Question

One of the possible functions $f(x)$ which satisfies $\int_{-2}^{2} f(x) dx = 0$ is

• A. $\log\left(\frac{2+x}{2-x}\right)$
• B. $\sin(2+x)$
• C. $2x^3 + 2x + 1$
• D. $2x \tan x$

Hint:

Any definite integral of the form $\int_{-a}^{a} f(x) dx$ should immediately trigger a check for whether the function $f(x)$ is odd or even. For odd functions, the integral is always zero, saving you from complex integration. Functions of the form $\log\left(\frac{a+x}{a-x}\right)$ are classic examples of odd functions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks for a function that integrates to zero over a symmetric interval $[-a, a]$, where $a=2$. A key property of definite integrals states that if $f(x)$ is an odd function (i.e., $f(-x) = -f(x)$), then $\int_{-a}^{a} f(x) dx = 0$.

Step 2: Key Formula or Approach:
Test each given option to see if it is an odd function by evaluating $f(-x)$ and checking if it equals $-f(x)$.

Step 3: Detailed Explanation:
Let's test the options for the property $f(-x) = -f(x)$: Option (1): $f(x) = \log\left(\frac{2+x}{2-x}\right)$ Evaluate $f(-x)$: \[ f(-x) = \log\left(\frac{2+(-x)}{2-(-x)}\right) = \log\left(\frac{2-x}{2+x}\right) \] Using the property of logarithms $\log(a/b) = -\log(b/a)$: \[ f(-x) = \log\left( \left(\frac{2+x}{2-x}\right)^{-1} \right) = -1 \cdot \log\left(\frac{2+x}{2-x}\right) = -f(x) \] Since $f(-x) = -f(x)$, this is an odd function. Therefore, its integral over $[-2, 2]$ will be zero. This is a possible function. Option (2): $f(x) = \sin(2+x)$ Evaluate $f(-x)$: \[ f(-x) = \sin(2-x) \] This is neither $\sin(2+x)$ nor $-\sin(2+x)$. It's neither even nor odd. The integral will not generally be zero. Option (3): $f(x) = 2x^3 + 2x + 1$ Evaluate $f(-x)$: \[ f(-x) = 2(-x)^3 + 2(-x) + 1 = -2x^3 - 2x + 1 \] This is not equal to $-f(x)$ which would be $-2x^3 - 2x - 1$. The $+1$ constant term prevents it from being odd. The integral of the odd parts ($2x^3+2x$) will be zero, but the integral of the constant $1$ will be $1 \cdot (2 - (-2)) = 4$, so the total integral is not zero. Option (4): $f(x) = 2x \tan x$ Evaluate $f(-x)$: \[ f(-x) = 2(-x) \tan(-x) = -2x (-\tan x) = 2x \tan x = f(x) \] This is an even function. Its integral over $[-2, 2]$ will be $2 \int_{0}^{2} f(x) dx$, which is generally not zero.

Step 4: Final Answer:
The function that satisfies the condition is $\log\left(\frac{2+x}{2-x}\right)$.