1. **Problem Statement:** Differentiate and analyze given functions, calculate statistics, and factorize polynomials as per the questions. --- ### Question Four **(a) Differentiate from first principles the function $f(x) = 3x^2$** 1. The definition of derivative from first principles is: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 2. Substitute $f(x) = 3x^2$: $$f'(x) = \lim_{h \to 0} \frac{3(x+h)^2 - 3x^2}{h}$$ 3. Expand numerator: $$3(x^2 + 2xh + h^2) - 3x^2 = 3x^2 + 6xh + 3h^2 - 3x^2 = 6xh + 3h^2$$ 4. So, $$f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2}{h} = \lim_{h \to 0} (6x + 3h) = 6x$$ **Answer:** $f'(x) = 6x$ **(b) Differentiate with respect to $x$:** (i) $y = 3x^3 - 4x^2 + 5x$ 1. Use power rule: $\frac{d}{dx} x^n = nx^{n-1}$ 2. Differentiate term-wise: $$\frac{dy}{dx} = 3 \times 3x^{2} - 4 \times 2x^{1} + 5 = 9x^2 - 8x + 5$$ (ii) $y = (2x - 3)^2 (x^2 + 4)^4$ 1. Use product rule: $\frac{d}{dx}(uv) = u'v + uv'$ 2. Let $u = (2x - 3)^2$, $v = (x^2 + 4)^4$ 3. Differentiate $u$: $$u' = 2(2x - 3) \times 2 = 4(2x - 3)$$ 4. Differentiate $v$: $$v' = 4(x^2 + 4)^3 \times 2x = 8x(x^2 + 4)^3$$ 5. Apply product rule: $$\frac{dy}{dx} = u'v + uv' = 4(2x - 3)(x^2 + 4)^4 + (2x - 3)^2 \times 8x (x^2 + 4)^3$$ 6. Factor common terms: $$(2x - 3)(x^2 + 4)^3 [4(x^2 + 4) + 8x(2x - 3)]$$ 7. Simplify inside bracket: $$4x^2 + 16 + 16x^2 - 24x = 20x^2 - 24x + 16$$ **Answer:** $$\frac{dy}{dx} = (2x - 3)(x^2 + 4)^3 (20x^2 - 24x + 16)$$ (iii) $y = \frac{x^2 + 2}{(x + 2)^2}$ 1. Use quotient rule: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$$ 2. Let $u = x^2 + 2$, $v = (x + 2)^2$ 3. Differentiate: $$u' = 2x, \quad v' = 2(x + 2)$$ 4. Apply quotient rule: $$\frac{dy}{dx} = \frac{2x (x + 2)^2 - (x^2 + 2) 2(x + 2)}{(x + 2)^4}$$ 5. Factor numerator: $$2(x + 2)[x(x + 2) - (x^2 + 2)] = 2(x + 2)(x^2 + 2x - x^2 - 2) = 2(x + 2)(2x - 2) = 4(x + 2)(x - 1)$$ 6. Simplify denominator: $$(x + 2)^4$$ **Answer:** $$\frac{dy}{dx} = \frac{4(x + 2)(x - 1)}{(x + 2)^4} = \frac{4(x - 1)}{(x + 2)^3}$$ **(c) Find gradient of $y = 9x - x^3$ at $x = -1$** 1. Differentiate: $$\frac{dy}{dx} = 9 - 3x^2$$ 2. Substitute $x = -1$: $$9 - 3(-1)^2 = 9 - 3 = 6$$ **Answer:** Gradient at $x = -1$ is 6 --- ### Question Five **(a) Find standard deviation of numbers 2, 3, 5, 6, 8 with mean 4.8** 1. Standard deviation formula: $$\sigma = \sqrt{\frac{1}{n} \sum (x_i - \bar{x})^2}$$ 2. Calculate squared deviations: $$(2 - 4.8)^2 = 7.84$$ $$(3 - 4.8)^2 = 3.24$$ $$(5 - 4.8)^2 = 0.04$$ $$(6 - 4.8)^2 = 1.44$$ $$(8 - 4.8)^2 = 10.24$$ 3. Sum: $$7.84 + 3.24 + 0.04 + 1.44 + 10.24 = 22.8$$ 4. Divide by $n=5$: $$\frac{22.8}{5} = 4.56$$ 5. Take square root: $$\sigma = \sqrt{4.56} \approx 2.136$$ **Answer:** Standard deviation is approximately 2.136 **(b) Frequency table analysis:** | Marks | Frequency | |-------|-----------| | 40-43 | 5 | | 44-47 | 7 | | 48-51 | 3 | | 52-55 | 8 | | 56-59 | 5 | | 60-63 | 2 | Total frequency $n=30$ (i) **Mode:** - Mode is the class with highest frequency. - Highest frequency is 8 for class 52-55. **Answer:** Mode class is 52-55 (ii) **Standard deviation:** 1. Find midpoints $x_i$: $$41.5, 45.5, 49.5, 53.5, 57.5, 61.5$$ 2. Calculate mean: $$\bar{x} = \frac{\sum f_i x_i}{n} = \frac{5\times41.5 + 7\times45.5 + 3\times49.5 + 8\times53.5 + 5\times57.5 + 2\times61.5}{30}$$ $$= \frac{207.5 + 318.5 + 148.5 + 428 + 287.5 + 123}{30} = \frac{1513}{30} = 50.433$$ 3. Calculate $\sum f_i (x_i - \bar{x})^2$: - $(41.5 - 50.433)^2 = 79.8$, times 5 = 399 - $(45.5 - 50.433)^2 = 24.3$, times 7 = 170.1 - $(49.5 - 50.433)^2 = 0.87$, times 3 = 2.61 - $(53.5 - 50.433)^2 = 9.41$, times 8 = 75.28 - $(57.5 - 50.433)^2 = 49.5$, times 5 = 247.5 - $(61.5 - 50.433)^2 = 122.1$, times 2 = 244.2 4. Sum: $$399 + 170.1 + 2.61 + 75.28 + 247.5 + 244.2 = 1138.69$$ 5. Variance: $$\sigma^2 = \frac{1138.69}{30} = 37.956$$ 6. Standard deviation: $$\sigma = \sqrt{37.956} \approx 6.16$$ **Answer:** Standard deviation is approximately 6.16 (iii) **Median mark:** 1. Median position: $$\frac{n+1}{2} = \frac{31}{2} = 15.5$$ 2. Cumulative frequencies: - 40-43: 5 - 44-47: 12 - 48-51: 15 - 52-55: 23 3. Median class is 52-55 (since 15.5 lies between 15 and 23) 4. Use median formula: $$L = 51.5, f = 8, F = 15, h = 4$$ $$\text{Median} = L + \frac{\frac{n}{2} - F}{f} \times h = 51.5 + \frac{15 - 15}{8} \times 4 = 51.5$$ **Answer:** Median mark is 51.5 --- ### Question Six **(a) Show $x=\frac{2}{3}$ is root of $3x^3 + x^2 - 8x + 4$ and factorize completely** 1. Substitute $x=\frac{2}{3}$: $$3\left(\frac{2}{3}\right)^3 + \left(\frac{2}{3}\right)^2 - 8\left(\frac{2}{3}\right) + 4 = 3 \times \frac{8}{27} + \frac{4}{9} - \frac{16}{3} + 4$$ $$= \frac{24}{27} + \frac{4}{9} - \frac{16}{3} + 4 = \frac{8}{9} + \frac{4}{9} - \frac{16}{3} + 4$$ $$= \frac{12}{9} - \frac{16}{3} + 4 = \frac{4}{3} - \frac{16}{3} + 4 = -4 + 4 = 0$$ 2. Since $x=\frac{2}{3}$ is root, factor out $(3x - 2)$: 3. Use polynomial division or synthetic division: $$3x^3 + x^2 - 8x + 4 \div (3x - 2) = x^2 + x - 2$$ 4. Factor quadratic: $$x^2 + x - 2 = (x + 2)(x - 1)$$ **Answer:** $$3x^3 + x^2 - 8x + 4 = (3x - 2)(x + 2)(x - 1)$$ **(b) Find $a$ and $b$ given remainders when dividing $x^4 + 3x^3 + ax^2 + bx - 1$ by $x-1$ and $x+2$** 1. Remainder theorem: remainder when divided by $x - c$ is $f(c)$ 2. For $x - 1$, remainder is 4: $$f(1) = 1 + 3 + a + b - 1 = 3 + a + b = 4 \Rightarrow a + b = 1$$ 3. For $x + 2$, remainder is 19: $$f(-2) = (-2)^4 + 3(-2)^3 + a(-2)^2 + b(-2) - 1 = 16 - 24 + 4a - 2b - 1 = -9 + 4a - 2b = 19$$ 4. Rearrange: $$4a - 2b = 28$$ 5. From $a + b = 1$, $b = 1 - a$ 6. Substitute into second equation: $$4a - 2(1 - a) = 28 \Rightarrow 4a - 2 + 2a = 28 \Rightarrow 6a = 30 \Rightarrow a = 5$$ 7. Then $b = 1 - 5 = -4$ **Answer:** $a = 5$, $b = -4$ **(c) Factorize completely:** (i) $x^2(x - 1) - 16(x - 1)$ 1. Factor out $(x - 1)$: $$(x - 1)(x^2 - 16)$$ 2. Recognize difference of squares: $$x^2 - 16 = (x - 4)(x + 4)$$ **Answer:** $$(x - 1)(x - 4)(x + 4)$$ (ii) $xt - 3xt^2 - y + 3yt$ 1. Group terms: $$(xt - 3xt^2) - (y - 3yt)$$ 2. Factor each group: $$xt(1 - 3t) - y(1 - 3t)$$ 3. Factor out $(1 - 3t)$: $$(1 - 3t)(xt - y)$$ **Answer:** $$(1 - 3t)(xt - y)$$