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T1, T2, T3 2, -9, -20 .... -141 l = a + (n - 1)d first term, a = 2 common difference d = T3 - T2 = T2 - T1 = -20 - (-9) = -9 -2 = -20 + 9 = -9 -2 = -20 + 9 = -11 -11 = -11 d = -1 last term l = -141 -141 = 2 + (\(\cap\) - 1) (-11) -141 = 2 + (-11 \(\cap\) + 11) = 2 - 11\(\cap\) + 11 -141 = 13 - 11\(\cap\) -141 - 13 = -11\(\cap\) -154 = -11\(\cap\) \(\cap\) = \(\frac{-154}{-11}\) \(\cap\) = 14

(a) \(\frac{1}{x} + \frac{1}{y} = 5 ..... (1)\) \(\frac{1}{y} - \frac{1}{x} = 1 ........ (2)\) Substitute p and q for \(\frac{1}{x}\) and \(\frac{1}{y}\) respectively. \(p + q = 5 ......... (1a)\) \(q - p = 1 ....... (2a)\) From (2a), q = 1 + p. Putting into (1a), we have: \(p + (1 + p) = 5\) \(2p + 1 = 5 \implies 2p = 5 - 1 = 4\) \(p = 2\) \(q = 1 + p = 1 + 2 = 3\) \(p = \frac{1}{x} = 2\) \(\implies x = \frac{1}{2}\) \(q = \frac{1}{y} = 3\) \(\implies y = \frac{1}{3}\) Hence, \(x = \frac{1}{2} ; y = \frac{1}{3}\). (b) Total distance travelled = 48 km Total time taken = 45 minutes = \(\frac{45}{60} = 0.75 hr\) Let the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface. Let the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs. Using \(speed = \frac{distance}{time}\) in each case On good surface : \(72 = \frac{x}{t}\) \(\implies x = 72t ... (1)\) On bad surface : \(48 = \frac{48 - x}{0.75 - t}\) \(48(0.75 - t) = 48 - x\) \(\implies 36 - 48t = 48 - x ...... (2)\) Substitute 72t = x in (2) : \(36 - 48t = 48 - 72t \implies 72t - 48t = 48 - 36\) \(24t = 12 \implies t = \frac{12}{24} = 0.5 hours\) From (1), x = 72t \(x = 72 \times 0.5 = 36 km\) Hence, the man travels a distance of 36 km on good road surface.

|x - 2| < 3 implies -(x - 2) < 3 .... or .... +(x - 2) < 3 -x + 2 < 3 .... or .... x - 2 < 3 -x < 3 - 2 .... or .... x < 3 + 2 x > -1 .... or .... x < 5 combining the two inequalities results, we get; -1 < x < 5

Using Tn = \(\frac{3n + 1}{n + 1}\), T1 = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\) T2 = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\) T3 = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)

2P + Q = 2\(\begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\) = \(\begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix}\)

Let A = \(\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}\) Then |A| = \(\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}\) = 20 - 18 = 2 Hence A-1 = \(\frac{1}{|A|}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}\) = \(\frac{1}{2}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}\) = \(\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \\ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}\) = \(\begin{pmatrix} 2 & -\frac{3}{2} \\ -3 & \frac{5}{2} \end{pmatrix}\)

\(\frac{2x - 5}{2} < \frac{(2 - x)}{1}\) 2x - 5 < 4 - 2x 2x + 2x < 4 + 5 4x < 9 x < \(\frac{9}{4}\) x < 2\(\frac{1}{4}\)

\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2) \(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2) x + 4y = 6......(iii) 3x - 4y = 2.....(iv) add (iii) and (iv) 4x = 8, x = \(\frac{8}{4}\) = 2 substitute x = 2 into equation (iii) x + 4y = 6 2 + 4y = 6 4y = 6 - 2 4y = 4 y = \(\frac{4}{4}\) = 1(x + y) 2 + 1 = 3

\(p^{\frac{1}{3}}\) = \(\frac{\sqrt[3]{q}}{r}\) cross multiply \(p^{\frac{1}{3}}\) = \(\frac{q^{1/3}}{r}\) r\(p^{\frac{1}{3}}\) = \(q^{\frac{1}{3}}\) take the cube of both sides \((rp^{\frac{1}{3}})^3\) = \((q^{\frac{1}{3}})^3\) \(r^3p^{\frac{3}{3}}\) = \(q^{\frac{3}{3}}\) \(r^3\)p = q ∴ q = p\(r^3\)

6k2 = 5k + 6 6k2 - 5k - 6 = 0 6k2 - 0k + 4k - 6 = 0 3k(2k - 3) + 2(2k - 3) = 0 (3k + 2)(2k - 3) = 0 3k + 2 = 0 or 2k - 3 = 0 3k = -2 or 2k = 3 k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\) k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))

(a) x -3 -2 -1 0 1 2 3 4 y 35 15 1 -7 -9 -5 5 21 (b) see graph above (c) (i) The roots of the equation is where the graph cuts the x- axis = -0.9, 2.6. (ii) Minimum value of y = -9.21 (iii) At x = 2, gradient = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) = \(\frac{4 - (-8.5)}{3.3 - 1.5}\) = \(\frac{12.5}{1.8}\) = 6.9444 \(\approxeq\) 6.9

Given P(2, -3) and Q(-5, 1) Midpoint = \((\frac{2 + (-5)}{2}, \frac{-3 + 1}{2})\) = \((\frac{-3}{2}, -1)\) Slope of the line PQ = \(\frac{1 - (-3)}{-5 - 2}\) = \(-\frac{4}{7}\) The slope of the perpendicular line to PQ = \(\frac{-1}{-\frac{4}{7}}\) = \(\frac{7}{4}\) The equation of the perpendicular line: \(y = \frac{7}{4}x + b\) Using a point on the line (in this case, the midpoint) to find the value of b (the intercept). \(-1 = (\frac{7}{4})(\frac{-3}{2}) + b\) \(-1 + \frac{21}{8} = \frac{13}{8} = b\) \(\therefore\) The equation of the perpendicular bisector of the line PQ is \(y = \frac{7}{4}x + \frac{13}{8}\) \(\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0\)

(a)(i) R is the required region. (iii) Coordinates of the point of intersection of the two graphs = (1.1, 3.8) (b) Assuming \(y - \frac{3x}{4} = 3\) If x = 0, \(y - \frac{3}{4}(0) = 3 \implies y = 3\) If y = 0, \(0 - \frac{3}{4}x = 3 \implies -3x = 12 ; x = - 4\) x 0 -4 y 3 0 Assuming \(y + 2x = 6\) If x = 0, \(y + 2(0) = 6 \implies y = 6\) If y = 0, \(0 + 2x = 6 \implies 2x = 6 ; x = 3\) x 0 3 y 6 0 (b) \((x + 2)(x + c) = x^{2} + 2x + cx + 2c\) \(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\) \( 2c = 18 ; c = 9\) \(b = 2 + c = 2 + 9 = 11\) (b, c) = (11, 9).

x2 - y2 = 4 .... (1) x + y = 2 .... (2) Simplify eqn (1) (x + y)(x - y) = 4 From eqn (2) x + y = 2 so substitute it into simplified eqn (1), we have 2 (x - y) = 4 therefore, x - y = 2 ... (1) x + y = 2 --------- 2x = 4 --------- x = 2, when y = 0

n2 - 6n - 4 For the 3rd term, 32 - 6(3) - 4 9 - 18 -4 = -13 For the 4th term, 42 - 6(4) - 4 16 - 24 - 4 = -12 Sum of both terms -13 - 12 = -25