zero.

Animated Solution for Mathematics - Definite Integration: Let $f (x) = \frac{x}{( 1 + x ^{n} ) ^{1/ n}}, x \in R - {- 1}, n \in N, n > 2$ . If $f_{n} (x) = (f \circ f \circ \dots upto n times) (x)$ , then $lim_{n \to \infty} \int_{0}^{1} x^{n - 2} (f_{n} (x)) d x$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Function $f (x)$

Calculate the Second Composition $f_{2} (x)$

Simplify $f_{2} (x)$

Generalize to $f_{n} (x)$ via Induction

Set up the Integral $I_{n}$

Apply Substitution Method

Update the Limits of Integration

Evaluate the Definite Integral

Perform the Integration

Simplify the Expression for $I_{n}$

Apply the Limit $n \to \infty$

Conclusion and Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Pattern Hunt

The Integral Setup

The Grand Finale

Similar Questions

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

For $n > 0$ , $\int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x = \dots\dots\dots$

The value of the integral $I = \int_{0}^{1} x (1 - x)^{n} d x$ is

Let $α \in (0, 1)$ and $β = lo g_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3} + ... + \frac{x ^{n}}{n}, x \in (0, 1)$ . Then the integral $\int_{0}^{α} \frac{t ^{50}}{1 - t} d t$ is equal to

The value of $\int_{1}^{a} [x] f^{'} (x) d x, a > 1$ where $[x]$ denotes the greatest integer not exceeding $x$ is

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

If $I_{n} = \int_{- π}^{π} \frac{s i n n x}{( 1 + π ^{x} ) s i n x} d x, n = 0, 1, 2, ..., t h e n$

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

Let $f : R \to R$ be a function defined by $f (x) = {[x], 0, x \leq 2 x > 2$ where $[x]$ is the greatest integer less than or equal to $x$ , if $I = \int_{- 1}^{2} \frac{x f ( x ^{2} )}{2 + f ( x + 1 )} d x$ , then the value of $(4 I - 1)$ is

Enter Numerical Value:

Visualized Solution

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

Calculate the Second Composition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​(x)

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​(x)

Generalize to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fn​(x) via Induction

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

Apply Substitution Method

Update the Limits of Integration

Evaluate the Definite Integral

Perform the Integration

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

Apply the Limit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n→∞

Conclusion and Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Pattern Hunt

The Integral Setup

The Grand Finale

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b+1−x)=f(x)∀x, where a and b are fixed positive real numbers, then (a+b)1​∫ab​x(f(x)+f(x+1))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For n>0, ∫02π​sin2nx+cos2nxxsin2nx​dx=………

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the integral I=∫01​x(1−x)ndx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α∈(0,1) and β=loge​(1−α). Let Pn​(x)=x+2x2​+3x3​+...+nxn​,x∈(0,1). Then the integral ∫0α​1−tt50​dt is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∫1a​[x]f′(x)dx,a>1 where [x] denotes the greatest integer not exceeding x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a+b+1−x)=f(x), for all x, where a and b are fixed positive real numbers, then (a+b)1​∫ab​x(f(x)+f(x+1))dx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If In​=∫−ππ​(1+πx)sinxsinnx​dx,n=0,1,2,...,then

Analyze the Function $f (x)$

Calculate the Second Composition $f_{2} (x)$

Simplify $f_{2} (x)$

Generalize to $f_{n} (x)$ via Induction

Set up the Integral $I_{n}$

Simplify the Expression for $I_{n}$

Apply the Limit $n \to \infty$

If $f (a + b + 1 - x) = f (x) \forall x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

For $n > 0$ , $\int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x = \dots\dots\dots$

The value of the integral $I = \int_{0}^{1} x (1 - x)^{n} d x$ is

Let $α \in (0, 1)$ and $β = lo g_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3} + ... + \frac{x ^{n}}{n}, x \in (0, 1)$ . Then the integral $\int_{0}^{α} \frac{t ^{50}}{1 - t} d t$ is equal to

The value of $\int_{1}^{a} [x] f^{'} (x) d x, a > 1$ where $[x]$ denotes the greatest integer not exceeding $x$ is

If $f (a + b + 1 - x) = f (x)$ , for all $x$ , where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{( a + b )} \int_{a}^{b} x (f (x) + f (x + 1)) d x$ is equal to

If $I_{n} = \int_{- π}^{π} \frac{s i n n x}{( 1 + π ^{x} ) s i n x} d x, n = 0, 1, 2, ..., t h e n$