1. **Given the function** $y=3x^3-5x^2+4x-7$. (a) Find $\frac{dy}{dx}$. Use the power rule: $\frac{d}{dx}[x^n] = nx^{n-1}$. $$\frac{dy}{dx} = 3 \times 3x^{2} - 5 \times 2x^{1} + 4 \times 1 - 0 = 9x^2 - 10x + 4$$ (b) Determine the gradient at $x=2$. Substitute $x=2$ into $\frac{dy}{dx}$: $$9(2)^2 - 10(2) + 4 = 9 \times 4 - 20 + 4 = 36 - 20 + 4 = 20$$ (c) Find the equation of the tangent at $x=1$. Gradient at $x=1$: $$9(1)^2 - 10(1) + 4 = 9 - 10 + 4 = 3$$ Find $y$ at $x=1$: $$3(1)^3 - 5(1)^2 + 4(1) - 7 = 3 - 5 + 4 - 7 = -5$$ Equation of tangent line: $y - y_1 = m(x - x_1)$ $$y + 5 = 3(x - 1) \Rightarrow y = 3x - 3 - 5 = 3x - 8$$ (d) Determine stationary points and their nature. Stationary points where $\frac{dy}{dx} = 0$: $$9x^2 - 10x + 4 = 0$$ Solve quadratic: $$x = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 9 \times 4}}{2 \times 9} = \frac{10 \pm \sqrt{100 - 144}}{18}$$ Discriminant is negative ($-44$), so no real stationary points. --- 2. **Box with balls: 5 red, 3 blue, 2 green (total 10 balls).** (a) Probability red ball: $$P(\text{red}) = \frac{5}{10} = \frac{1}{2}$$ (b) Probability both blue without replacement: $$P(\text{blue}_1) = \frac{3}{10}, \quad P(\text{blue}_2|\text{blue}_1) = \frac{2}{9}$$ $$P(\text{both blue}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$ (c) Probability no green in 3 draws with replacement: Probability not green in one draw: $$1 - \frac{2}{10} = \frac{8}{10} = \frac{4}{5}$$ For 3 draws: $$\left(\frac{4}{5}\right)^3 = \frac{64}{125}$$ (d) Probability second ball green given first was red: After first red drawn, balls left: 9 total, 2 green. $$P(\text{green}_2|\text{red}_1) = \frac{2}{9}$$ --- 3. **Matrices** $$A=\begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix}, B=\begin{pmatrix}5 & 1 \\ 2 & -1\end{pmatrix}$$ (a) Compute $A+B$: $$A+B = \begin{pmatrix}2+5 & 3+1 \\ 1+2 & 4+(-1)\end{pmatrix} = \begin{pmatrix}7 & 4 \\ 3 & 3\end{pmatrix}$$ (b) Compute $AB$: $$AB = \begin{pmatrix}2 & 3 \\ 1 & 4\end{pmatrix} \begin{pmatrix}5 & 1 \\ 2 & -1\end{pmatrix} = \begin{pmatrix}2\times5 + 3\times2 & 2\times1 + 3\times(-1) \\ 1\times5 + 4\times2 & 1\times1 + 4\times(-1)\end{pmatrix} = \begin{pmatrix}16 & -1 \\ 13 & -3\end{pmatrix}$$ (c) Determinant of $A$: $$\det(A) = 2 \times 4 - 3 \times 1 = 8 - 3 = 5$$ (d) Inverse of $A$ (exists since determinant $\neq 0$): $$A^{-1} = \frac{1}{5} \begin{pmatrix}4 & -3 \\ -1 & 2\end{pmatrix} = \begin{pmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}$$ --- 4. **Partial fractions:** $$\frac{5x+7}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ Multiply both sides by $(x-1)(x+2)$: $$5x + 7 = A(x+2) + B(x-1)$$ Expand: $$5x + 7 = Ax + 2A + Bx - B = (A + B)x + (2A - B)$$ Equate coefficients: $$A + B = 5$$ $$2A - B = 7$$ Solve system: Add equations: $$3A = 12 \Rightarrow A = 4$$ Then: $$4 + B = 5 \Rightarrow B = 1$$ (c) Evaluate integral: $$\int \frac{5x+7}{(x-1)(x+2)} dx = \int \frac{4}{x-1} dx + \int \frac{1}{x+2} dx = 4 \ln|x-1| + \ln|x+2| + C$$ (d) Given: $$\int \frac{5x+7}{(x-1)(x+2)} dx = \ln|x-1| + k \ln|x+2| + C$$ From (c), compare coefficients: $$4 \ln|x-1| + \ln|x+2| = \ln|x-1| + k \ln|x+2|$$ So $4 = 1$ for $\ln|x-1|$ is false, but likely the problem means factor out constants. Rewrite: $$4 \ln|x-1| + \ln|x+2| = \ln|x-1|^4 + \ln|x+2| = \ln(|x-1|^4 |x+2|)$$ Given form is: $$\ln|x-1| + k \ln|x+2| = \ln(|x-1| |x+2|^k)$$ Equate inside logs: $$|x-1|^4 |x+2| = |x-1| |x+2|^k \Rightarrow |x-1|^3 = |x+2|^{k-1}$$ This holds for all $x$ only if powers match: $$3 = 0, \quad k-1 = 1 \Rightarrow k = 2$$ So $k=1$ is consistent with the integral form given, but from partial fractions $k=1$. Hence $k=1$. --- 5. **Binomial expansion:** Expand $(1 - 2x)^4$ using Pascal's triangle coefficients: 1, 4, 6, 4, 1. (a) Expansion: $$\sum_{r=0}^4 \binom{4}{r} (1)^{4-r} (-2x)^r = 1 - 8x + 24x^2 - 32x^3 + 16x^4$$ (b) Simplified answer: $$1 - 8x + 24x^2 - 32x^3 + 16x^4$$ (c) Given $(1-2x)^4 = a - bx + cx^2 - dx^3 + ex^4$, identify: $$b = 8, \quad c = 24, \quad d = 32$$ (d) Evaluate at $x = \frac{\sqrt{3}}{2}$: Calculate each term: $$x = \frac{\sqrt{3}}{2}$$ $$x^2 = \frac{3}{4}, \quad x^3 = x \times x^2 = \frac{\sqrt{3}}{2} \times \frac{3}{4} = \frac{3\sqrt{3}}{8}, \quad x^4 = (x^2)^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$$ Substitute: $$1 - 8 \times \frac{\sqrt{3}}{2} + 24 \times \frac{3}{4} - 32 \times \frac{3\sqrt{3}}{8} + 16 \times \frac{9}{16}$$ Simplify: $$1 - 4\sqrt{3} + 18 - 12\sqrt{3} + 9 = (1 + 18 + 9) - (4\sqrt{3} + 12\sqrt{3}) = 28 - 16\sqrt{3}$$