Integral Calculus JEE 2007: Iterated Function Solution

What is the Integral Calculus JEE 2007 iterated-function question?

Option A is correct in this Integral Calculus JEE 2007 question: the substitution introduces two factors of the exponent, not one. This is Paper 2, Mathematics, Question 52, a single-correct MCQ rated hard on the question bank’s scale.

For an integer exponent, define the function and its repeated composition:

n \geq 2, f (x) = \frac{x}{(1 + x^{n})^{1 / n}}, g (x) = \underset{n applications}{\underset{⏟}{(f \circ f \circ \dots \circ f)}} (x) .

Each output becomes the next input. This is composition, not multiplication of identical function values. Find an antiderivative of: $x^{n - 2} g (x) .$

The supplied choices are:

\frac{(1 + n x^{n})^{1 - 1 / n}}{n (n - 1)} + K

\frac{(1 + n x^{n})^{1 - 1 / n}}{n - 1} + K

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

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

The final constant is arbitrary. Separate composition from integration so that you can check the nesting and the coefficient independently.

How does taking the nth power simplify the composition?

Taking the nth power removes the root and turns each application into a rational transformation. The official method tracks the nth power of the current input rather than expanding nested radicals:

[f (x)]^{n} = \frac{x^{n}}{1 + x^{n}} .

Define the transformed variable and function:

y = x^{n}, T (y) = \frac{y}{1 + y} .

One application of the original function corresponds to one application of this transformation to the nth power of the input. Applying it twice gives:

T (T (y)) = \frac{\frac{y}{1 + y}}{1 + \frac{y}{1 + y}} .

Simplify the denominator before cancelling:

1 + \frac{y}{1 + y} = \frac{1 + 2 y}{1 + y}, T^{2} (y) = \frac{y}{1 + 2 y} .

The superscript here counts applications, not multiplication. Composing the transformation twice and squaring its output are different operations:

T^{2} (y) = T (T (y)), [T (y)]^{2} = \frac{y^{2}}{(1 + y)^{2}} .

How do you prove the formula for repeated applications?

Induction proves that each application increases the denominator’s coefficient by one. Keep the application count separate from the original exponent:

T^{k} (y) = \frac{y}{1 + k y}, k = number of applications, n = original exponent .

The base case is exactly the definition of the transformation. For one application:

k = 1, T^{1} (y) = \frac{y}{1 + y} .

For the induction step, assume the proposed formula holds after the current number of applications. Applying the transformation once more gives:

T^{k + 1} (y) = \frac{\frac{y}{1 + k y}}{1 + \frac{y}{1 + k y}} = \frac{y}{1 + (k + 1) y} .

This proves the formula for every positive application count. Set the count equal to the original exponent and restore the original variable:

k = n, y = x^{n}, [g (x)]^{n} = \frac{x^{n}}{1 + n x^{n}} .

On intervals where the original real-valued compositions are defined, this gives the official result:

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

Retain the signed numerator. For even exponents, each original application preserves the input’s sign, so replacing the numerator by its absolute value would be wrong.

How do you integrate the simplified expression and verify option A?

One substitution gives option A, provided its differential retains both factors of the exponent. First combine the powers in the numerator:

I = \int x^{n - 2} \frac{x}{(1 + n x^{n})^{1 / n}} d x = \int x^{n - 1} (1 + n x^{n})^{- 1 / n} d x .

Choose the inner expression as the new variable. Differentiation keeps its existing coefficient and multiplies by the exponent:

u = 1 + n x^{n}, d u = n \cdot n x^{n - 1} d x = n^{2} x^{n - 1} d x .

Therefore,

x^{n - 1} d x = \frac{d u}{n^{2}}, I = \frac{1}{n^{2}} \int u^{- 1 / n} d u .

Apply the power rule, then simplify the coefficient separately:

I = \frac{1}{n^{2}} \frac{u^{1 - 1 / n}}{1 - 1 / n} + K .

1 - \frac{1}{n} = \frac{n - 1}{n}, \frac{1}{n^{2}} \cdot \frac{n}{n - 1} = \frac{1}{n (n - 1)} .

The given lower bound on the exponent ensures that the power-rule denominator is nonzero. Substituting back gives option A:

I = \frac{(1 + n x^{n})^{1 - 1 / n}}{n (n - 1)} + K .

Differentiate to check both the coefficient and the power. All constant factors must cancel:

\frac{d I}{d x} = \frac{1}{n (n - 1)} \cdot \frac{n - 1}{n} \cdot n^{2} x^{n - 1} (1 + n x^{n})^{- 1 / n} = x^{n - 1} (1 + n x^{n})^{- 1 / n} .

This is the required integrand. C and D have the wrong power, as their derivatives show:

\frac{d}{d x} [\frac{(1 + n x^{n})^{1 + 1 / n}}{n (n + 1)} + K] = x^{n - 1} (1 + n x^{n})^{1 / n} .

\frac{d}{d x} [\frac{(1 + n x^{n})^{1 + 1 / n}}{n + 1} + K] = n x^{n - 1} (1 + n x^{n})^{1 / n} .

Both produce a positive exponent on the inner expression. The required integrand has a negative exponent, so neither coefficient can fix them.

How does a missing chain-rule factor produce option B?

Starting from the correctly simplified integral, dropping one factor in the differential produces option B. This isolates the error to integration, not composition:

I = \int x^{n - 1} (1 + n x^{n})^{- 1 / n} d x, u = 1 + n x^{n} .

Compare the differentials:

Incorrect: d u = n x^{n - 1} d x Correct: d u = n^{2} x^{n - 1} d x .

Differentiation must retain the existing coefficient and multiply by the exponent. Missing the existing coefficient gives:

\frac{1}{n} \int u^{- 1 / n} d u = \frac{u^{1 - 1 / n}}{n - 1} + K,

which is exactly option B. Differentiating that choice exposes the error:

\frac{d}{d x} [\frac{(1 + n x^{n})^{1 - 1 / n}}{n - 1} + K] = n x^{n - 1} (1 + n x^{n})^{- 1 / n} .

Its derivative is the required integrand multiplied by the original exponent. Write the differential explicitly; when choices differ only by a coefficient, differentiate your selected answer.

Which three practice questions check the same method?

These are original practice variants, not verified past-paper questions. The first two separate the exponent from the number of compositions; the third checks definite-integral bounds. Use the same transformed-variable method, and attempt each question before reading its solution.

1. How do you integrate the function after three compositions?

Track the square of the input through three applications. The denominator coefficient becomes three, while the root remains a square root.

Question:

f (x) = \frac{x}{\sqrt{1 + x^{2}}}, h = f \circ f \circ f .

Find \int h (x) d x .

Worked answer:

y = x^{2}, T^{3} (y) = \frac{y}{1 + 3 y}, h (x) = \frac{x}{\sqrt{1 + 3 x^{2}}} .

u = 1 + 3 x^{2}, d u = 6 x d x .

\int h (x) d x = \frac{1}{6} \int u^{- 1 / 2} d u = \frac{\sqrt{1 + 3 x^{2}}}{3} + K .

2. How do you integrate the weighted function after two compositions?

Track cubes through two applications. The extra factor in the integrand supplies the power needed for substitution, but the denominator coefficient comes from the composition count.

Question:

f (x) = \frac{x}{(1 + x^{3})^{1 / 3}}, h = f \circ f .

Find \int x h (x) d x .

Worked answer:

y = x^{3}, T^{2} (y) = \frac{y}{1 + 2 y}, h (x) = \frac{x}{(1 + 2 x^{3})^{1 / 3}} .

u = 1 + 2 x^{3}, d u = 6 x^{2} d x .

\int x h (x) d x = \frac{1}{6} \int u^{- 1 / 3} d u = \frac{(1 + 2 x^{3})^{2 / 3}}{4} + K .

3. How do you handle the bounds in the definite integral?

Change both bounds when changing variables. Once the integral uses the transformed variable, its limits must use that variable too.

Question:

f (x) = \frac{x}{\sqrt{1 + x^{2}}}, g = f \circ f .

Evaluate \int_{0}^{1} g (x) d x .

Worked answer:

y = x^{2}, T^{2} (y) = \frac{y}{1 + 2 y}, g (x) = \frac{x}{\sqrt{1 + 2 x^{2}}} .

u = 1 + 2 x^{2}, d u = 4 x d x, x = 0 \Rightarrow u = 1, x = 1 \Rightarrow u = 3 .

\int_{0}^{1} g (x) d x = \frac{1}{4} \int_{1}^{3} u^{- 1 / 2} d u = \frac{\sqrt{3} - 1}{2} .

Cover the solutions and redo the differentials, coefficients and transformed bounds. Then differentiate the first two answers to check your work.

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Frequently asked questions

What is the correct answer to JEE 2007 Paper 2 Maths Question 52?

Option A is correct: the antiderivative is (1+nx^n)^(1-1/n)/[n(n-1)] + K. Repeated composition gives g(x) = x/(1+nx^n)^(1/n), so the integrand becomes x^(n-1)(1+nx^n)^(-1/n). Substitute u = 1+nx^n and use du = n^2 x^(n-1) dx.

How do you simplify repeated composition in the JEE 2007 integral question?

For f(x) = x/(1+x^n)^(1/n), taking the nth power gives [f(x)]^n = x^n/(1+x^n). Set y = x^n and T(y) = y/(1+y); induction shows that k applications give y/(1+ky). After n applications, g(x) = x/(1+nx^n)^(1/n) wherever the original real-valued compositions are defined, retaining the signed numerator.

Why is option B wrong in the JEE 2007 integral calculus question?

Option B results from incorrectly writing du = nx^(n-1) dx when u = 1+nx^n. The correct differential is du = n^2 x^(n-1) dx because differentiation retains the existing coefficient n and multiplies by the exponent n. Differentiating option B gives n times the required integrand.

Integral Calculus JEE 2007: Iterated Function Solution

What is the Integral Calculus JEE 2007 iterated-function question?

How does taking the nth power simplify the composition?

How do you prove the formula for repeated applications?

How do you integrate the simplified expression and verify option A?

How does a missing chain-rule factor produce option B?

Which three practice questions check the same method?

1. How do you integrate the function after three compositions?

2. How do you integrate the weighted function after two compositions?

3. How do you handle the bounds in the definite integral?

Frequently asked questions

What is the correct answer to JEE 2007 Paper 2 Maths Question 52?

How do you simplify repeated composition in the JEE 2007 integral question?

Why is option B wrong in the JEE 2007 integral calculus question?

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