1. **Problem:** Darren buys a car for 35000. The value decreases by 15% in the first year. **Step 1:** Calculate the value at the end of the first year using the formula for depreciation: $$V = P(1 - r)$$ where $P = 35000$ and $r = 0.15$. **Step 2:** Substitute values: $$V = 35000(1 - 0.15) = 35000 \times 0.85 = 29750$$ **Answer (a):** The value at the end of the first year is 29750. 2. **Problem:** After the first year, the value decreases by 11% each subsequent year. Find the value after 10 years. **Step 1:** The value after 1 year is 29750. For each subsequent year, multiply by $1 - 0.11 = 0.89$. **Step 2:** Value after 10 years (which is 9 more years after the first): $$V = 29750 \times 0.89^9$$ **Step 3:** Calculate: $$0.89^9 \approx 0.3894$$ $$V \approx 29750 \times 0.3894 = 11579.55$$ **Answer (b):** The value after 10 years is approximately 11580. 3. **Problem:** Find least $n$ such that value is less than 10% of original. **Step 1:** Original value is 35000, 10% is 3500. **Step 2:** Value after $n$ years: $$V = 35000 \times 0.85 \times 0.89^{n-1} < 3500$$ **Step 3:** Divide both sides by 35000: $$0.85 \times 0.89^{n-1} < 0.1$$ **Step 4:** Divide by 0.85: $$0.89^{n-1} < \frac{0.1}{0.85} \approx 0.1176$$ **Step 5:** Take natural logarithm: $$ (n-1) \ln(0.89) < \ln(0.1176)$$ **Step 6:** Calculate logs: $$\ln(0.89) \approx -0.1165, \quad \ln(0.1176) \approx -2.1401$$ **Step 7:** Solve for $n$: $$n-1 > \frac{-2.1401}{-0.1165} = 18.37$$ **Step 8:** So, $$n > 19.37$$ **Answer (c):** The least integer $n$ is 20. 4. **Problem:** Quadratic equation $(k-1)x^2 + 2x + (2k-3) = 0$ has real distinct roots. **Step 1:** For real distinct roots, discriminant $\Delta > 0$. **Step 2:** Discriminant formula: $$\Delta = b^2 - 4ac$$ where $a = k-1$, $b=2$, $c=2k-3$. **Step 3:** Calculate: $$\Delta = 2^2 - 4(k-1)(2k-3) = 4 - 4(k-1)(2k-3) > 0$$ **Step 4:** Expand: $$(k-1)(2k-3) = 2k^2 - 3k - 2k + 3 = 2k^2 - 5k + 3$$ **Step 5:** Substitute: $$4 - 4(2k^2 - 5k + 3) > 0$$ $$4 - 8k^2 + 20k - 12 > 0$$ $$-8k^2 + 20k - 8 > 0$$ **Step 6:** Divide by -4 (reverse inequality): $$2k^2 - 5k + 2 < 0$$ **Step 7:** Factor quadratic: $$2k^2 - 5k + 2 = (2k - 1)(k - 2)$$ **Step 8:** Inequality: $$(2k - 1)(k - 2) < 0$$ **Step 9:** Roots are $k=\frac{1}{2}$ and $k=2$. The product is negative between roots. **Answer:** $$\frac{1}{2} < k < 2$$ 5. **Problem:** Function $f(x) = x^2 - 8x$ passes through $A(3, -15)$ with tangent gradient -2. (a) Gradient of normal at $A$ is negative reciprocal: $$m_{normal} = -\frac{1}{-2} = \frac{1}{2}$$ (b) Equation of normal line at $A$: $$y - y_1 = m(x - x_1)$$ $$y + 15 = \frac{1}{2}(x - 3)$$ $$y = \frac{1}{2}x - \frac{3}{2} - 15 = \frac{1}{2}x - \frac{33}{2}$$ (c) Find point $B$ where normal intersects $f$ again. **Step 1:** Set $f(x) = y$ and normal line equal: $$x^2 - 8x = \frac{1}{2}x - \frac{33}{2}$$ **Step 2:** Multiply both sides by 2: $$2x^2 - 16x = x - 33$$ **Step 3:** Rearrange: $$2x^2 - 16x - x + 33 = 0$$ $$2x^2 - 17x + 33 = 0$$ **Step 4:** Solve quadratic: $$\Delta = (-17)^2 - 4 \times 2 \times 33 = 289 - 264 = 25$$ $$x = \frac{17 \pm 5}{4}$$ **Step 5:** Roots: $$x = \frac{22}{4} = 5.5, \quad x = \frac{12}{4} = 3$$ **Step 6:** $x=3$ is point $A$, so $B$ is at $x=5.5$. **Step 7:** Find $y$: $$y = (5.5)^2 - 8(5.5) = 30.25 - 44 = -13.75$$ **Answer (c):** $B(5.5, -13.75)$ 6. **Problem:** Curve $y = a(x-p)(x-q)$ with roots $A(-2,0)$ and $B(4,0)$. (a)(i) $p = -2$, $q = 4$. (ii) Given point $(6,8)$ on curve: $$8 = a(6 + 2)(6 - 4) = a(8)(2) = 16a$$ $$a = \frac{8}{16} = \frac{1}{2}$$ (iii) Expand: $$y = \frac{1}{2}(x + 2)(x - 4) = \frac{1}{2}(x^2 - 2x - 8) = \frac{1}{2}x^2 - x - 4$$ 7. **Problem:** Tangent gradient at $P$ is 7, $P$ lies on $y=20$. **Step 1:** Derivative of $y$: $$y' = \frac{d}{dx} \left( \frac{1}{2}x^2 - x - 4 \right) = x - 1$$ **Step 2:** Set gradient equal to 7: $$x - 1 = 7 \Rightarrow x = 8$$ **Step 3:** Check if $y=20$ at $x=8$: $$y = \frac{1}{2}(8)^2 - 8 - 4 = 32 - 8 - 4 = 20$$ **Answer:** $x=8$ 8. **Problem:** Line $L$ passes through $B(4,0)$ with gradient $-\frac{1}{3}$. (i) Equation of $L$: $$y - 0 = -\frac{1}{3}(x - 4)$$ $$y = -\frac{1}{3}x + \frac{4}{3}$$ (ii) Find $x$ where $L$ intersects curve again: Set curve equal to line: $$\frac{1}{2}x^2 - x - 4 = -\frac{1}{3}x + \frac{4}{3}$$ Multiply both sides by 6: $$3x^2 - 6x - 24 = -2x + 8$$ Rearrange: $$3x^2 - 6x - 24 + 2x - 8 = 0$$ $$3x^2 - 4x - 32 = 0$$ Solve quadratic: $$\Delta = (-4)^2 - 4 \times 3 \times (-32) = 16 + 384 = 400$$ $$x = \frac{4 \pm 20}{6}$$ Roots: $$x = 4, \quad x = -\frac{8}{3}$$ $x=4$ is point $B$, so other intersection is at $x = -\frac{8}{3}$. **Answer:** $x = -\frac{8}{3}$