1. State the domain and range of each relation.
For \(A = \{1, 2, 3, 4\}\), find \(A \times A\) first, then solve other parts.
Find \(A \times A\)
\[A \times A = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)\}\]
(i) Relation \( \{(x, y) \, | \, y = x\} \)
\[R_1 = \{(1, 1), (2, 2), (3, 3), (4, 4)\}\]
Domain:
\[\{1, 2, 3, 4\}\]
Range:
\[\{1, 2, 3, 4\}\]
(ii) Relation \( \{(x, y) \, | \, y + x = 5\} \)
\[R_2 = \{(1, 4), (2, 3), (3, 2), (4, 1)\}\]
Domain:
\[\{1, 2, 3, 4\}\]
Range:
\[\{1, 2, 3, 4\}\]
(iii) Relation \( \{(x, y) \, | \, x + y < 5\} \)
\[R_3 = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\}\]
Domain:
\[\{1, 2, 3, 4\}\]
Range:
\[\{1, 2, 3, 4\}\]
(iv) Relation \( \{(x, y) \, | \, x + y > 5\} \)
\[R_4 = \{(2, 4), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)\}\]
Domain:
\[\{2, 3, 4\}\]
Range:
\[\{2, 3, 4\}\]
2. Which of the following diagrams represent functions and of which type?
Fig (1):
Not a function. Each element in the domain is not mapped to exactly one element in the range.
Fig (2):
Function. It is one-to-one.
Fig (3):
Function. It is many-to-one.
Fig (4):
Not a function. An element in the domain is mapped to multiple elements in the range.
3. If \(g(x) = 3x + 2\) and \(h(x) = x^2 + 1\), then find:
(i) \(g(0)\)
\[g(0) = 3(0) + 2 = 2\]
(ii) \(g(-3)\)
\[g(-3) = 3(-3) + 2 = -9 + 2 = -7\]
(iii) \(g\left(\frac{2}{3}\right)\)
\[g\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right) + 2 = 2 + 2 = 4\]
(iv) \(h(1)\)
\[h(1) = (1)^2 + 1 = 1 + 1 = 2\]
(v) \(h(-4)\)
\[h(-4) = (-4)^2 + 1 = 16 + 1 = 17\]
(vi) \(h\left(-\frac{1}{2}\right)\)
\[h\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4}\]
4. Given \(f(x) = ax + b + 1\), where \(a\) and \(b\) are constant numbers, \(f(3) = 8\) and \(f(6) = 14\). Find the values of \(a\) and \(b\).
\[f(x) = ax + b + 1\]
Substitute \(f(3) = 8\):
\[8 = 3a + b + 1\]
\[3a + b = 7 \quad \text{(1)}\]
Substitute \(f(6) = 14\):
\[14 = 6a + b + 1\]
\[6a + b = 13 \quad \text{(2)}\]
Subtract (1) from (2):
\[(6a + b) – (3a + b) = 13 – 7\]
\[3a = 6 \]
\[ a = 2\]
Substitute \(a = 2\) in (1):
\[3(2) + b = 7\]
\[6 + b = 7 \]
\[ b = 1\]
\[a = 2, \, b = 1\]
5. Given \(g(x) = ax + b + 5\), where \(a\) and \(b\) are constant numbers, \(g(-1) = 0\) and \(g(2) = 10\). Find the values of \(a\) and \(b\).
\[g(x) = ax + b + 5\]
Substitute \(g(-1) = 0\):
\[0 = -a + b + 5\]
\[-a + b = -5 \quad \text{(1)}\]
Substitute \(g(2) = 10\):
\[10 = 2a + b + 5\]
\[2a + b = 5 \quad \text{(2)}\]
Subtract (1) from (2):
\[(2a + b) – (-a + b) = 5 – (-5)\]
\[3a = 10 \]
\[ a = \frac{10}{3}\]
Substitute \(a = \frac{10}{3}\) in (1):
\[-\frac{10}{3} + b = -5\]
\[b = -5 + \frac{10}{3} = -\frac{15}{3} + \frac{10}{3} = -\frac{5}{3}\]
\[a = \frac{10}{3}, \, b = -\frac{5}{3}\]
6. Consider the function \(f(x) = 5x + 1\). If \(f(x) = 32\), find the \(x\) value.
\[f(x) = 5x + 1\]
Substitute \(f(x) = 32\):
\[32 = 5x + 1\]
\[5x = 31 \]
\[ = \frac{31}{5}\]
\[x = \frac{31}{5}\]
7. Consider the function \(f(x) = cx^2 + d\), where \(c\) and \(d\) are constant numbers. If \(f(1) = 6\) and \(f(-2) = 10\), find the values of \(c\) and \(d\).
\[f(x) = cx^2 + d\]
Substitute \(f(1) = 6\):
\[6 = c(1)^2 + d\]
\[c + d = 6 \quad \text{(1)}\]
Substitute \(f(-2) = 10\):
\[10 = c(-2)^2 + d\]
\[4c + d = 10 \quad \text{(2)}\]
Subtract (1) from (2):
\[(4c + d) – (c + d) = 10 – 6\]
\[3c = 4 \]
\[ c = \frac{4}{3}\]
Substitute \(c = \frac{4}{3}\) in (1):
\[\frac{4}{3} + d = 6\]
\[d = 6 – \frac{4}{3} = \frac{18}{3} – \frac{4}{3} = \frac{14}{3}\]
\[c = \frac{4}{3}, \, d = \frac{14}{3}\]
