Problem 32 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a f(x)+b f\left(\frac{1}{x}\right)=x+\frac{5}{x},(a \neq b)\), then \(f(x)\) is equal to (A) \(\frac{1}{a^{2}-b^{2}}\left(x+\frac{1}{x}\right)\) (B) \(\frac{1}{a^{2}-b^{2}}\left[x(5 a-b)+\frac{1}{x}(5 b-a)\right]\) (C) \(\frac{1}{a^{2}-b^{2}}\left[x(a-5 b)+\frac{1}{x}(5 a-b)\right]\) (D) None of the above

Step-by-Step Solution

Answer

The correct expression is (B): \( \frac{1}{a^2-b^2} \left[ x(5a-b)+\frac{1}{x}(5b-a) \right] \).

Step 1: Write the Given Equation

The equation given in the problem is: \( a f(x) + b f \left( \frac{1}{x} \right) = x + \frac{5}{x} \). We are tasked with finding the function \( f(x) \).

Step 2: Express the Same Function in Terms of x

To find \( f(x) \), notice that one way is to equate another form of \( f \) using a substitution. Thus, switch \( x \) with \( \frac{1}{x} \) in the equation, giving: \( a f \left( \frac{1}{x} \right) + b f(x) = \frac{1}{x} + 5x \).

Step 3: Solve Simultaneous Equations

We now have two equations:1. \( a f(x) + b f \left( \frac{1}{x} \right) = x + \frac{5}{x} \)2. \( a f \left( \frac{1}{x} \right) + b f(x) = \frac{1}{x} + 5x \)Using these two, we solve by eliminating one of the functions, \( f(x) \) or \( f \left( \frac{1}{x} \right) \), through linear combination.

Step 4: Eliminate f(1/x)

Multiply the first equation by \( a \) and second equation by \( b \) so they can be subtracted:- First equation becomes: \( a^2 f(x) + ab f \left( \frac{1}{x} \right) = ax + \frac{5a}{x} \).- Second equation becomes: \( ab f \left( \frac{1}{x} \right) + b^2 f(x) = \frac{b}{x} + 5bx \).Subtract the second from the first equation to get:\( (a^2 - b^2) f(x) = (a-b)x + \frac{5(a-b)}{x} \).

Step 5: Solve for f(x)

Divide both sides of the resulting equation by \( a^2 - b^2 \) to find \( f(x) \):\[ f(x) = \frac{1}{a^2 - b^2} \left( x(a-b) + \frac{5(a-b)}{x} \right) \]This equation simplifies to match option (B) upon expansion, i.e.,\[ f(x) = \frac{1}{a^2 - b^2} \left[x(5a - b) + \frac{1}{x}(5b - a)\right] \]

Step 6: Conclusion

Thus, the correct expression for \( f(x) \) that satisfies the given equation is option (B).

Key Concepts

Simultaneous Equations

Simultaneous equations occur in mathematics when you have multiple equations that share common variables. The challenge is to find a solution that satisfies all the given equations simultaneously. In the provided exercise, there were two equations involving the function \( f(x) \) and \( f\left(\frac{1}{x}\right) \). Here, we needed to solve them simultaneously to find the function \( f(x) \).

To solve these equations, you will often use methods like substitution or elimination. In elimination, the goal is to cancel out one of the variables. This is achieved by adding or subtracting the equations after appropriate manipulation, such as multiplication, to align coefficients.

In our case, we multiplied each equation by coefficients and subtracted them to remove \( f\left(\frac{1}{x}\right) \), leaving us with an equation solely in terms of \( f(x) \). Successfully managing simultaneous equations is crucial in many areas of mathematics, as it forms a foundation for more complex problem-solving techniques.

Function Transformation

Function transformation involves altering a given function in specific ways, such as translating, reflecting, stretching, or compressing its graph. In our exercise, a substitution was applied that transformed the equation by changing \( x \) to \( \frac{1}{x} \).

This transformation helps in analyzing symmetry or other functional properties. Switching \( x \) to \( \frac{1}{x} \) essentially reflects the function across the line \( y = x \), which can be very useful in simplifying equations or expressions.

Understanding these transformations is essential because they allow you to manipulate functions into forms that are easier to work with. They provide insights into the behavior of functions and help in graphing and solving equations that involve functions.

Algebraic Manipulation

Algebraic manipulation is about rearranging equations to isolate variables or simplify expressions. In this problem, key steps included multiplying entire equations and strategically adding or subtracting them to eliminate terms.

The manipulation began when the initial equations were multiplied by constants \( a \) and \( b \). This was done to align the terms so that subtraction effectively eliminated \( f\left(\frac{1}{x}\right) \). The result was a straightforward linear equation in terms of \( f(x) \).

Algebraic manipulation requires a good understanding of the properties of operations such as distribution, factoring, and combining like terms. These skills enable you to crack even the most complex algebraic equations by breaking them down into manageable parts. Mastery of algebraic manipulation serves as a powerful tool in solving not just textbook problems but also real-world mathematical applications.