Problem 86 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\frac{x-1}{x+1}, f^{2}(x)=f(f(x)), \ldots f_{(x)}^{k+1}=f\left(f^{k}(x)\right)\), \(k=1,2,3, \ldots\) and \(\phi(x)=f^{1998}(x)\), then \(\int_{1 / e}^{1} \phi(x) d x=\) (A) 1 (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The answer is (A) 1.

Step 1: Understand the function composition

First, observe the function \( f(x) = \frac{x - 1}{x + 1} \). Our goal is to understand the composition of this function on itself multiple times. Let's compute a few iterations. We recognize that \( f^1(x) = f(x) \) is the first application.

Step 2: Compute the second function composition

To compute \( f^2(x) = f(f(x)) \), substitute \( f(x) \) back into itself. Doing this, we get: \[f(f(x)) = f\left( \frac{x - 1}{x + 1} \right) = \frac{\left(\frac{x-1}{x+1}\right) - 1}{\left(\frac{x-1}{x+1}\right) + 1} \]After simplifying:\[= \frac{\frac{x-1-x-1}{x+1}}{\frac{x-1+x+1}{x+1}} = \frac{-2}{2x} = -\frac{1}{x}\]

Step 3: Compute the third function composition

Next, compute \( f^3(x) = f(f^2(x)) \). Using \( f^2(x) = -\frac{1}{x} \), we find:\[f^3(x) = f\left(-\frac{1}{x}\right) = \frac{\left(-\frac{1}{x}\right) - 1}{\left(-\frac{1}{x}\right) + 1} \]Simplify to get:\[= \frac{-\frac{1+x}{x}}{-\frac{1-x}{x}} = \frac{1+x}{1-x}\]

Step 4: Compute the fourth function composition

Compute \( f^4(x) = f(f^3(x)) \). Substitute \( f^3(x) = \frac{1+x}{1-x} \):\[f^4(x) = f\left(\frac{1+x}{1-x}\right) = \frac{\left(\frac{1+x}{1-x}\right) - 1}{\left(\frac{1+x}{1-x}\right) + 1} \]Simplify:\[= \frac{\frac{1+x-(1-x)}{1-x}}{\frac{1+x+1-x}{1-x}} = \frac{\frac{2x}{1-x}}{\frac{2}{1-x}} = x \]Thus, \( f^4(x) = x \), indicating that the function composition returns to its original form by the fourth application.

Step 5: Find the composition for \(f^{1998}(x)\)

Since \( f^4(x) = x \), the function composition pattern repeats every 4 steps. To find \( f^{1998}(x) \), we calculate the remainder of 1998 divided by 4, which is 2.Hence, \( f^{1998}(x) = f^2(x) = -\frac{1}{x} \).

Step 6: Evaluate the definite integral

Now, the integral becomes:\[\int_{1/e}^{1} \phi(x) \, dx = \int_{1/e}^{1} -\frac{1}{x} \, dx \]Use the antiderivative of \(-\frac{1}{x}\), which is \(-\ln|x|\):\[= \left[ -\ln|x| \right]_{1/e}^{1} = \left(-\ln|1|\right) - \left(-\ln\left|\frac{1}{e}\right|\right)\]Simplifying gives:\[= 0 - (-(-1)) = -(-1) = 1 \]

Step 7: Choose the correct answer

The final result of the integral is 1, which matches with option (A).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that help us find the total accumulation of a quantity. In simpler terms, they allow us to calculate the exact area under a curve within a specific interval on the x-axis.

Consider a function, like the one we have in the exercise, and we are tasked to integrate it within certain limits, say from \(1/e\) to \(1\). This is done by evaluating the integral \[ \int_{1/e}^{1} \, -\frac{1}{x} \, dx \] using an antiderivative.

Here's a quick breakdown of the process:

• Find the antiderivative of the given function. For \( -\frac{1}{x} \), the antiderivative is \(-\ln|x|\).
• Evaluate this antiderivative at the upper limit of integration, which in this case is 1.
• Then, subtract the evaluation of the antiderivative at the lower limit, \(1/e\).

The result will give us the exact area under the curve between our specified limits, reflecting how the accumulation or change happens over that interval.

Function Iteration

Function iteration is the process of applying a function repeatedly. This means feeding the output of a function back into the same function. In our task, we begin with a simple rational function \( f(x) = \frac{x-1}{x+1} \), and we apply it repeatedly.

Function composition like this can often reveal interesting patterns or simplify complex relationships. In this case, applying the function repeatedly led us to find that after four iterations, \(f^4(x)\), the result returns to the original input, \(x\). This indicates a cycle in function outputs, often seen in such iterative processes.

By understanding the recurring pattern:

• \(f(x)\)
• \(f(f(x)) = -\frac{1}{x}\)
• \(f(f(f(x))) = \frac{1+x}{1-x}\)
• \(f(f(f(f(x)))) = x\)

we could predict the outcome for very large iterations by identifying cycles or periodicity in the iteration results. Since our exercise asked for \(f^{1998}(x)\), knowing that the cycle length is 4, reveals that \(f^{1998}(x) = f^2(x) = -\frac{1}{x}\). Understanding this iterative process is crucial for simplifying complex function compositions over many iterations.

Calculus Problem Solving

Calming the chaos of calculus problems often requires looking at a problem from multiple angles, using our toolbox of knowledge to peel back complex layers. Various strategies - such as understanding function composition and tackling integrals - rotate around patient step-by-step work.

In this exercise, we started by breaking down the function iterate, understanding how the function behaves when it's applied repeatedly. Noticing its cyclical behavior allows us to simplify what could be a daunting 1998th iteration computation.

Having simplified \(f^{1998}(x)\) using iteration patterns, we leveraged integral calculus to find the area under the curve. Calculus demanded careful attention through evaluation of the integral, crucially applying the property of logarithms in our final steps to avoid mistakes.

Key takeaways for calculus problem solving include:

• Don't shy away from breaking a problem into smaller parts.
• Look for patterns and cycles in functions, especially when facing iteration.
• Apply calculus concepts like integration meticulously, step by step.

By remaining methodical and patient, even the most complex calculus problems become approachable, leading to the correct solution.