We are given two functions and a specific condition. Let's break this down to find the value of $a$ when $a > 0$.

Step 1: Function Definition

• We are told that $g(x) = 3f(ax) + 1$.
• Additionally, we know $f(x) = x^2 - 3x$.

Thus, the function $g(x)$ becomes: $g(x) = 3f(ax) + 1 = 3\left( (ax)^2 - 3(ax) \right) + 1 = 3\left( a^2x^2 - 3ax \right) + 1.$ Simplifying this expression: $g(x) = 3a^2x^2 - 9ax + 1.$

Step 2: Use the Condition $g(2) = 1$

We are also given that $g(2) = 1$. Substituting $x = 2$ into the equation for $g(x)$: $g(2) = 3a^2(2)^2 - 9a(2) + 1 = 1.$ Simplifying this: $g(2) = 3a^2(4) - 18a + 1 = 1,$ $12a^2 - 18a + 1 = 1.$ Subtract 1 from both sides: $12a^2 - 18a = 0.$

Step 3: Solve the Equation for $a$

Factor the equation: $6a(2a - 3) = 0.$ This gives two possible solutions: $a = 0 \quad \text{or} \quad a = \frac{3}{2}.$

Step 4: Consider $a > 0$

Since we are told that $a > 0$, we reject $a = 0$ and conclude that: $a = \frac{3}{2}.$

Thus, the value of $a$ is $\frac{3}{2}$ for all $a > 0$.

Would you like further details or have any questions about this solution?

Here are 5 related questions to expand your understanding:

1. How do you factor a quadratic expression like $12a^2 - 18a = 0$?
2. Can you explain why $a = 0$ is not a valid solution in this case?
3. How would the solution change if $f(x)$ was a different quadratic function?
4. What happens to $g(x)$ if $f(x)$ has an added constant term?
5. Can you derive the general form of $g(x)$ for any function $f(x)$?

Tip: Always verify your solution by substituting it back into the original equation to check its validity.