Problem 129 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The integral is independent of \( g \). So, option (B) is correct.

Step 1: Understand Even and Odd Functions

Recall that an even function satisfies the property \( f(x) = f(-x) \) for all \( x \), while an odd function satisfies \( g(x) = -g(-x) \). This will help us determine the symmetry of the integrand.

Step 2: Identify the Symmetry in the Integrand

Observe the integrand \( \frac{f(x)}{b^{g(x)}+1} \). Note that \( f(x) \) being even implies \( f(-x) = f(x) \), and \( g(x) \) being odd implies \( g(-x) = -g(x) \). Hence, \( b^{g(-x)} = b^{-g(x)} = \frac{1}{b^{g(x)}} \). Thus, the expression becomes \( \frac{f(-x)}{b^{-g(x)}+1} = \frac{f(x)}{\frac{1}{b^{g(x)}}+1} = \frac{f(x)b^{g(x)}}{1+b^{g(x)}} \).

Step 3: Symmetry of the Integrand's Reciprocal Part

Notice now that if we add the integrand \( \frac{f(x)}{b^{g(x)}+1} \) and the transformed form \( \frac{f(x)b^{g(x)}}{1+b^{g(x)}} \). Sum of these two expressions is \( f(x) \), which is independent of \( g(x) \).

Step 4: Evaluate the Integral Over Symmetric Limits

The integral of an even function over a symmetric interval (\([-a, a]\)) is equal to twice the integral from 0 to \( a \). The form of the integral indicates it depends symmetrically only on \( f(x) \), showing \( \int_{-a}^{a} f(x) dx \) results in a cancelling out of the dependence on the odd function \( g(x) \). Thus, it indicates computationally independence from \( g(x) \).

Step 5: Determine the Effect of b and Other Constants

Ensure the constant \( b \) affects the dynamics only in conjunction with \( g(x) \), affecting only the alternating form, thus symmetry result in no net change as previously argued to be aligned.

Key Concepts

Even and Odd Functions

In mathematics, understanding the nature of even and odd functions helps in analyzing the symmetry properties of functions. An **even function** is a function where the graph is symmetrical with respect to the y-axis. These functions satisfy the condition \( f(x) = f(-x) \) for every value of \( x \). Common examples of even functions include \( x^2 \), \( \cos(x) \), and any polynomial that only contains even-powered terms.

An **odd function**, on the other hand, has symmetry with respect to the origin of the coordinate system. This means that an odd function satisfies \( g(x) = -g(-x) \) for all \( x \). Examples of odd functions are \( x^3 \), \( \sin(x) \), and other polynomials with only odd-powered terms.

Knowing whether a function is even or odd can greatly simplify the process of integration, especially when working with definite integrals over symmetric limits such as \([-a, a]\). This symmetry often leads to simplifications since certain terms can cancel out.

Symmetry in Integrals

Symmetry can make evaluating integrals much easier. For integrals over symmetric limits (\([-a, a]\)), symmetry can lead to simplifications. Let's consider a function \( h(x) \). If \( h(x) \) is even, the integral over \([-a, a]\) simplifies because the curve above the x-axis is mirrored below, leading to the integral \( \int_{-a}^{a} h(x) \, dx = 2 \int_{0}^{a} h(x) \, dx \). This relation comes from both halves contributing equally because of symmetry.

In our exercise problem, the integrand is \( \frac{f(x)}{b^{g(x)}+1} \), with \( f(x) \) being even and \( g(x) \) odd. Combining these symmetrically in the integrand results in an expression that cancels the effects of \( g(x) \) over the symmetric interval. Thus, the integral of the whole expression becomes independent of the odd function \( g(x) \), highlighting the importance of symmetry in simplifying integrals.

Continuous Functions

A function is continuous if it doesn't have any breaks, jumps, or holes in its graph. In more rigorous terms, a continuous function at a point means that the limit of the function as it approaches that point equals the function's value at that point. The notion of continuity is essential in calculus because many of the theorems, especially those concerning integration, assume that the functions are continuous over their domains.

For definite integrals, continuity ensures that the integral can be computed as the accumulated area under the function's curve without any unexpected jumps or discontinuities. This makes the properties of even and odd functions reliable and applicable over any continuous range. In the problem at hand, the functions \( f(x) \) and \( g(x) \) are both continuous, which means they behave predictably over the interval \([-a, a]\). This assumption of continuity is crucial for leveraging the symmetry properties to simplify the integral.