1. **Problem 11 (a):** Given $\sin p = -\frac{3}{5}$ and $\tan q = -2$, with angles $p$ and $q$ in the same quadrant. Find $m$ if $\cos(p - q) = \frac{m\sqrt{5}}{5}$. 2. **Step 1:** Determine the quadrant of $p$ and $q$. Since $\sin p$ is negative and $\tan q$ is negative, and both angles are in the same quadrant, they must be in the third quadrant where sine and tangent are negative. 3. **Step 2:** Find $\cos p$ using $\sin^2 p + \cos^2 p = 1$. $$\sin p = -\frac{3}{5} \Rightarrow \sin^2 p = \frac{9}{25}$$ $$\cos^2 p = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow \cos p = -\frac{4}{5}$$ (negative in third quadrant). 4. **Step 3:** Find $\cos q$ and $\sin q$ from $\tan q = -2 = \frac{\sin q}{\cos q}$. Let $\cos q = x$, then $\sin q = -2x$. Using $\sin^2 q + \cos^2 q = 1$: $$(-2x)^2 + x^2 = 1 \Rightarrow 4x^2 + x^2 = 1 \Rightarrow 5x^2 = 1 \Rightarrow x^2 = \frac{1}{5}$$ Since $q$ is in the third quadrant, $\cos q < 0$, so $$\cos q = -\frac{1}{\sqrt{5}}, \sin q = -2 \times -\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$$ but sine must be negative in third quadrant, so $$\sin q = -\frac{2}{\sqrt{5}}$$ 5. **Step 4:** Use the cosine difference formula: $$\cos(p - q) = \cos p \cos q + \sin p \sin q$$ Substitute values: $$= \left(-\frac{4}{5}\right) \left(-\frac{1}{\sqrt{5}}\right) + \left(-\frac{3}{5}\right) \left(-\frac{2}{\sqrt{5}}\right) = \frac{4}{5\sqrt{5}} + \frac{6}{5\sqrt{5}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$$ 6. **Step 5:** Express $\cos(p - q)$ in the form $\frac{m\sqrt{5}}{5}$: $$\frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$ So, $m = 2$. --- 7. **Problem 11 (b):** Given $\tan(\theta - 60) = -1$, $\tan 60 = \sqrt{3}$, and $\tan \theta = k$. Find $k$ in surd form. 8. **Step 1:** Use the tangent subtraction formula: $$\tan(\theta - 60) = \frac{\tan \theta - \tan 60}{1 + \tan \theta \tan 60} = -1$$ Substitute $\tan 60 = \sqrt{3}$ and $\tan \theta = k$: $$\frac{k - \sqrt{3}}{1 + k \sqrt{3}} = -1$$ 9. **Step 2:** Cross multiply and solve for $k$: $$k - \sqrt{3} = -1 - k \sqrt{3}$$ $$k + k \sqrt{3} = -1 + \sqrt{3}$$ $$k(1 + \sqrt{3}) = \sqrt{3} - 1$$ 10. **Step 3:** Solve for $k$: $$k = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$$ Multiply numerator and denominator by the conjugate of denominator $1 - \sqrt{3}$: $$k = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{1 - 3} = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{-2}$$ 11. **Step 4:** Expand numerator: $$ (\sqrt{3} - 1)(1 - \sqrt{3}) = \sqrt{3} \times 1 - \sqrt{3} \times \sqrt{3} - 1 \times 1 + 1 \times \sqrt{3} = \sqrt{3} - 3 - 1 + \sqrt{3} = 2\sqrt{3} - 4$$ 12. **Step 5:** Substitute back: $$k = \frac{2\sqrt{3} - 4}{-2} = -\frac{2\sqrt{3} - 4}{2} = -\sqrt{3} + 2 = 2 - \sqrt{3}$$ --- 13. **Problem 12 (a)(i):** Given the mapping $g$ from $x$ to $y$ with pairs $(-1, 2), (1, 10), (3, 10), (5, k)$. Express $y$ in terms of $x$. 14. **Step 1:** Observe the mapping: - $x = -1 \to y = 2$ - $x = 1 \to y = 10$ - $x = 3 \to y = 10$ - $x = 5 \to y = k$ 15. **Step 2:** Notice $y=10$ for $x=1$ and $x=3$, so $y$ is not a function of $x$ in a simple linear way unless $y$ is piecewise or constant for some values. Since $g$ is a function, and $y$ repeats for different $x$, it suggests $y$ depends on $x$ in a way that can be expressed as: Try $y = ax^2 + bx + c$. Use points to find $a,b,c$. 16. **Step 3:** Use points $(-1,2)$, $(1,10)$, $(3,10)$: $$a(-1)^2 + b(-1) + c = 2 \Rightarrow a - b + c = 2$$ $$a(1)^2 + b(1) + c = 10 \Rightarrow a + b + c = 10$$ $$a(3)^2 + b(3) + c = 10 \Rightarrow 9a + 3b + c = 10$$ 17. **Step 4:** Solve the system: From first two equations: $$a - b + c = 2$$ $$a + b + c = 10$$ Add: $$2a + 2c = 12 \Rightarrow a + c = 6$$ Subtract: $$2b = 8 \Rightarrow b = 4$$ 18. **Step 5:** Use $a + c = 6$ and third equation: $$9a + 3b + c = 10$$ Substitute $b=4$: $$9a + 12 + c = 10 \Rightarrow 9a + c = -2$$ From $a + c = 6$, $c = 6 - a$, substitute: $$9a + 6 - a = -2 \Rightarrow 8a = -8 \Rightarrow a = -1$$ Then $c = 6 - (-1) = 7$. 19. **Step 6:** So, $$y = -x^2 + 4x + 7$$ 20. **Problem 12 (a)(ii):** Find $k = y$ when $x=5$: $$k = -(5)^2 + 4(5) + 7 = -25 + 20 + 7 = 2$$ 21. **Problem 12 (b)(i):** Given $f'(x) = \frac{x+7}{3}$, find $f(x)$. 22. **Step 1:** Integrate $f'(x)$: $$f(x) = \int \frac{x+7}{3} dx = \frac{1}{3} \int (x+7) dx = \frac{1}{3} \left( \frac{x^2}{2} + 7x \right) + C = \frac{x^2}{6} + \frac{7x}{3} + C$$ 23. **Problem 12 (b)(ii):** Find $h$ if $f\left(\frac{4h}{3}\right) = 20$. 24. **Step 1:** Substitute $x = \frac{4h}{3}$ into $f(x)$: $$f\left(\frac{4h}{3}\right) = \frac{(\frac{4h}{3})^2}{6} + \frac{7(\frac{4h}{3})}{3} + C = 20$$ 25. **Step 2:** Simplify: $$\frac{\frac{16h^2}{9}}{6} + \frac{28h}{9} + C = 20$$ $$\frac{16h^2}{54} + \frac{28h}{9} + C = 20$$ $$\frac{8h^2}{27} + \frac{28h}{9} + C = 20$$ 26. **Step 3:** Without initial condition, assume $C=0$ for simplicity: $$\frac{8h^2}{27} + \frac{28h}{9} = 20$$ Multiply both sides by 27: $$8h^2 + 84h = 540$$ 27. **Step 4:** Rearrange: $$8h^2 + 84h - 540 = 0$$ Divide by 2: $$4h^2 + 42h - 270 = 0$$ 28. **Step 5:** Use quadratic formula: $$h = \frac{-42 \pm \sqrt{42^2 - 4 \times 4 \times (-270)}}{2 \times 4} = \frac{-42 \pm \sqrt{1764 + 4320}}{8} = \frac{-42 \pm \sqrt{6084}}{8}$$ $$\sqrt{6084} = 78$$ 29. **Step 6:** So, $$h = \frac{-42 \pm 78}{8}$$ Two solutions: $$h = \frac{36}{8} = 4.5$$ or $$h = \frac{-120}{8} = -15$$ **Final answers:** (a) $m = 2$ (b) $k = 2 - \sqrt{3}$ (12a)(i) $y = -x^2 + 4x + 7$ (12a)(ii) $k = 2$ (12b)(i) $f(x) = \frac{x^2}{6} + \frac{7x}{3} + C$ (12b)(ii) $h = 4.5$ or $h = -15$