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### Q5. **If \( f \) and \( g \) are real-valued functions defined by \( f(x) = x^2 + 7 \) and \( g(x) = 3x + 5 \), find the values of \( f(-2) + g(-1) \).**

- \( f(x) = x^2 + 7 \)
- \( g(x) = 3x + 5 \)

Now let's evaluate \( f(-2) \) and \( g(-1) \):

1. \( f(-2) = (-2)^2 + 7 = 4 + 7 = 11 \)
2. \( g(-1) = 3(-1) + 5 = -3 + 5 = 2 \)

So, \( f(-2) + g(-1) = 11 + 2 = 13 \).

---

### Q6. **Evaluate \( \int \frac{4x + 3}{x^2 - x - 2} \, dx \)**

This is an integral involving partial fraction decomposition.

First, factor the denominator:

\[
x^2 - x - 2 = (x - 2)(x + 1)
\]

Now, express the fraction as a sum of simpler fractions:

\[
\frac{4x + 3}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}
\]

Solve for \( A \) and \( B \) by equating:

\[
4x + 3 = A(x + 1) + B(x - 2)
\]

Expand and solve for \( A \) and \( B \).

---

### Q7. **Define the modulus function and write its domain and range.**

The modulus function, also known as the absolute value function, is defined as:

\[
f(x) = |x| = 
\begin{cases} 
x, & \text{if } x \geq 0 \\
-x, & \text{if } x < 0 
\end{cases}
\]

- **Domain**: The set of all real numbers, \( (-\infty, \infty) \).
- **Range**: Non-negative real numbers, \( [0, \infty) \).

---

### Q8. **A ball is drawn from a bag containing 5 white and 7 black balls. What is the probability of drawing one white and one black ball?**

This question assumes two balls are drawn. The total number of balls is \( 5 + 7 = 12 \). To find the probability of drawing one white and one black:

\[
P(\text{white then black}) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}
\]

Similarly, \( P(\text{black then white}) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132} \).

Thus, the total probability is:

\[
P(\text{one white, one black}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}
\]

---

### Q9. **Define Bayes' theorem for mutually exclusive events.**

Bayes' theorem is used for conditional probabilities, not mutually exclusive events. However, for conditional probability, it states:

\[
P(A|B) = \frac{P(B|A)P(A)}{P(B)}
\]

For mutually exclusive events \( A \) and \( B \), \( P(A \cap B) = 0 \), so Bayes' theorem does not apply.

---

Would you like more details or assistance on any specific question? Let me know if you'd like to focus on a particular problem.

Here are 5 related questions to explore further:
1. How would you solve partial fraction decomposition for the integral in Q6?
2. What are some applications of Bayes' theorem in real-world scenarios?
3. How does the domain and range of other functions compare to the modulus function?
4. Can you find the probability of drawing 2 white or 2 black balls in Q8?
5. What happens to the probability if balls are drawn with replacement in Q8?

**Tip**: Always double-check conditions (like replacement or mutual exclusivity) in probability problems to avoid common mistakes.

A set of math problems including algebraic functions, calculus, probability, Bayes' theorem, and others

This math problem set covers various topics including algebraic functions, calculus (integration), modulus function, Bayes' theorem, and probability. These problems are designed for students in Grades 10-12, with solutions including key formulas and theorems like Bayes' theorem and the fundamental theorem of calculus.

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