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\(\begin{vmatrix} 2 & 3 \\ 5 & 3x \end{vmatrix}\) = \(\begin{vmatrix} 4 & 1 \\ 3 & 2x \end{vmatrix}\) (2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1) 6x - 15 = 8x - 3 6x - 8x = 15 - 3 -2x = 12 x = \(\frac{12}{-2}\) = -6

\(\begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\) 4 \(\begin{vmatrix} 3 & -1 \\ 1 & 3 \end{vmatrix}\) -2 \(\begin{vmatrix} 2 & -1 \\ -1 & 3\end{vmatrix}\) -1 \(\begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix}\) 4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)] = 4[9 + 1] -2 [6 - 1] -1 [2 + 3] = 4(10) - 2(5) - 1(5) = 40 - 10 - 5 = 25

N = [2 3] N-1 = \(\frac{adj N}{|N|}\) adj N = \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\) |N| = (2 x4) - (1 x 3) = 8 - 3 =5 N-1 = \(\frac {1}{5}\) \(\begin{vmatrix} 4 & -3 \\ -1 & 2 \end{vmatrix}\)

2y = 5x + 4 (4, 2) y = \(\frac{5x}{2}\) + 4 comparing with y = mx + e m = \(\frac{5}{2}\) Since they are perpendicular m1m2 = -1 m2 = \(\frac{-1}{m_1}\) = -1 \(\frac{5}{2}\) = -1 x \(\frac{2}{5}\) The equator of the line is thus y = mn + c (4, 2) 2 = -\(\frac{2}{5}\)(4) + c \(\frac{2}{1}\) + \(\frac{8}{5}\) = c c = \(\frac{18}{5}\) y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\) 5y = -2x + 18 or 5y + 2x - 18 = 0

\(\frac{3}{5}\)(2x - 1) = \(\frac{1}{4}\)(5x - 3) \(\frac{6x}{5} - \frac{3}{5} = \frac{5x}{4} - \frac{3}{4}\) \(\frac{6x}{5} - \frac{5x}{4} = \frac{3}{5} - \frac{3}{4}\) \(\frac{24x - 25x}{20} = \frac{12 - 15}{20}\) \(\frac{-x}{20} = \frac{-3}{20}\) -20x = -60 x = \(\frac{-60}{-20}\) x = 3

3x - 2y = 7; Solve by elimination method x + 2y = -3 3x - 2y = 7.....(i) -3x + 6y = -9.....(ii) -8y = 16 y = \(\frac{16}{-8} = -2\) Put y = -2 into equation (i); 3x - 2y = 7 3x - 2(-2) = 7 3x + 4 = 7 3x = 7 - 4 3x = 3 x = \(\frac{3}{3}\) = 1

x = \(\frac{1}{2}\) and x = \(\frac{-2}{3}\) expand (x - \(\frac{1}{2}\))(x + \(\frac{2}{3}\)) = 0 x(x + \(\frac{2}{3}\)) - \(\frac{1}{2}(x + \frac{2}{3}\)) = 0 x2 + \(\frac{4x - 3x}{6} - \frac{2}{6} = 0\) \(x^2 + \frac{x}{6} - 2 = 0\) 6x2 + x - 2 = 0

x2 - 3x = 7 x2 - 3x - 7 = 0 What can you add to both sides of the equation to give the same value of y = x2 - 3x - 3 The number is 4 x2 - 3x - 7 + 4 = 4 x2 - 3x - 3 = 4 but y = x2 - 3x - 3 y = 4; So are y = 4 draw a line parallel to x axis, to cut or intersect the graph. At these points look down to see the corresponding values on x axis This give -1.55 and 4.55

x\(^2\) + 7x + 10 = 0 x2 + 5x + 2x + 10 = 0 x(x + 5) + 2(x + 5) = 0 (x + 2)(x + 5) = 0 x + 2 = 0 or x + 5 = 0 x = -2 or x = -5 the smaller value x = -5

let the number of question = n n(n + 50) = 600 n2 + 50n - 600 - 0 by quadratic formular x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2}\) n = \(\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)}\) = \(\frac{-50 \pm \sqrt{4900}}{2}\) = \(\frac{-50 + 70}{2}\) = \(\frac{20}{2}\) = 10

No expanded working has been attached for this question yet. The verified answer is Option C.

(a) \(\frac{x^{2} - 4}{x^{2} + x} \div \frac{x^{2} - 4x + 4}{x + 1}\) = \(\frac{x^{2} - 4}{x^{2} + x} \times \frac{x + 1}{x^{2} - 4x + 4}\) = \(\frac{(x - 2)(x + 2)}{x(x + 1)} \times \frac{x + 1}{x^{2} - 2x - 2x + 4}\) (Using difference of two squares) = \(\frac{(x - 2)(x + 2)}{x} \times \frac{1}{(x - 2)^{2}}\) = \(\frac{(x + 2)}{x(x - 2)}. (b) Roots of \(ax^{2} +bx + c = mx + k\) For graph \(y = ax^{2} + bx + c\), \(x = -1\) and \(x = 2\) \(x + 1 = 0\) and \(x - 2 = 0\) \((x + 1)(x - 2) = 0\) \(x^{2} - 2x + x - 2 = 0\) \(x^{2} - x - 2 = 0 ..... (i)\) \(-x^{2} + x + 2 = 0 ...... (ii)\) For graph \(y = mx + k\) \(m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 - (-1)}{1.3 - (-1.3)}\) \(m = \frac{2}{2.6} = 0.769\) \(k = 0.25\) \(x^{2} - x - 2 = 0.769x + 0.25\) \(x^{2} - x - 0.769x - 2 - 0.25 = 0\) \(x^{2} - 1.769x - 2.25 = 0\) Using formula method, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-1.769) \pm \sqrt{(-1.769)^{2} - 4(1)(-2.25)}}{2(1)}\) \(x = \frac{1.769 \pm \sqrt{3.1294 + 9}}{2}\) \(x = \frac{1.769 \pm 3.483}{2}\) \(x = \frac{1.769 + 3.483}{2}\) or \(x = \frac{1.769 - 3.483}{2}\) \(x = \frac{5.252}{2} \approxeq 2.626\) or \(x = \frac{-1.714}{2} \approxeq -0.857\) (ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively. Using coordinate for L: \(0 = a(-1^{2}) + b(-1) + c \implies 0 = a - b + c ..... (1)\) Using M : \(2 = a(0^{2}) + b(0) + c \implies c = 2 ...... (2)\) Using N : \(0 = a(2^{2}) + b(2) + c \implies 0 = 4a + 2b + c .... (3)\) From (2), c = 2 \(\therefore (1) : a - b + 2 = 0 \implies a - b = -2 .... (1a)\) \(\therefore (3) : 4a + 2b + 2 = 0 \implies 4a + 2b = -2 ..... (3a)\) Solving equations (1a) and (3a) simultaneously, we get \(a = -1 , b = 1 , c = 2\) \(\therefore\) Equation of the curve : \(-x^{2} + x + 2 = 0\) (iii) Line of symmetry = \(x = \frac{-b}{2a}\) \(y = -x^{2} + x + 2 \) Line of symmetry = \(\frac{-1}{2(-1)}\) \(x = \frac{1}{2}\)

\(\begin{vmatrix} x & 3 \\ 2 & 7 \end{vmatrix} = 15\) x x 7 - 2 x 3 = 15 7x - 6 = 15 7x = 15+6 7x = 21 x = 21/7 x = 3

Locus of a point equidistant from two fixed points is the perpendicular bisector of the straight line joining the points Gradients of PQ = (5-3 / 3-1) = 2/2 = 1 Gradient bisector = -1 Equation of perpendicular bisector y - y1 = m(χ - χ1 y - 4 = -χ + 2 y = -χ + 6

\(\begin{vmatrix} 2 & 0 & 5 \\ 4 & 6 & 3 \\ 8 & 9 & 1 \end{vmatrix}\) = 2(6 - 27) - 0(4 - 24) + 5(36 - 48) = 2(-21) - 0 + 5(-12) = -42 + 5(-12) = -42 - 60 = -102