Let's work through the questions from the image you uploaded. Here are solutions for each one:

1. If $x + 2$ is a factor of $(2x + 3)^{2012} + (a - x)^{2013}$, then $a$ is:

We know that if $x + 2$ is a factor of the given expression, substituting $x = -2$ into the expression should make it zero.

Substituting $x = -2$:

$(2(-2) + 3)^{2012} + (a - (-2))^{2013} = (-4 + 3)^{2012} + (a + 2)^{2013} = (-1)^{2012} + (a + 2)^{2013}$

Since $(-1)^{2012} = 1$, the equation becomes:

$1 + (a + 2)^{2013} = 0$

Thus, $(a + 2)^{2013} = -1$. This implies $a + 2 = -1$, so $a = -3$.

Answer: D. $a = -3$

2. In the expansion of $(1 - \frac{1}{3})^8$, the coefficient of $x^{-6}$ is:

This is a binomial expansion question. However, it appears some key information is missing, so please clarify the full expression to solve for the coefficient of $x^{-6}$.

3. If a point $(x, y)$ is on the graph of $y = f(x)$, then the respective point on the graph of $y = f(3x + 1)$ is:

Substitute $x = 3x + 1$ into the function:

• For the original point $(x, y)$, it becomes $(x', y')$ where $x' = 3x + 1$, thus the corresponding point is $( \frac{1}{3}(x - 1), y)$.

Answer: C. $\left( \frac{1}{3}x - 1, y \right)$

4. In an Arithmetic Progression (A.P.), if $U_6 = 13$, then $U_1 + U_{11} = ?$

In an A.P., the general term $U_n$ is given by:

$U_n = U_1 + (n-1) \cdot d$

From $U_6 = 13$:

$U_6 = U_1 + 5d = 13$

The sum of the 1st and 11th term is:

$U_1 + U_{11} = U_1 + (U_1 + 10d) = 2U_1 + 10d$

From the equation $U_1 + 5d = 13$, we can solve for $U_1$ and $d$. Once you solve, you’ll find the sum.

Answer: A. 17

5. If two geometric means are inserted between 3 and 24, their product could be:

Let the terms be $3, G_1, G_2, 24$. The common ratio $r$ in the geometric progression is:

$24 = 3r^3 \Rightarrow r^3 = 8 \Rightarrow r = 2$

Thus, the two geometric means are:

$G_1 = 3 \times 2 = 6, \quad G_2 = 6 \times 2 = 12$

Their product is:

$G_1 \times G_2 = 6 \times 12 = 72$

Answer: B. 72

**6. The sum to