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The minimum value is the lowest value of the curve on y axis which gives a value of -5.3

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\) If x = 3 and y = 100, then, \(\frac{100}{1} = \frac{k}{3^2}\) \(\frac{100}{1} = \frac{k}{9}\) k = 100 x 9 = 900 Substitute 900 for k in y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\) = \(yx^2 = 900\)

Since the curve cuts the x-axis at x = -2 and x = 3, (x + 2)(x - 3) = 0 \(x^2 - 3x + 2x - 6\) = 0 \(x^2 - x - 6\) = 0 Hence, the equation of the curve is y = \(x^2 - x - 6\)

Using the two - point from \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\) \(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\) \(\frac{y - 2}{-4} = \frac{x - 4}{-12}\) \(\frac{-12(y -2)}{-4}\) = x - 4 3(y -2) = x -4 3y - 6 = x - 4 3y = x - 4 + 6 3y = x + 2... By comparing the equations; 3y = px + , p = 1

\(2x^2 - x - 6\) = 0 \(2x^2 - 4x + 3x - 6\) = 0 2x(x - 2) + 3(x - 2) = 0 (2x + 3) (x - 2) = 0 Either; 2x + 3 = 0 or x - 2 = 0 x = \(\frac{-3}{2}\) or x = 2

2 x 3, 4 x 6, 8 x 9, 16 x 12,... 2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,.... 2\(^n\) x 3n

(a) Using the two- point form, \(\frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}}\) \(\frac{y - 5}{-7 - 5} = \frac{x - 2}{-4 - 2}\) \(\frac{y - 5}{- 12} = \frac{x - 2}{- 6}\) \(y - 5 = 2(x - 2)\) \(y - 5 = 2x - 4 \implies y - 2x = -4 + 5 = 1\) \(Equation : y = 2x + 1\) (b) (i)see figure above In \(\Delta{\text PQR,}\) (ii) (1) In \(\Delta PQR, |QR|^{2} = 5^{2} + 8^{2}\) \(25 + 64 = 89\) \(|QR| = \sqrt{89} = 9.434 km \approxeq 9.43 km\) (2) In the diagram above, the bearing of R from Q is the obtuse angle NQR. But \(\tan \alpha = \frac{8}{5} = 1.6\) \(\alpha = \tan^{-1} (1.6) = 58^\circ\) Hence, angle NQR = 360^\circ - (a + 60^\circ + 90^\circ) = 360^\circ - (58^\circ + 60^\circ + 90^\circ) = 360^\circ - 208^\circ = 152^\circ

(a) S.P = \(C.P + \frac{% profit}{100} \times C.P\) = \(300 + (\frac{x}{100} \times 300)\) S.P = N(300 + 3x) Therefore, Bola bought it at N(300 + 3x). James paid \(N(6x + \frac{3}{4})\) extra from what Lamin paid, therefore Bola's S.P = \(N(300 + 6x + \frac{3}{4})\) = N(300.75 + 6x). Profit for Bola = \(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\) \(\frac{x}{100} \times N(300 + 3x) = N(0.75 + 3x)\) \(300x + 3x^{2} = 75 + 300x\) \(\implies 3x^{2} = 75\) \(x^{2} = 25 \therefore x = 5\) (b) \(3x - 2 < 10 + x < 2 + 5x\) \(3x - 2 < 10 + x \implies 3x - x < 10 + 2\) \(2x < 12 \implies x < 6\) \(10 + x < 2 + 5x\) \(x - 5x < 2 - 10\) \(-4x < -8 \implies x > 2\) The range = \(2 < x < 6\)

(a) x -5 -4 -3 -2 -1 0 1 2 3 4 y 35 18 5 -4 -9 -10 -7 0 11 26 (b) (c)(i) 2x\(^{2}\) + x = 10 \(\equiv\) 2x\(^{2}\) + x - 10 = 0 Thus, the solution of 2x\(^{2}\) + x - 10 = 0 are the values of x at which the curve cuts the x- axis. x = -2.5 or x = 2. (ii) 2x\(^{2}\) + x - 10 = 2x To solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6. The values of x at which y = 2x\(^{2}\) + x - 10 and y = 2x intersect give the solution of 2x\(^{2}\) + x - 10 = 2x. These are x = -2 or x = 2.5.

(a) px + qy = z \(p \begin{pmatrix} 2 \\ 3 \end{pmatrix} + q \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\) \(\begin{pmatrix} 2p + 5q \\ 3p - 2q \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\) Thus, \(2p + 5q = -4 .....(1)\) \(3p - 2q = 13 ......(2)\) Solving the equations, we have p = 3 and q = -2. (b)(i) (ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2). (iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90^\circ is \(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\) and \(Z_{1}(-2, -3)\).

(a) Marks(x) Frequency (f) fx 10 1 10 20 1 20 30 x 30x 40 5 200 50 y 50y 60 1 60 70 4 280 80 3 240 90 1 90 Total 16 + x + y 900 + 30x + 50y \(\sum f = 16 + x + y = 30\) \(\implies x + y = 14 ... (1)\) \(\bar{x} = \frac{\sum fx}{\sum f}\) \(52 = \frac{900 + 30x + 50y}{30}\) \(1560 = 900 + 30x + 50y \implies 30x + 50y = 660\) \(3x + 5y = 66 .... (2)\) Solving equation (1) and (2), From (1), x = 14 - y \(3(14 - y) + 5y = 42 - 3y + 5y = 66\) \(2y = 24 \implies y = 12\) \(x = 14 - y = 14 - 12 = 2\) (x, y) = (2, 12). (b) Class interval Frequency Upper class boundary 1 - 10 1 10.5 11 - 20 1 20.5 21 - 30 2 30.5 31 - 40 5 40.5 41 - 50 12 50.5 51 - 60 1 60.5 61 - 70 4 70.5 71 - 80 3 80.5 81 - 90 1 90.5 (c) (d) From the histogram, Mode = 44.

\(\begin{vmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{vmatrix}\) = \(3(-3 - (-10)) - 5(6 - 10) + (-4)(12 - 6)\) = \(21 + 20 - 24\) = 17

(a) If Musa's age = x and Manya's age = y, then x = y + 3.... (1) and (x — 7) = 2(y— 7) .......(2) Solving both equations will yield Musa's age = 13 years and Manya's age = 10 years. (b) Let n be the year when the sum of their ages will be 45, so that (10 + n) + (13 + n) = 45 and solving will give n = 11 years.

(a) By the use of Pythagoras theorem, the hypotenuse side is 13 and also sin x = \(\frac{5}{13}\) and cos x = \(\frac{12}{13}\) Substituting and simplifying yields \(\frac{65}{161}\) (b) we have (11 + 2x)(14 + 2x) - (11 x 14) =186. Expanding and simplifying to get the quadratic equation 2x\(^2\) + 25x - 93 = 0. Solving the quadratic equation results in x = 3 m as the width of the path.

(a) it was expected that they substitute into the formula for sum of an Arithmetic Progression to have 130 = 1\(\frac{1}{2}\) [2a+(10 - 1)d] and simplifying will yield 2a + 9d = 26 -------(1). Also, U5 = a + (5 – 1)d = 3a and simplifying will yield a = 2d --------------------------------(2) Substituting for a in equation (1) will yield 2(2d) + 9d = 26 and solving for d, d = 2. (b), Substituting for d in equation (2) to get the first term:a = 2(2) = 4 (c) The number of terms required can be obtained by solving 28 = 4 + (n – 1)2 which resulted to n = 13.