Key Concepts and Formulas:

1. Solving Functional Equations: This problem involves a specific type of functional equation where $f(x)$ and $f(1/x)$ are related. The standard technique is to substitute $x$ with $1/x$ to generate a second equation, forming a system of linear equations that can be solved for $f(x)$.
2. Definite Integration: The Fundamental Theorem of Calculus is used to evaluate definite integrals. Key integration rules needed are:
   • $\int \frac{1}{x} \, dx = \ln|x| + C$
   • $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (\text{for } n \neq -1)$
   • $\int c \, dx = cx + C$
   • $\int_a^b g(x) \, dx = G(b) - G(a), \text{ where } G'(x) = g(x)$

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Step-by-Step Solution:

Step 1: Generating a System of Functional Equations We are given the functional equation: $5 f(x) + 4 f\left(\frac{1}{x}\right) = \frac{1}{x} + 3 \quad \text{for } x > 0 \quad \text{...(1)}$ To solve for $f(x)$, we need another equation involving $f(x)$ and $f(1/x)$. We can obtain this by substituting $x$ with $1/x$ in Equation (1). This is a common strategy for functional equations of this form. Replacing $x$ with $\frac{1}{x}$ in Equation (1): $5 f\left(\frac{1}{x}\right) + 4 f\left(\frac{1}{1/x}\right) = \frac{1}{1/x} + 3$ Simplifying the terms: $5 f\left(\frac{1}{x}\right) + 4 f(x) = x + 3 \quad \text{...(2)}$ Now we have a system of two linear equations with $f(x)$ and $f(1/x)$ as variables:

1. $5 f(x) + 4 f\left(\frac{1}{x}\right) = \frac{1}{x} + 3$
2. $4 f(x) + 5 f\left(\frac{1}{x}\right) = x + 3$

Step 2: Solving for $f(x)$ We will now solve the system of linear equations obtained in Step 1 to find an explicit expression for $f(x)$. We can use the method of elimination. To eliminate $f(1/x)$, we can multiply Equation (1) by 5 and Equation (2) by 4, so that the coefficients of $f(1/x)$ become 20 in both equations. Multiply Equation (1) by 5: $5 \left( 5 f(x) + 4 f\left(\frac{1}{x}\right) \right) = 5 \left( \frac{1}{x} + 3 \right)$ $25 f(x) + 20 f\left(\frac{1}{x}\right) = \frac{5}{x} + 15 \quad \text{...(3)}$ Multiply Equation (2) by 4: $4 \left( 4 f(x) + 5 f\left(\frac{1}{x}\right) \right) = 4 (x + 3)$ $16 f(x) + 20 f\left(\frac{1}{x}\right) = 4x + 12 \quad \text{...(4)}$ Now, subtract Equation (4) from Equation (3) to eliminate $f(1/x)$: $(25 f(x) + 20 f\left(\frac{1}{x}\right)) - (16 f(x) + 20 f\left(\frac{1}{x}\right)) = \left(\frac{5}{x} + 15\right) - (4x + 12)$ $(25 - 16) f(x) = \frac{5}{x} + 15 - 4x - 12$ $9 f(x) = \frac{5}{x} - 4x + 3$ Divide by 9 to get the expression for $f(x)$: $f(x) = \frac{1}{9} \left( \frac{5}{x} - 4x + 3 \right)$

Step 3: Evaluating the Definite Integral We need to find the value of 18 \int_\limits{1}^{2} f(x) d x. Substitute the expression for $f(x)$ we found: 18 \int_\limits{1}^{2} \frac{1}{9} \left( \frac{5}{x} - 4x + 3 \right) d x We can take the constants outside the integral: \frac{18}{9} \int_\limits{1}^{2} \left( \frac{5}{x} - 4x + 3 \right) d x = 2 \int_\limits{1}^{2} \left( \frac{5}{x} - 4x + 3 \right) d x Now, we integrate each term with respect to $x$: = 2 \left[ 5 \int_\limits{1}^{2} \frac{1}{x} \, dx - 4 \int_\limits{1}^{2} x \, dx + 3 \int_\limits{1}^{2} 1 \, dx \right] Using the integration formulas: = 2 \left[ 5 [\ln|x|]_\limits{1}^{2} - 4 \left[\frac{x^2}{2}\right]_\limits{1}^{2} + 3 [x]_\limits{1}^{2} \right] Since the interval of integration is $[1, 2]$, $x > 0$, so $\ln|x| = \ln x$. $= 2 \left[ 5 (\ln 2 - \ln 1) - 4 \left(\frac{2^2}{2} - \frac{1^2}{2}\right) + 3 (2 - 1) \right]$

Step 4: Applying Limits of Integration and Final Calculation Evaluate the terms within the brackets:

• $\ln 2 - \ln 1 = \ln 2 - 0 = \ln 2$
• $\frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}$
• $2 - 1 = 1$ Substitute these values back into the expression: $= 2 \left[ 5 (\ln 2) - 4 \left(\frac{3}{2}\right) + 3 (1) \right]$ $= 2 \left[ 5 \ln 2 - 6 + 3 \right]$ $= 2 \left[ 5 \ln 2 - 3 \right]$ Distribute the 2: $= 10 \ln 2 - 6$

This result matches option (A).

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Common Mistakes & Tips:

• System of Equations: Ensure you correctly substitute $1/x$ for $x$ and simplify the resulting equation. The coefficients in the system are crucial.
• Algebraic Errors: Be very careful with algebraic manipulations, especially when multiplying equations and subtracting them. Sign errors are common.
• Integration of $1/x$: Remember that the integral of $1/x$ is $\ln|x|$, not $x$. Also, recall that $\ln 1 = 0$.
• Constant Factor: Do not forget to multiply the final integrated result by the constant factor that was initially outside the integral ($18$ in this case).

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Summary: The problem required solving a functional equation to find an explicit form for $f(x)$. This was achieved by creating a system of two linear equations by substituting $x$ with $1/x$. Once $f(x)$ was determined, it was substituted into the definite integral. Standard integration techniques and the Fundamental Theorem of Calculus were then applied to evaluate the integral, yielding the final result.

The final answer is $\boxed{10 \log _{\mathrm{e}} 2+6}$.