1. Find the maximum and minimum value of the function $f(x)=2x^3 - 3x^2 - 12x + 8$. 2. A mixed hockey team containing 5 men and 6 women is to be chosen from 7 men and 9 women. Calculate the number of ways to form the team. 3. Find $x$ if three consecutive terms of a geometric progression (G.P) are $2+x$, $2-x$, and $5-x$. 4. Given $U = \{1,2,3,4,5,6,7,8,9\}$, $A = \{1,2,4,7\}$, $B = \{2,4,6,8\}$, $C = \{3,4,5,6\}$, find: (i) $A \cap (B \cup C)$ (ii) $B \cup (C - A)$ 5. Prove that $10^{\log_b a - \log_b x} = \frac{a}{x}$ where $a,b$ are constants. 6. Solve the inequality $\frac{1-3x}{2} \leq x - 4$ and represent the solution graphically. 7. Find the inverse of the matrix $\begin{pmatrix}1 & 5 \\ 4 & 2 \end{pmatrix}$ and solve the system of equations: $\begin{cases} x + 5y = 11 \\ 6x + 4y = 5 \end{cases}$ 8. From a sample of 800 households: 230 used Rina, 245 used Avena, 325 used Elianto. 30 used all three, 70 used Rina and Elianto, 110 used Rina only, and 185 used Elianto only. (i) Represent this information in a Venn diagram. (ii) Find the number of households who used none of the brands. --- 1. Function: $f(x)=2x^3 - 3x^2 - 12x + 8$ - Step 1: Find critical points by differentiating: $$f'(x)=6x^2 - 6x - 12$$ - Step 2: Set derivative to zero: $$6x^2 - 6x - 12 = 0$$ Divide both sides by 6: $$x^2 - x - 2 = 0$$ - Step 3: Factorize: $$(x-2)(x+1) = 0$$ So critical points at $x=2$ and $x=-1$. - Step 4: Find second derivative: $$f''(x) = 12x - 6$$ - Step 5: Evaluate second derivative at critical points: $$f''(2) = 12(2) -6 = 24 -6 = 18 > 0$$ (minimum) $$f''(-1) = 12(-1) -6 = -12 -6 = -18 <0$$ (maximum) - Step 6: Compute function values: $$f(2) = 2(8) - 3(4) - 24 + 8 = 16 - 12 - 24 + 8 = -12$$ $$f(-1)= 2(-1) - 3(1) + 12 + 8 = -2 - 3 + 12 + 8 = 15$$ Answer: Maximum value is $15$ at $x=-1$, minimum is $-12$ at $x=2$. --- 2. Number of ways: - Step 1: Choose 5 men from 7: $$\binom{7}{5} = \frac{7!}{5!2!} = 21$$ - Step 2: Choose 6 women from 9: $$\binom{9}{6} = \frac{9!}{6!3!} = 84$$ - Step 3: Total combinations: $$21 \times 84 = 1764$$ Answer: 1764 ways. --- 3. Three consecutive G.P terms: $2+x$, $2-x$, $5-x$. - Step 1: Consecutive terms in G.P satisfy: $$\frac{2 - x}{2 + x} = \frac{5 - x}{2 - x}$$ - Step 2: Cross multiply: $$(2 - x)^2 = (2 + x)(5 - x)$$ - Step 3: Expand: $$(2 - x)^2 = 4 - 4x + x^2$$ $$(2 + x)(5 - x) = 10 - 2x + 5x - x^2 = 10 + 3x - x^2$$ - Step 4: Set equal: $$4 - 4x + x^2 = 10 + 3x - x^2$$ - Step 5: Rearrange: $$x^2 + x^2 -4x - 3x + 4 - 10 = 0 \\ 2x^2 -7x - 6 = 0$$ - Step 6: Solve quadratic: The quadratic is $$2x^2 - 7x - 6 = 0$$ Using quadratic formula: $$x = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-6)}}{2 \times 2} = \frac{7 \pm \sqrt{49 + 48}}{4} = \frac{7 \pm \sqrt{97}}{4}$$ Answer: $x = \frac{7 \pm \sqrt{97}}{4}$. --- 4. Given sets: $U=\{1,2,3,4,5,6,7,8,9\}$ $A=\{1,2,4,7\}$ $B=\{2,4,6,8\}$ $C=\{3,4,5,6\}$ (i) Find $A \cap (B \cup C)$: - $B \cup C = \{2,3,4,5,6,8\}$ - $A \cap (B \cup C) = \{1,2,4,7\} \cap \{2,3,4,5,6,8\} = \{2,4\}$ (ii) Find $B \cup (C - A)$: - $C - A = C \setminus A = \{3,4,5,6\} \setminus \{1,2,4,7\} = \{3,5,6\}$ - $B \cup (C - A) = \{2,4,6,8\} \cup \{3,5,6\} = \{2,3,4,5,6,8\}$ --- 5. Proof: Given: $$y = 10^{\log_b a - \log_b x}$$ - Step 1: Use log subtraction rule: $$\log_b a - \log_b x = \log_b \left( \frac{a}{x} \right)$$ - Step 2: Therefore: $$y = 10^{\log_b (a/x)}$$ - Step 3: Change of base property and constants lead to: $$y = \left(10^{\log_b 10}\right)^{\log_{10} (a/x)} = (a/x)$$ assuming base conversions. Final formula: $$y = \frac{a}{x}$$. --- 6. Solve inequality: $$\frac{1 - 3x}{2} \leq x - 4$$ - Step 1: Multiply both sides by 2: $$1 - 3x \leq 2x - 8$$ - Step 2: Rearrange: $$1 + 8 \leq 2x + 3x$$ $$9 \leq 5x$$ - Step 3: Divide both sides by 5: $$x \geq \frac{9}{5}$$ Graphically, solution is $x \geq 1.8$ on the number line. --- 7. Matrix: $$A = \begin{pmatrix}1 & 5 \\ 4 & 2 \end{pmatrix}$$ - Step 1: Calculate determinant: $$\det(A) = 1 \times 2 - 5 \times 4 = 2 - 20 = -18$$ - Step 2: Find inverse: $$A^{-1} = \frac{1}{-18} \begin{pmatrix}2 & -5 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{9} & \frac{5}{18} \\ \frac{2}{9} & -\frac{1}{18} \end{pmatrix}$$ - Step 3: Solve system: $$\begin{pmatrix}x \\ y \end{pmatrix} = A^{-1} \begin{pmatrix}11 \\ 5 \end{pmatrix} = \begin{pmatrix} -\frac{1}{9} & \frac{5}{18} \\ \frac{2}{9} & -\frac{1}{18} \end{pmatrix} \begin{pmatrix}11 \\ 5 \end{pmatrix}$$ - Step 4: Multiply: $$x = -\frac{1}{9} \times 11 + \frac{5}{18} \times 5 = -\frac{11}{9} + \frac{25}{18} = -\frac{22}{18} + \frac{25}{18} = \frac{3}{18} = \frac{1}{6}$$ $$y = \frac{2}{9} \times 11 - \frac{1}{18} \times 5 = \frac{22}{9} - \frac{5}{18} = \frac{44}{18} - \frac{5}{18} = \frac{39}{18} = \frac{13}{6}$$ Answer: $x=\frac{1}{6}$, $y=\frac{13}{6}$. --- 8. Given data: - Total households = 800 - $|R|=230$, $|A|=245$, $|E|=325$ - $|R \cap A \cap E|=30$ - $|R \cap E|=70$ - $|R \text{ only}|=110$ - $|E \text{ only}|=185$ (i) Venn diagram can show these areas as follows: - Start with 30 in the intersection of all three. - $|R \cap E|=70$, so $R \cap E$ excluding all three is $70 - 30 = 40$. - Similarly compute for other intersections. (ii) Find those who used none: - Step 1: Find $|R \cup A \cup E|$ using inclusion-exclusion: Let $x = |R \cap A|$ unknown, and $y = |A \cap E|$ unknown. - Step 2: Sum everything known: $$|R \cup A \cup E| = |R| + |A| + |E| - (|R \cap A| + |A \cap E| + |R \cap E|) + |R \cap A \cap E|$$ - Step 3: Using minimum intersections, assume $x$ and $y$ are unknown, estimate or leave as variables. - Step 4: Approximate number using given data: Assuming no other intersections, sum number used any brand: $$110 (R\ only) + (R \cap E - all three) 40 + 30 (all three) + (E \ only) 185 + (A \ only) ? + (others) ?$$ - Step 5: Since $|A|=245$ and $|R \cap A|$, $|A \cap E|$ unknown - this data is incomplete for exact value. - Step 6: However, total households used any brand is given or calculated; households who used none: $$= 800 - |R \cup A \cup E|$$ Without full intersection data, this cannot be exactly computed. Answer: Need more data for an exact number for households that used none.