Math problems covering functions, sets, and algebraic expressions.

Let's solve each problem one by one with detailed explanations.

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1. Let $ R = \{(x, y) : y = x + 1\} $ and $ y \in \{0, 1, 2, 3, 4, 5\} $. List the elements of $ R $.

We are given a relation $ R $ defined as:
$$
R = \{(x, y) : y = x + 1\}
$$
with the condition that $ y \in \{0, 1, 2, 3, 4, 5\} $.

So we need to find all pairs $ (x, y) $ such that $ y = x + 1 $ and $ y $ is in the set $ \{0, 1, 2, 3, 4, 5\} $.

Let’s go through possible values of $ y $:

- If $ y = 0 $, then $ x = y - 1 = -1 $
- If $ y = 1 $, then $ x = 0 $
- If $ y = 2 $, then $ x = 1 $
- If $ y = 3 $, then $ x = 2 $
- If $ y = 4 $, then $ x = 3 $
- If $ y = 5 $, then $ x = 4 $

So the elements of $ R $ are:
$$
R = \{(-1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)\}
$$

✔ Answer:
$$
\boxed{R = \{(-1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)\}}
$$

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2. Let $ f $ be the subset of $ \mathbb{Q} \times \mathbb{Z} $ defined by

$$
f = \left\{ \left( \frac{m}{n}, m \right) : m, n \in \mathbb{Z}, n \neq 0 \right\}
$$
Is $ f $ a function from $ \mathbb{Q} $ to $ \mathbb{Z} $? Justify your answer.

We need to check if $ f $ defines a function from $ \mathbb{Q} $ to $ \mathbb{Z} $. A function must assign exactly one output in $ \mathbb{Z} $ to each input in $ \mathbb{Q} $.

But consider:
The rational number $ \frac{m}{n} $ can have multiple representations.

For example:
Take $ \frac{1}{2} $. This can be written as $ \frac{1}{2} $ or $ \frac{2}{4} $, etc.

In $ f $, we define:
- For $ \frac{1}{2} $: $ f\left(\frac{1}{2}\right) = 1 $
- For $ \frac{2}{4} $: $ f\left(\frac{2}{4}\right) = 2 $

But $ \frac{1}{2} = \frac{2}{4} $, so the same input maps to different outputs: 1 and 2.

This violates the definition of a function (not well-defined).

Thus, $ f $ is not a function because it assigns multiple values to the same rational number.

✔ Answer:
$$
\boxed{\text{No, } f \text{ is not a function because it is not well-defined — different representations of the same rational number give different outputs.}}
$$

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3. The function $ f $ which maps temperature in Celsius into temperature in Fahrenheit is defined by $ f(c) = \frac{9}{5}c + 32 $. Find $ f(0) $.

Just plug in $ c = 0 $:
$$
f(0) = \frac{9}{5}(0) + 32 = 0 + 32 = 32
$$

✔ Answer:
$$
\boxed{f(0) = 32}
$$

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4. If $ f(x) = x^3 - \frac{1}{x^3} $, prove that $ f(x) + f\left(\frac{1}{x}\right) = 0 $.

Given:
$$
f(x) = x^3 - \frac{1}{x^3}
$$

Compute $ f\left(\frac{1}{x}\right) $:
$$
f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^3 - \frac{1}{\left(\frac{1}{x}\right)^3} = \frac{1}{x^3} - x^3
$$

Now add:
$$
f(x) + f\left(\frac{1}{x}\right) = \left(x^3 - \frac{1}{x^3}\right) + \left(\frac{1}{x^3} - x^3\right) = x^3 - \frac{1}{x^3} + \frac{1}{x^3} - x^3 = 0
$$

✔ Answer:
$$
\boxed{f(x) + f\left(\frac{1}{x}\right) = 0 \quad \text{is proven.}}
$$

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5. If $ y = \frac{6x - 5}{x - 6} $, prove that $ f(y) = x $, $ x \ne \frac{6}{5} $.

Wait — this seems to suggest that $ f $ is defined such that $ f(y) = x $. But we are given $ y $ in terms of $ x $. So we are to solve for $ x $ in terms of $ y $ and show that $ x = f(y) $, i.e., inverse function.

Let:
$$
y = \frac{6x - 5}{x - 6}
$$

Solve for $ x $ in terms of $ y $:

Multiply both sides by $ x - 6 $:
$$
y(x - 6) = 6x - 5
\Rightarrow yx - 6y = 6x - 5
$$

Bring all terms to one side:
$$
yx - 6x = 6y - 5
\Rightarrow x(y - 6) = 6y - 5
\Rightarrow x = \frac{6y - 5}{y - 6}
$$

So $ x = f(y) = \frac{6y - 5}{y - 6} $

Thus, $ f(y) = x $, provided $ y \ne 6 $, and also $ x \ne 6 $, but original expression has $ x \ne 6 $.

Also, $ x \ne \frac{6}{5} $? Wait — let's check when $ y $ is undefined.

Actually, the restriction $ x \ne \frac{6}{5} $ might be a typo or misstatement.

Wait — maybe it's saying that $ x \ne \frac{6}{5} $ because $ y $ could be undefined? No.

From the original function: denominator $ x - 6 \ne 0 \Rightarrow x \ne 6 $

But $ x = \frac{6}{5} $ is fine unless it causes issues.

Let’s see: when $ x = \frac{6}{5} $,
$$
y = \frac{6 \cdot \frac{6}{5} - 5}{\frac{6}{5} - 6} = \frac{\frac{36}{5} - 5}{\frac{6 - 30}{5}} = \frac{\frac{11}{5}}{-\frac{24}{5}} = -\frac{11}{24}
$$
Defined.

So why $ x \ne \frac{6}{5} $? Maybe a mistake.

Alternatively, perhaps the problem meant to say "for $ x \ne 6 $"?

But anyway, we have shown:
$$
x = \frac{6y - 5}{y - 6}
$$

So if $ f(y) = \frac{6y - 5}{y - 6} $, then $ f(y) = x $

Hence, $ f(y) = x $, as required.

Note: $ y \ne 6 $, so $ x \ne \frac{6}{5} $? Not necessarily.

Wait — if $ y = 6 $, then from earlier:
$$
6 = \frac{6x - 5}{x - 6} \Rightarrow 6(x - 6) = 6x - 5 \Rightarrow 6x - 36 = 6x - 5 \Rightarrow -36 = -5 \quad \text{Contradiction}
$$
So $ y \ne 6 $, so $ x \ne ? $

But no restriction on $ x = \frac{6}{5} $.

Possibly the condition $ x \ne \frac{6}{5} $ is a typo.

Anyway, we’ve shown:
$$
\boxed{f(y) = x = \frac{6y - 5}{y - 6}, \quad y \ne 6}
$$

✔ Answer:
$$
\boxed{f(y) = x \text{ where } x = \frac{6y - 5}{y - 6}, \text{ hence } f(y) = x \text{ is verified.}}
$$

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6. Let $ f: X \to Y $ be defined by $ f(x) = x^2 $ for all $ x \in X $, where $ X = \{-2, -1, 0, 1, 3\} $ and $ Y = \{0, 1, 4, 7, 9, 10\} $. Write the relation $ f $ in roster form. Is $ f $ a function?

First, compute $ f(x) = x^2 $ for each $ x \in X $:

- $ f(-2) = (-2)^2 = 4 $
- $ f(-1) = (-1)^2 = 1 $
- $ f(0) = 0^2 = 0 $
- $ f(1) = 1^2 = 1 $
- $ f(3) = 3^2 = 9 $

So the relation $ f $ in roster form is:
$$
f = \{(-2, 4), (-1, 1), (0, 0), (1, 1), (3, 9)\}
$$

Now, is $ f $ a function?

Yes — every element in $ X $ appears exactly once as the first component, and maps to one value in $ Y $. Even though $ -1 $ and $ 1 $ map to same value (1), that's allowed in functions.

Also, $ f(x) \in Y $? Check:
- 4 ∈ Y ✔
- 1 ∈ Y ✔
- 0 ∈ Y ✔
- 9 ∈ Y ✔

All outputs are in $ Y $.

✔ Answer:
$$
\boxed{
\begin{aligned}
&\text{Roster form: } f = \{(-2, 4), (-1, 1), (0, 0), (1, 1), (3, 9)\} \\
&\text{Yes, } f \text{ is a function.}
\end{aligned}
}
$$

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7. Determine a quadratic function $ f(x) = ax^2 + bx + c $ if $ f(0) = 6 $, $ f(2) = 11 $, $ f(-3) = 6 $.

We are given:
- $ f(0) = 6 \Rightarrow a(0)^2 + b(0) + c = 6 \Rightarrow c = 6 $
- $ f(2) = 11 \Rightarrow a(4) + b(2) + 6 = 11 \Rightarrow 4a + 2b = 5 $ → (1)
- $ f(-3) = 6 \Rightarrow a(9) + b(-3) + 6 = 6 \Rightarrow 9a - 3b = 0 $ → (2)

Now solve equations:

From (2): $ 9a - 3b = 0 \Rightarrow 3a = b \Rightarrow b = 3a $

Plug into (1):
$$
4a + 2(3a) = 5 \Rightarrow 4a + 6a = 5 \Rightarrow 10a = 5 \Rightarrow a = \frac{1}{2}
$$

Then $ b = 3a = \frac{3}{2} $, $ c = 6 $

So:
$$
f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6
$$

✔ Answer:
$$
\boxed{f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6}
$$

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8. Find the domain and range of the function $ f(x) = \frac{x+2}{|x+2|} $.

Note: $ |x+2| $ is in the denominator, so $ x + 2 \ne 0 \Rightarrow x \ne -2 $

So domain: $ \mathbb{R} \setminus \{-2\} $

Now simplify:
$$
f(x) = \frac{x+2}{|x+2|}
$$

This is the sign function of $ x+2 $:

- If $ x + 2 > 0 $ ⇒ $ x > -2 $, then $ f(x) = \frac{x+2}{x+2} = 1 $
- If $ x + 2 < 0 $ ⇒ $ x < -2 $, then $ f(x) = \frac{x+2}{-(x+2)} = -1 $

So:
- $ f(x) = 1 $ for $ x > -2 $
- $ f(x) = -1 $ for $ x < -2 $
- Undefined at $ x = -2 $

So range = $ \{-1, 1\} $

✔ Answer:
$$
\boxed{
\begin{aligned}
\text{Domain: } & \mathbb{R} \setminus \{-2\} \\
\text{Range: } & \{-1, 1\}
\end{aligned}
}
$$

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9. Find the domain and range of $ f(x) = \frac{x^2}{1 + x^2} $.

Domain: Denominator $ 1 + x^2 \ge 1 > 0 $ for all real $ x $, so defined everywhere.

⇒ Domain: $ \mathbb{R} $

Range: Let $ y = \frac{x^2}{1 + x^2} $

Note: $ x^2 \ge 0 $, so $ y \ge 0 $

Also:
$$
y = \frac{x^2}{1 + x^2} = 1 - \frac{1}{1 + x^2}
$$

Since $ x^2 \ge 0 $, $ 1 + x^2 \ge 1 $, so $ \frac{1}{1 + x^2} \le 1 $, and $ > 0 $

So:
$$
0 < \frac{1}{1 + x^2} \le 1 \Rightarrow -1 \le -\frac{1}{1 + x^2} < 0 \Rightarrow 0 \le y < 1
$$

As $ x \to 0 $, $ y \to 0 $

As $ x \to \infty $, $ y \to 1 $, but never reaches 1.

So $ y \in [0, 1) $

✔ Answer:
$$
\boxed{
\begin{aligned}
\text{Domain: } & \mathbb{R} \\
\text{Range: } & [0, 1)
\end{aligned}
}
$$

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10. $ A = \{1, 2, 3\}, B = \{1, 2, 3, 4\} $, and $ R = \{(x, y) : x \in A, y \in B, y = x + 1\} $

#### (i) Find $ A \times B $

Cartesian product:
$$
A \times B = \{(x, y) : x \in A, y \in B\}
$$

List all combinations:
- $ x = 1 $: $ (1,1), (1,2), (1,3), (1,4) $
- $ x = 2 $: $ (2,1), (2,2), (2,3), (2,4) $
- $ x = 3 $: $ (3,1), (3,2), (3,3), (3,4) $

So:
$$
A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}
$$

#### (ii) Write domain and range of $ R $

Given $ R = \{(x, y) : x \in A, y \in B, y = x + 1\} $

Find valid pairs:
- $ x = 1 $, $ y = 2 $ → $ (1,2) $
- $ x = 2 $, $ y = 3 $ → $ (2,3) $
- $ x = 3 $, $ y = 4 $ → $ (3,4) $

So $ R = \{(1,2), (2,3), (3,4)\} $

- Domain of $ R $: set of all first components = $ \{1, 2, 3\} $
- Range of $ R $: set of all second components = $ \{2, 3, 4\} $

✔ Answer:
$$
\boxed{
\begin{aligned}
(i)\quad & A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\} \\
(ii)\quad & \text{Domain of } R = \{1, 2, 3\}, \quad \text{Range of } R = \{2, 3, 4\}
\end{aligned}
}
$$

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✔ Final Answers Summary:

1. $ R = \{(-1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)\} $

2. No, $ f $ is not a function because it is not well-defined due to multiple representations of rationals.

3. $ f(0) = 32 $

4. Proven: $ f(x) + f(1/x) = 0 $

5. $ f(y) = x $ where $ x = \frac{6y - 5}{y - 6} $, so $ f(y) = x $

6. $ f = \{(-2, 4), (-1, 1), (0, 0), (1, 1), (3, 9)\} $; Yes, it's a function.

7. $ f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6 $

8. Domain: $ \mathbb{R} \setminus \{-2\} $, Range: $ \{-1, 1\} $

9. Domain: $ \mathbb{R} $, Range: $ [0, 1) $

10. (i) $ A \times B $: all 12 pairs listed above
(ii) Domain: $ \{1,2,3\} $, Range: $ \{2,3,4\} $

Let me know if you'd like these formatted nicely!