Problem 98 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)\) be a continuous function in \([-2,2]\) such that \(f(x)+f(y)=f(x+y)\), then \(\int_{-2}^{2} f(x) d x=\) (A) \(2 \int_{0}^{2} f(x) d x\) (B) 0 (C) \(2 f(2)\) (D) None of these

Step-by-Step Solution

Answer

The integral is 0, matching option (B).

Step 1: Analyzing functional equation

Given that the function \( f(x) \) satisfies \( f(x) + f(y) = f(x+y) \), we can see that this is a functional equation. Let \( x = 0\), then \( f(0) + f(y) = f(y) \), which implies \( f(0) = 0 \).

Step 2: Identifying a possible form for the function

The functional equation \( f(x) + f(y) = f(x+y) \) resembles Cauchy's functional equation, whose general solution over the reals is \( f(x) = cx \), where \( c \) is a constant. Since \( f(0) = 0 \), we consider \( f(x) = cx \).

Step 3: Determine if this form holds

If \( f(x) = cx \), then \( f(x) + f(y) = cx + cy = c(x+y) = f(x+y) \), which satisfies the given functional equation. Thus, \( f(x) = cx \).

Step 4: Calculating the integral

Given \( f(x) = cx \), the integral \( \int_{-2}^{2} f(x) \, dx = \int_{-2}^{2} cx \, dx \). This simplifies to \( c \int_{-2}^{2} x \, dx \).

Step 5: Evaluate the integral

The integral \( \int_{-2}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{-2}^{2} = \frac{2^2}{2} - \frac{(-2)^2}{2} = 2 - 2 = 0 \). Hence, \( c \int_{-2}^{2} x \, dx = c \cdot 0 = 0 \).

Step 6: Conclusion on the integral value

Thus, \( \int_{-2}^{2} f(x) \, dx = 0 \), which matches option (B).

Key Concepts

Continuous Function

A continuous function is one where small changes in the input result in small changes in the output. In other words, if you have a continuous function, you can draw its graph without lifting your pen from the paper. This property is crucial for understanding the behavior of functions and integral calculus. In the given problem, the function \( f(x) \) is continuous over the interval \([-2, 2]\).

When a function is continuous over an interval, it doesn't have any gaps, jumps, or discontinuities in that range. This is significant because it means the function behaves predictably, a key condition when evaluating integrals. In continuous functions, you can reasonably analyze, calculate, and predict the total area under the curve in a specified interval, making integration possible.

Cauchy's Functional Equation

Cauchy's Functional Equation is a classic form of a functional equation, expressed as \( f(x) + f(y) = f(x + y) \). This equation suggests that the function \( f \) behaves like an additive function. A basic solution to this equation over the real numbers is linear, meaning that \( f(x) = cx \), where \( c \) is a constant.

In the exercise, the functional equation given, \( f(x) + f(y) = f(x+y) \), indicates that any function satisfying this must align with Cauchy's solution unless additional constraints are given. Since we are also told that \( f(x) \) is continuous, this excludes non-linear types of solutions that are always erratic unless further explored with advanced techniques. Here, with \( f(0) = 0 \), it's logical to conclude \( f(x) \) takes the form of \( cx \), fulfilling our functional equation and the condition given by continuity.

Definite Integral

A definite integral is a way to calculate the area under the curve of a function over a specified interval, from a point \( a \) to a point \( b \). The definite integral of a function \( f \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \).

In this exercise, we compute \( \int_{-2}^{2} f(x) \, dx \) for the function \( f(x) = cx \). Calculating this involves integrating over the specified interval, a key step in determining how our function behaves across this range. In our problem, this resulted in the integral of \( cx \) simplifying to \( c \int_{-2}^{2} x \, dx \). The immediate simplification—\( \int_{-2}^{2} x \, dx \)—ends up being zero because the areas under the line \( y = x \) from \(-2\) to \(0\) cancels out the area from \(0\) to \(2\). It essentially shows that no net area is enclosed within the bounds, leading to a solution of zero for this integral.