Problem 61 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(2 f(x)+3 f\left(\frac{1}{x}\right)=x^{2}-1\), then \(f(x)\) is (A) a periodic function (B) an even function (C) an odd function (D) None of these

Step-by-Step Solution

Answer

\( f(x) \) is (D) None of these.

Step 1: Understand the Function

We need to find a property of the function \( f(x) \) such that it satisfies the equation \( 2f(x) + 3f\left(\frac{1}{x}\right) = x^2 - 1 \). To solve for the property, we can try substituting and transforming \( x \).

Step 2: Substitute \( x = \frac{1}{x} \)

Substitute \( x \) with \( \frac{1}{x} \) in the equation: \[ 2f\left(\frac{1}{x}\right) + 3f(x) = \left(\frac{1}{x}\right)^2 - 1 = \frac{1}{x^2} - 1 \]

Step 3: Write the System of Equations

Now we have two equations: 1. \( 2f(x) + 3f\left(\frac{1}{x}\right) = x^2 - 1 \) 2. \( 2f\left(\frac{1}{x}\right) + 3f(x) = \frac{1}{x^2} - 1 \)

Step 4: Solve the System of Equations

Multiply the first equation by 2 and the second equation by 3 to align the coefficients:1. \( 4f(x) + 6f\left(\frac{1}{x}\right) = 2(x^2 - 1) \)2. \( 6f\left(\frac{1}{x}\right) + 9f(x) = 3\left(\frac{1}{x^2} - 1\right) \) Subtract one from the other to eliminate \( f\left(\frac{1}{x}\right) \):\( 4f(x) - 9f(x) = 2(x^2 - 1) - 3\left(\frac{1}{x^2} - 1\right) \)

Step 5: Simplify the Equation

This simplifies to:\[ -5f(x) = 2x^2 - 2 - \frac{3}{x^2} + 3 \]\[ -5f(x) = 2x^2 + 1 - \frac{3}{x^2} \]Solve for \( f(x) \):\[ f(x) = -\frac{1}{5}\left( 2x^2 + 1 - \frac{3}{x^2} \right) \]

Step 6: Check for Parity

Determine whether \( f(x) \) is even or odd:\( f(x) \) is neither even nor odd because no combination of functions and substitutions results in a simplification to \( f(-x) = f(x) \) or \( f(-x) = -f(x) \) for all \( x \).

Step 7: Conclusion

\( f(x) \) does not manifest periodicity or symmetry with respect to the y-axis or origin, given its rich structure involving \( x^2 \), \( \frac{1}{x^2} \), and linear constants.

Key Concepts

Even Function

An even function is a type of mathematical function where the function's graph is symmetric with respect to the y-axis. This means that the left side of the graph mirrors the right side. Formally, a function \( f(x) \) is considered even if the equation \( f(-x) = f(x) \) holds true for all values of \( x \). This symmetry arises because when you replace \( x \) with \(-x\), the output remains unchanged.
To determine if a function is even, you simply substitute \( -x \) for \( x \) in the function. If the simplified expression equals the original function, the function is even.
If we take the specific example provided, where \( f(x) = -\frac{1}{5}(2x^2 + 1 - \frac{3}{x^2}) \), substituting \( -x \) into this function doesn’t yield the original function. Thus, it lacks the symmetry needed to be classified as even.

Odd Function

Odd functions exhibit a different type of symmetry compared to even functions. A function \( f(x) \) is called odd if \( f(-x) = -f(x) \) for every \( x \) in the domain of \( f \). Graphically, an odd function has rotational symmetry around the origin. This means if you rotate the graph 180 degrees about the origin, it would look the same as the original graph.
An easy way to verify whether a function is odd is to substitute \( -x \) into the function. If multiplying this result by \(-1\) returns the original function, then the function is odd.
In the example we analyzed, substituting \( -x \) into \( f(x) = -\frac{1}{5}(2x^2 + 1 - \frac{3}{x^2}) \) does not result in \(-f(x)\). Therefore, the function isn’t characterized as odd, which means it doesn’t have the necessary rotational symmetry.

Periodic Function

A periodic function is one that repeats its values at regular intervals or periods. The most straightforward examples of periodic functions are the sine and cosine functions, which repeat every \( 2\pi \). For a function \( f(x) \) to be periodic, there must exist some positive value \( T \) such that \( f(x+T) = f(x) \) for all \( x \) in the domain.
Periodic functions maintain consistency over their intervals and demonstrate repeating behavior.
In evaluating the function from the given problem, \( f(x) = -\frac{1}{5}(2x^2 + 1 - \frac{3}{x^2}) \), it becomes clear that no such constant \( T \) exists where the function repeats its values consistently over the entire domain. The components \( x^2 \) and \( \frac{1}{x^2} \) within the formula do not permit such repetition. Thus, \( f(x) \) isn't a periodic function.