We are given the function $f(x)$ defined as:

$f(x) = \frac{\sin(ax^2) - 9a}{x - 3} + 5 \quad \text{if} \quad x \neq 3$ $f(x) = 8a - 13 \quad \text{if} \quad x = 3$

We are also told that $f(x)$ is continuous at every point on the real line $\mathbb{R}$, and we need to determine the value of $a$.

Step 1: Continuity Condition

For $f(x)$ to be continuous at $x = 3$, we must have: $\lim_{x \to 3} f(x) = f(3)$ We already know that $f(3) = 8a - 13$. So, we need to find $\lim_{x \to 3} f(x)$ and set it equal to $f(3)$.

Step 2: Evaluate the Limit $\lim_{x \to 3} f(x)$

We start with the expression for $f(x)$ when $x \neq 3$: $f(x) = \frac{\sin(ax^2) - 9a}{x - 3} + 5$ To compute $\lim_{x \to 3} f(x)$, we first focus on the term $\frac{\sin(ax^2) - 9a}{x - 3}$. We apply L'Hôpital's Rule because substituting $x = 3$ directly results in an indeterminate form $\frac{0}{0}$.

Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and denominator with respect to $x$:

• The derivative of the numerator $\sin(ax^2) - 9a$ is: $\frac{d}{dx}[\sin(ax^2) - 9a] = 2ax \cos(ax^2)$
• The derivative of the denominator $x - 3$ is: $\frac{d}{dx}[x - 3] = 1$

Thus, by L'Hôpital's Rule: $\lim_{x \to 3} \frac{\sin(ax^2) - 9a}{x - 3} = \lim_{x \to 3} \frac{2ax \cos(ax^2)}{1} = 6a \cos(9a)$

Step 4: Set the Limit Equal to $f(3)$

The limit of $f(x)$ as $x \to 3$ is: $\lim_{x \to 3} f(x) = 6a \cos(9a) + 5$ For continuity at $x = 3$, this must be equal to $f(3) = 8a - 13$. Therefore, we set up the equation: $6a \cos(9a) + 5 = 8a - 13$ Simplifying: $6a \cos(9a) = 8a - 18$ $6a \cos(9a) = 2a - 18$

Step 5: Solve for $a$

If $a = 0$, both sides of the equation are satisfied because: $0 = 0$ Thus, $a = 0$ is a solution.

Step 6: Verify Continuity

If $a = 0$, the function simplifies as follows:

• For $x \neq 3$: $f(x) = \frac{\sin(0) - 0}{x - 3} + 5 = 5$
• For $x = 3$: $f(3) = 8(0) - 13 = -13$ Thus, the function is not continuous at $x = 3$ for $a = 0$. Therefore, we must solve $6a \cos(9a) = 2a - 18$ further.

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