Problem 129 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f\) and \(g\) are two continuous functions being even and odd, respectively, then \(\int_{-a}^{a} \frac{f(x)}{b^{g(x)}+1} d x\) is equal to \((a\) being any non-zero number and \(b\) is positive real number, \(b \neq 1\) ) (A) Independent of \(f\) (B) Independent of \(g\) (C) Independent of both \(f\) and \(g\) (D) None of these

Step-by-Step Solution

Answer

D: None of these

Step 1: Identify the Properties of Even and Odd Functions

An even function satisfies the condition \( f(x) = f(-x) \), and an odd function satisfies \( g(x) = -g(-x) \). These properties will be essential in evaluating the definite integral.

Step 2: Analyze the Integrand

Our integrand is \( \frac{f(x)}{b^{g(x)}+1} \). Since \( f(x) \) is even, \( f(x) = f(-x) \). The function \( g(x) \) is odd, meaning \( g(x) = -g(-x) \). We need to investigate if this affects the integrand.

Step 3: Substitute and Simplify Using Symmetry

Consider \( \frac{f(x)}{b^{g(x)}+1} \). Applying substitutions: \( x \to -x \), we get \( \frac{f(-x)}{b^{g(-x)}+1} = \frac{f(x)}{b^{-g(x)}+1} = \frac{f(x)}{\frac{1}{b^{g(x)}}+1} \). Simplifying further gives: \( \frac{f(x)}{b^{g(x)}+1} \). This shows the integrand's symmetry is dependent on both functions.

Step 4: Evaluate the Integral Over Symmetric Limits

Considering the interval \([-a, a]\), the symmetric properties of the limits imply \( \int_{-a}^{a} \) over an odd function without a matching even function component is zero. Here both terms in the fraction are critical, resulting in non-zero, aiding in calculating the integral.

Step 5: Determine the Independence

Since both the numerator and denominator involve explicit function forms of \( f(x) \) and \( g(x) \), and given no term zero out due to symmetry or simplification, the expression is not independent of either function because both exert influence.

Key Concepts

Even and odd functions

Even and odd functions are fundamental concepts in calculus, significantly affecting integrals and symmetry. An **even function** is one where the function value mirrors across the y-axis, meaning if you plug in a negative x value, you get the same result as the positive one. The mathematical representation of this is:

• Even function: \[ f(x) = f(-x) \]

For example, the function \( f(x) = x^2 \) is even, because \( f(2) = 4 \) and \( f(-2) = 4 \). The graph is symmetric about the y-axis.
An **odd function** is one where the function value is the opposite when you plug in a negative x; it's symmetric about the origin. The representation of an odd function is:

• Odd function: \[ g(x) = -g(-x) \]

For instance, \( g(x) = x^3 \) is odd because \( g(2) = 8 \) and \( g(-2) = -8 \). This characteristic makes analyzing integrands involving these functions interesting, especially over symmetric limits like \([-a, a]\).

Properties of symmetry

In calculus, the **properties of symmetry** play a pivotal role in integrating functions over symmetric intervals. If a function exhibits symmetry, it can simplify calculations.
For even functions, integral over a symmetric interval \([-a, a]\) depends purely on the even component, often allowing simplification if combined with odd components. The symmetry and even function properties indicate:

• The integral of an even function over symmetric limits often results in simplifying expressions.

On the other hand, the integral of an odd function over \([-a, a]\) is zero because the areas under the curve are equal and opposite on both sides of the y-axis. Calculating definite integrals exploits these properties:

• If only odd function: \[ ext{If } g(x) ext{ is odd, then } \ \ \int_{-a}^{a}{{g(x) \, dx}} = 0 \]

By evaluating functions with these symmetry properties, one can predict integral behavior without detailed calculation, especially when assessing complex expressions like \( \frac{f(x)}{b^{g(x)} + 1} \).

Continuous functions

The property of **continuous functions** is crucial in calculus, especially when dealing with definite integrals. A function is considered continuous if it has no breaks, jumps, or holes. You can visualize it as drawing a curve without lifting your pencil from the paper.

• Continuous functions: No gaps, breaks, or jumps.

For a definite integral, if the function is continuous on a closed interval \([a, b]\), this assures the existence of the integral, meaning you can compute the area under the curve over that interval. Calculus students often leverage the continuity of functions to:

• Ensure integrability over finite intervals.
• Predict the smooth behavior of function evaluations.

In problems like \( \int_{-a}^{a} \frac{f(x)}{b^{g(x)}+1} \, dx \), the continuous nature of \(f\) and \(g\) allows for smooth calculations across symmetric and critical points. This ensures that no abrupt changes affect integration results.