Let f be a function such that 3f(x) + 2f(m/(19x)) = 5x, x != 0, where m = Σ(i=1 to 9) (i)^2. Then f(

Q: Wait, why do we replace $x$ with $\frac{15}{x}$? How did we know to do that?

This is a classic technique for solving functional equations involving reciprocal arguments. By replacing $x$ with $\frac{15}{x}$, we transform the term $f(\frac{15}{x})$ into $f(x)$, effectively creating a system of two linear equations. This allows us to eliminate the unknown function term and solve for $f(x)$ directly.

Q: Can't we just plug in $x=5$ and $x=2$ directly into the first equation?

If you plug in $x=5$, you get $3f(5) + 2f(3) = 25$. If you plug in $x=2$, you get $3f(2) + 2f(7.5) = 10$. You have four unknowns ($f(5), f(3), f(2), f(7.5)$) and only two equations. You cannot solve for the values individually without first finding the general form of $f(x)$.

Q: Is there a shortcut to finding $f(5) - f(2)$ without finding the full function $f(x)$?

While you could potentially manipulate the system to find the difference directly, finding the explicit function $f(x)$ is the most reliable and robust method. It prevents algebraic errors and gives you a clear path to the final answer.

Question

Let $f$ be a function such that $3f(x) + 2f\left(\frac{m}{19x}ight) = 5x, x 
eq 0$, where $m = \sum_{i=1}^{9} (i)^2$. Then $f(5) - f(2)$ is equal to

Options:

(A) 18

(B) 36

(C) 9

(D) -9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given functional equation is $3f(x) + 2f(\frac{m}{19x}) = 5x$. To proceed, we must first determine the value of the constant $m$, which is defined as the sum of squares: $m = \sum_{i=1}^{9} i^2$.

Using the standard formula for the sum of squares, $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$, we substitute $n=9$:

$$m = \frac{9(10)(19)}{6} = 285$$

Now, we calculate the ratio required in the functional equation: $\frac{m}{19} = \frac{285}{19} = 15$. Consequently, our functional equation simplifies to:

$$3f(x) + 2f\left(\frac{15}{x}ight) = 5x \quad \dots 	ext{(i)}$$

### The Symmetry Trick

To isolate $f(x)$, we employ the method of substitution. We replace $x$ with $\frac{15}{x}$ in Equation (i).

The term $f(x)$ transforms into $f(\frac{15}{x})$, and the term $f(\frac{15}{x})$ transforms into $f(\frac{15}{15/x})$, which simplifies back to $f(x)$. The right-hand side becomes $5(\frac{15}{x}) = \frac{75}{x}$.

This yields our second equation:

$$2f(x) + 3f\left(\frac{15}{x}ight) = \frac{75}{x} \quad \dots 	ext{(ii)}$$

### The Algebraic Dance

We now have a system of two linear equations in terms of $f(x)$ and $f(\frac{15}{x})$. To eliminate $f(\frac{15}{x})$, we multiply Equation (i) by 3 and Equation (ii) by 2:

$$9f(x) + 6f\left(\frac{15}{x}ight) = 15x \quad \dots 	ext{(iii)}$$

$$4f(x) + 6f\left(\frac{15}{x}ight) = \frac{150}{x} \quad \dots 	ext{(iv)}$$

Subtracting Equation (iv) from Equation (iii) gives:

$$(9-4)f(x) = 15x - \frac{150}{x}$$

$$5f(x) = 15x - \frac{150}{x}$$

Dividing by 5, we obtain the explicit form of the function:

$$f(x) = 3x - \frac{30}{x}$$

### Final Calculation

With the function $f(x)$ determined, we evaluate the required values. First, for $x=5$:

$$f(5) = 3(5) - \frac{30}{5} = 15 - 6 = 9$$

Next, for $x=2$:

$$f(2) = 3(2) - \frac{30}{2} = 6 - 15 = -9$$

Finally, we compute the difference:

$$f(5) - f(2) = 9 - (-9) = \mathbf{18}$$

Animated Solution for Mathematics - Functions: Let $f$ be a function such that $3 f (x) + 2 f (\frac{m}{19 x}) = 5 x, x \neq = 0$ , where $m = \sum_{i = 1}^{9} (i)^{2}$ . Then $f (5) - f (2)$ is equal to

Select Answer:

Visualized Solution

Understanding the Goal

Calculating $m$

Simplifying the Constant

The Simplified Equation

The Substitution Trick

Generating the Second Equation

Elimination Strategy

Subtracting the Equations

Finding the General Function $f (x)$

Calculating $f (5)$

Calculating $f (2)$

The Final Subtraction

Summary and Key Takeaways

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Symmetry Trick

The Algebraic Dance

Final Calculation

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a function such that 3f(x)+2f(19xm​)=5x,x=0, where m=∑i=19​(i)2. Then f(5)−f(2) is equal to

Select Answer:

Visualized Solution

Understanding the Goal

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Simplifying the Constant

The Simplified Equation

The Substitution Trick

Generating the Second Equation

Elimination Strategy

Subtracting the Equations

Finding the General Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(5)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

The Final Subtraction

Summary and Key Takeaways

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Symmetry Trick

The Algebraic Dance

Final Calculation

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