For the function $f(x)$ to be continuous at $x=1$, the left-hand limit (LHL) and the right-hand limit (RHL) at $x=1$ must be equal, and also equal to the value of the function at that point.
The left-hand limit at $x=1$ is given by: $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 2) $$ Substituting $x=1$, we get: $$ \lim_{x \to 1^-} f(x) = 3(1) - 2 = 1 $$
The right-hand limit at $x=1$ is given by: $$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2 + ax) $$ Substituting $x=1$, we get: $$ \lim_{x \to 1^+} f(x) = 2(1)^2 + a(1) = 2 + a $$
For $f(x)$ to be continuous at $x=1$, the LHL must equal the RHL. Therefore: $$ 1 = 2 + a $$
Solving for $a$, we get: $$ a = 1 - 2 = -1 $$
Final Answer: -1
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