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P = \(\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}\) |P| = 2 - 1 x -3 = 5 P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}\) = \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)

-6 \(\leq\) 4 - 2x < 5 - x split inequalities into two and solve each part as follows: -6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x -10 \(\leq\) -2x \(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\) giving 5 \(\geq\) x or x \(\leq\) 5 4 - 2x < 5 - x -2x + x < 5 - 4 -x < 1 \(\frac{-x}{-1}\) > \(\frac{1}{-1}\) giving x > -1 or -1 < x Combining the two results, gives -1 < x \(\leq\) 5

3rd term : a + 2d = -9 .......(1) 7th term : a + 6d = -29 ......(2) (2) - (1): 4d = -20 :. d = -20/4 = -5 From (1) : a + 2(-5) = -9 a - 10 = -9 :. a = -9 + 10 = 1 :. 10th term of A.P is a + 9d = 1 + 9 (-5) = 1 - 45 = -44

Using S\(\infty\) = \(\frac{a}{1 - r}\) r = \(\frac{0.05}{0.5}\) = \(\frac{1}{10}\) S\(\infty\) = \(\frac{0.5}{{\frac{1}{10}}}\) = \(\frac{0.5}{({\frac{9}{10}})}\) = \(\frac{0.5 \times 10}{9}\) = \(\frac{5}{9}\)

By comparing y = mx + c with y = -4x + 2, the gradient of y = -4x + 2 is m1 = -4 let the gradient of the line parallel to the given line be m2, then, m2 = m1 = -4 (condition for parallelism) using, y - y1 = m2(x - x1) Hence the equation of the parallel line is y - 3 = -4(x-2) y - 3 = -4 x + 8 y + 4x = 8 + 3 y + 4x = 11 y + 4x - 11 = 0

3(x + 1) \(\leq\) 5(x + 2) + 15 3x + 3 \(\leq\) 5x + 10 + 15 3x - 5x \(\leq\) 10 + 15 - 3 -2x \(\leq\) 22 x \(\geq\) \(\frac{-22}{2}\) x \(\geq\) -11

sequence: -2, 4, -8, 16........{GP} a = -2; r = \(\frac{4}{-2}\) = -2 nth term Tn = arn-1 Tn = (-2)(-2)^n-1 Tn = (-2)1 + n - 1 Tn = (-2)n

y = mx + c; when x = 0; y = 1 1 = m(0) + c; 1 = 0 + c; c = 1 when x = 2; y = 2 2 = m(2) + c; 2 = 2m + c; but c = 1 2 = 2m + 1 2 - 1 = 2m 2m = 1 m = \(\frac{1}{2}\) y = \(\frac{1}{2}\)x + 1

(x - p)(2x + 1) = 0 2x2 + x - 2px - p = 0 2x2 + x (1 - 2p) - p = 0 2x2 - (2p - 1)x - p = 0 divide through by 2 x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0 compare to x2 - (sum of roots)x + product of roots = 0 sum of roots = \(\frac{2p - 1}{2}\) But sum of roots = 1 Given; \(\frac{2p - 1}{2}\) = 1 2p - 1 = 2 x 1 2p - 1 = 2 2p = 2 + 1 = 3 p = \(\frac{3}{2}\) p = 1\(\frac{1}{2}\)

Using \(\frac{x}{a} + \frac{y}{b}\) < 1for the equation of the time where a = intercept on x-axis and b = intercept on y - axis \(\frac{x}{3} + \frac{y}{3} = 1\) = \(\frac{x + y}{3} = 1\) = x + y < 3

(a) x -3 -2 -1 0 1 2 3 y -10 -4 0 2 2 0 -4 (b) (c) (ii) roots of the equation : \(x^{2} - x - 2 = 0\) are x = -1 or x = 2. (iii) gradient of the curve at x = -0.5 is 2.1.

\(T_{n} = ar^{n - 1}\) (Terms of a G.P) \(T_{3} = ar^{2} =24 .... (1)\) \(T_{7} = ar^{6} = 4\frac{20}{27} = \frac{128}{27} .......... (2)\) Divide (2) by (1), \(\frac{ar^{6}}{ar^{2}} = \frac{\frac{128}{27}}{24}\) \(r^{4} = \frac{16}{81}\) \(r = \sqrt[4]{\frac{16}{81}}\) = \(\frac{2}{3}\) Putting r in equation (1), \(ar^{2} = a(\frac{2}{3})^{2} = 24\) \(\frac{4}{9} a = 24 \implies a = \frac{24 \times 9}{4}\) a = 54. (b) \(y \propto x \propto \frac{1}{z^{2}}\) \(\implies y = \frac{kx}{z^{2}}\) \(4 = \frac{3k}{1^{2}} \implies k = \frac{4}{3}\) \(\therefore y = \frac{4x}{3z^{2}}\) When x = 3, z = 2, y = ? \(y = \frac{4(3)}{3(2^{2})}\) = \(\frac{12}{12} = 1\).

If 4/3 and -3/5 are roots of a polynomial Imply x = 4/3 and - 3/5 3x = 4 and 5x = -3 ∴3x-4 = 0 and 5x+3 = 0 are factors (3x-4)(5x+3) = 0 product of the factors 15x2 + 9x – 20x – 12 = 0 By expansion 15x2 - 11x – 12 = 0

x = -2 and x = 1/4 x = -2 and 4x = 1 x+2 and 4x-1 (x+2)(4x-1) = 0 4x2 - x + 8x -2 = 0 4x2 + 7x – 2 = 0 but y intercept is positive. Multiply the equation by -1 -4x2 - 7x + 2 = 0 ∴y = 2 – 7x – 4x2

a = 5, d = 6, n = n Sn = n/2(2a + (n-1)d) = n/2(2(5) + (n-1)6) = n/2(10 + 6n-6) = n/2(6n+4) = 6n2/2 + 4n/2 = 32 + 2n = n(3n + 2)