1. (a) Expand \( (1+2 y)^{10} \) in ascending powers of \( y \), up to and including the term \( y^{3} \). Hence evaluate \( (1.002)^{10} \) correct to 6 decimal places. (b) Express each of the following in the form \( \frac{x}{y} \), where \( x, y \in Z, y \neq 0 \) (i) \( 1.6666 \ldots \) (ii) \( \quad 2 . \overline{4} \overline{5} \) (c) If \( \alpha \) and \( \beta \) are the roots of the equation \( 6 x^{2}+2 x-3=0 \), find the value of \( \frac{a}{\beta}+\frac{\beta}{\alpha} \). 2. (a) Given that \( f(x)=\frac{x-1}{2 x^{2}+2 x-3} \) find (i) \( f(-1) \) and \( f\left(\frac{1}{2}\right) \) (ii) what values of \( x \) must be excluded from the domain? (2) (b) Given that the determinant of the matrix \( \left(\begin{array}{cc}2 & -5 \\ 1 & 5\end{array}\right) \) is equal to the determinant of the matrix \( \left(\begin{array}{cc}5 & x \\ 5 & 8\end{array}\right) \) find the value of \( x \). (3) (c) \( (x-1) \) and \( (x+1) \) are the factors of the expression \( f(x)=x^{3}+a x^{2}+b x+c \)

### Problem 1 #### (a) Expand \( (1+2 y)^{10} \) in ascending powers of \( y \), up to and including the term \( y^{3} \). Hence evaluate \( (1.002)^{10} \) correct to 6 decimal places. **Solution:** To expand \( (1+2y)^{10} \) up to \( y^3 \), we use the binomial theorem: \[ (1+2y)^{10} = \sum_{k=0}^{10} \binom{10}{k} (2y)^k \] We only need the terms up to \( y^3 \): \[ (1+2y)^{10} = \binom{10}{0}(2y)^0 + \binom{10}{1}(2y)^1 + \binom{10}{2}(2y)^2 + \binom{10}{3}(2y)^3 \] \[ = 1 + 20y + 180y^2 + 960y^3 \] Now, to evaluate \( (1.002)^{10} \), we can use the binomial expansion for small \( y \): \[ (1 + 0.002)^{10} \approx 1 + 10 \cdot 0.002 + \frac{10 \cdot 9}{2} \cdot (0.002)^2 + \frac{10 \cdot 9 \cdot 8}{6 \cdot 2} \cdot (0.002)^3 \] \[ = 1 + 0.02 + 0.009 + 0.0000144 \] \[ = 1.0290144 \] Rounded to 6 decimal places: \[ 1.029014 \] #### (b) Express each of the following in the form \( \frac{x}{y} \), where \( x, y \in \mathbb{Z}, y \neq 0 \) (i) \( 1.6666 \ldots \) **Solution:** This is an infinite geometric series with the first term \( a = 1.6666 \) and the common ratio \( r = \frac{1}{6} \). The sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] \[ S = \frac{1.6666}{1 - \frac{1}{6}} = \frac{1.6666}{\frac{5}{6}} = \frac{1.6666 \cdot 6}{5} = \frac{10}{3} \] So, \( 1.6666 \ldots = \frac{10}{3} \). (ii) \( 2.\overline{45} \) **Solution:** This is a repeating decimal. We can express it as: \[ 2.\overline{45} = 2 + 0.0\overline{45} \] The repeating part \( 0.0\overline{45} \) is the same as \( \frac{45}{99} \) (since \( 0.0\overline{45} = \frac{45}{100} + \frac{45}{10000} + \ldots \)). \[ 0.0\overline{45} = \frac{45}{99} = \frac{5}{11} \] So, \( 2.\overline{45} = 2 + \frac{5}{11} = \frac{22}{11} + \frac{5}{11} = \frac{27}{11} \). #### (c) If \( \alpha \) and \( \beta \) are the roots of the equation \( 6x^{2}+2x-3=0 \), find the value of \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \). **Solution:** Using Vieta's formulas, we know that for a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \): \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a} \] For the equation \( 6x^2 + 2x - 3 = 0 \): \[ \alpha + \beta = -\frac{2}{6} = -\frac{1}{3}, \quad \alpha \beta = \frac{-3}{6} = -\frac{1}{2} \] We want to find: