Problem 43 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a, b(a

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \). Option (B).

Step 1: Find the Discontinuity Points

The function \( f(x) = \frac{1}{1-x} \) is discontinuous at \( x = 1 \). For \( f \circ f(x) \), first evaluate \( f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1-\frac{1}{1-x}} = \frac{1-x}{-x} \). This expression is undefined when \( x = 0 \). Similarly, evaluate \( f \circ f \circ f(x) \) and determine where it is undefined, leading to discontinuity points being \( x = 0 \) and \( x = 1 \). Hence, \( a = 0 \) and \( b = 1 \).

Step 2: Simplify the Integrand

In the integrand \( \frac{f(x)}{f(x) + f(1-x)} \), substitute \( f(x) = \frac{1}{1-x} \) and \( f(1-x) = \frac{1}{x} \). Hence, the integrand becomes \( \frac{\frac{1}{1-x}}{\frac{1}{1-x} + \frac{1}{x}} = \frac{\frac{1}{1-x}}{\frac{x + 1-x}{x(1-x)}} = \frac{x}{1-x+x} = \frac{x}{1}. \) Therefore, the integrand simplifies to \( x \).

Step 3: Evaluate the Integral

With the integrand simplified to \( x \), evaluate the integral between \( a = 0 \) and \( b = 1 \):\[\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.\]

Step 4: Conclude the Result

Since the integral evaluates to \( \frac{1}{2} \), the answer to the question is \( \frac{1}{2} \), which corresponds to option (B).

Key Concepts

Discontinuity

Discontinuity is a key concept in calculus, specifically when dealing with functions. A function is said to be discontinuous at a certain point if there is a "break" or "gap" in its graph at that point. In mathematical terms, a function \( f(x) \) has a point of discontinuity at \( x = c \) if at least one of the following is true:

• \( \lim_{x \to c^{-}} f(x) eq \lim_{x \to c^{+}} f(x) \)
• \( \lim_{x \to c} f(x) \) does not exist
• \( f(c) \) is not defined

For the function \( f(x) = \frac{1}{1-x} \), it is clear that a discontinuity occurs at \( x = 1 \), where the denominator becomes zero, causing the function to be undefined. Similarly, when finding the composition \( f \circ f \circ f(x) \), you will find discontinuities at additional points. For this particular exercise, these points are \( x = 0 \) and \( x = 1 \). Understanding a function's discontinuities is crucial before performing operations like integration over its range.

Integration

Integration is a fundamental concept in calculus that allows us to find the area under a curve, among other things. It utilizes a process called antiderivation, where we find a function \( F(x) \) known as the antiderivative or integral, whose derivative is the given function \( f(x) \). In definite integration, we find the total area under a curve from one point to another along the \( x \)-axis.In this problem, after simplifying the integrand function \( \frac{f(x)}{f(x) + f(1-x)} \) to simply \( x \), the exercise focuses on finding the definite integral from \( a = 0 \) to \( b = 1 \). This involves computing \( \int_{0}^{1} x \, dx \). By integrating, you find the antiderivative of \( x \), which is \( \frac{x^2}{2} \), and then evaluate it at the boundaries (upper and lower limits). This gives us the area under the curve, which in this case is \( \frac{x^2}{2} \bigg|_{0}^{1} = \frac{1}{2} \). This concept of integration is immensely useful in various fields such as physics, engineering, and beyond.

Composite Function

A composite function occurs when one function is applied inside another, written as \( f \circ g(x) = f(g(x)) \). It substitutes the output of the inner function \( g(x) \) into the outer function \( f(x) \). This concept is essential in advanced calculus as it allows the study of more complex behaviors of functions.In this problem, the function \( f(x) = \frac{1}{1-x} \) is nested multiple times: \( f(f(f(x))) \). Step-by-step, we substitute one function into the other repeatedly. For example, \( f(f(x)) \) results in \( f\left(\frac{1}{1-x}\right) = \frac{1-x}{-x} \). Finding the behavior and any points of discontinuity in these inward nested functions helps us determine the nature of the overall composite function. This understanding is crucial, particularly in tasks involving differentiation and integration of complicated functions in calculus.