Doubt: AL-QLE-017-20260607-0223-q4

Good thing to flag — the phrase "infinitely many common solutions" is exactly why $x$ never gets multiplied or computed here. When two lines coincide, $x$ takes infinitely many values, so the question can't be about $x$ at all. It's a condition on the coefficients $a$ and $b$.

Two lines have infinitely many common solutions exactly when they are the same line. For that, the coefficients of the two equations must be proportional:

$\frac{a+2}{1} = \frac{b^2-4}{1} = \frac{3b}{1}$

(The last ratio is $\frac{3b}{1}$ because the constant $3b$ is compared with the constant $1$ of $x + y = 1$.)

This gives two simple equations.

From the $x$-coefficients: $a + 2 = 3b \Rightarrow a = 3b - 2$.

From the $y$-coefficients: $b^2 - 4 = 3b \Rightarrow b^2 - 3b - 4 = 0 \Rightarrow (b-4)(b+1) = 0$, so $b = 4$ or $b = -1$.

Each value of $b$ gives one valid pair $(a, b)$:

$b = 4$: $a = 3(4) - 2 = 10$, so $ab = 10 \times 4 = 40$.

$b = -1$: $a = 3(-1) - 2 = -5$, so $ab = (-5) \times (-1) = 5$.

Now the multiplication question: the multiplication does happen — it's the $a \times b$ in each pair. What we add is the two resulting values of $ab$, because the question asks for the sum of all possible values of $ab$:

$40 + 5 = 45$

And we never touch $x$: with coincident lines, $x$ can be any real number, so "multiplying all values of $x$" isn't even a defined quantity.

Final answer: B