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a = 1, r =\(\frac{9}{10}\) S\(_n\) = \(\frac{a}{1 - r}\) S\(_n\) = \(\frac{1}{1-\frac{9}{10}}\) = 1\(\div \frac{1}{10}\) =1\(\times \frac{10}{1}\) = 10

x = \(\frac{-1}{2}\) x = 3 2x = -1, x = 3; 2x + 1 = 0 x - 3 = 0 (2x + 1)(x - 3) = 0 2x2 - 6x + x - 3 = 0 2x2 - 5x - 3 = 0

3x - 4 1 = 1< x; since x is an integer, and 1 < x3.33 x = {2, 3}

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

when y = 0, value of x = roots of the equation x2 - 5x + 4 = 0 x = 1, 4 (see graph)

(a) \(9 \cos x - 7 = 1 \implies 9 \cos x = 1 + 7 = 8\) \(\cos x = \frac{8}{9}\) \(x = \cos^{-1} (0.8889)\) \(x = 27.26^\circ\) (b) \(7x < 2x - 13\) \(7x - 2x < - 13\) \(5x < - 13\) \(x < \frac{-13}{5}\) \([-2.6] = -3, -4, -5\) Note : The greatest value function rounds up a real number to an integer that is less or equal to the real number.

(a) \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\) = \(\frac{(x - 1)(x + 2) - (x - 2)(x + 3)}{(x - 2)(x - 1)}\) = \(\frac{x^{2} + 2x - x - 2 - (x^{2} + 3x - 2x - 6)}{(x - 2)(x - 1)}\) = \(\frac{x^{2} - x^{2} + x - x - 2 + 6}{(x - 2)(x - 1)}\) = \(\frac{4}{(x - 2)(x - 1)}\) (b) (1, 4) and (2, 10) are supposed roots of the equation When x = 1, equation becomes \(A(1)^{2} + B(1) = 4\) When x = 2, \(A(2^{2}) + B(2) = 10\) \(\implies A + B = 4 ...... (1)\) \(4A + 2B = 10 ....... (2)\) Solving for A and B, we have A = 1 and B = 3. (iii) \(y = x^{2} + 3x\) \(y = x(x + 3)\) x = 0 or x = 3.

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ y 2.0 2.2 1.9 1.0 -0.1 -1.2 -2.0 -2.2 -1.9 (b) (c) (i) x = 117^\circ ; (ii) x = 9^\circ , 45^\circ. (d) When x = 171^\circ, y = -1.75.

(a) In the set of numbers from 75, 76, 77, ... , 497, 498, 499, 500, you will find that the first term which is the first number divisible by 7 = 77 and the last term = 497. Given the first and last terms of an A.P, you can use the formula \(n = \frac{l - a}{d} + 1\) to find the number of terms in the sequence. First term = a = 77 Last term = l = 497 Common difference = d = 7 \(n = \frac{497 - 77}{7} + 1\) \(n = \frac{420}{7} + 1\) = \(60 + 1\) = 61. There are 61 numbers between 75 and 500 which are divisible by 7. (b) \(T_{n} = a + (n - 1)d\) (for an A.P) \(T_{8} = a + 7d\) \(T_{3} = a + 2d\) \(\implies a + 7d = 5(a + 2d)\) \(T_{7} = a + 6d\) \(T_{4} = a + 3d\) \(\implies a + 6d = a + 3d + 9\) \(a + 7d = 5a + 10d \implies 5a - a + 10d - 7d = 0\) \(4a + 3d = 0 ... (1)\) \(a + 6d = a + 3d + 9 \implies a - a + 6d - 3d = 9\) \(3d = 9 \implies d = 3\) Putting d = 3 in (1), \(4a + 3(3) = 0 \implies 4a + 9 = 0\) \(4a = -9 \implies a = \frac{-9}{4}\) \(\therefore\) The second term of the sequence = \(\frac{-9}{4} + 3 = \frac{3}{4}\) Third term = \(\frac{3}{4} + 3 = \frac{15}{4}\) Fourth term = \(\frac{15}{4} + 3 = \frac{27}{4}\) Fifth term = \(\frac{27}{4} + 3 = \frac{39}{4}\) The first 5 terms of the sequence = \(\frac{-9}{4}, \frac{3}{4}, \frac{15}{4}, \frac{27}{4}, \frac{39}{4}\)

4X - 7 \(\leq\) 3X and 3X - 4 \(\leq\) 4X 4X - 3X \(\leq\) 7 and 3X - 4X \(\leq\) 4 X \(\leq\) 7 and -X \(\leq\) 4 = X \(\geq\) -4 Range -4 \(\leq\) x \(\leq\) 7

x\(^2\) - 5x + 6 = 0 (X-2)(X-3) = 0 X-2 = 0 implies X = 2 X-3 = 0 implies X = 3 ∴ x \le 2, x \ge 3

U5 = 24, n = 5 and U11 = 96, n = 11 Un = a + (n-1)d 24 = a + (5-1)d imply 24 = a+4d .....eqn1 96 = a + (11-1)d imply 96 = a+10d ...eqn2 eqn1 - eqn2 -72 = -6d d = 72/6 = 12 but 24 = a+4d 24 = a + 4(12) 24 = a + 48 a = 24-48 a = -24

3X + 2Y + 1 = 0 2Y = -3X - 1 \(\frac{-3}{2}X - \frac{1}{2}\) Gradient of 3X + 2Y +1 = 0 is -3/2 Gradient of a line perpendicular to 3X + 2Y + 1 = 0 \(=-1 \div \frac{3}{2}\) =-1 \(\times \frac{-2}{3} = \frac{2}{3}\)

\(\begin{pmatrix} 1 & 0 \\ -1 & -1\\ 2 & 2 \end{pmatrix}\) + \(\begin{pmatrix} x & 1 \\ -1 & 0\\ y & -2 \end{pmatrix}\) = \(\begin{pmatrix} -2 & 1 \\ -2 & -1\\ -3 & 0 \end{pmatrix}\) therefore, (x, y) = (-3, -5) respectively

-2p + r = 1.......(i) 2p + 3r = 0.......(ii) r - 1 + 2p ........(iii) 2p + 3(1 + 2p) = 0 2p + 3(1 + 2p) = 0 2p + 3 + 6p = 0 3 - 8p = 0 \(\to\) 8p = 3 p = \(\frac{3}{8}\) 6 = 1 - 2 \(\frac{3}{8}\) = 1 - \(\frac{6}{8}\) \(\frac{8 - 6}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)