Explanation

Since PQRM is a parallelogram,

\(\overrightarrow{PQ} = \overrightarrow{MR}\)

\((4 - 1)i + (-5 - 3)j = (x + 5)i + (y - 2)j\)

Equating components, we have

\(3 = x + 5 \implies x = 3 - 5 = -2\)

\(-8 = y - 2 \implies y = -8 + 2 = -6\)

\(\therefore M = (-2i - 6j)\)

(b) \(|\overrightarrow{PM}| = \sqrt{(-2 - 4)^{2} + (- 6 + 5)^{2}} = \sqrt{37}\)

\(|\overrightarrow{PQ}| = \sqrt{(1 - 4)^{2} + (3 + 5)^{2}} = \sqrt{73}\)

(c) Let the angle be \(\theta\).

\(\overrightarrow{PM} . \overrightarrow{PQ} = |PM| |PQ| \cos \theta\)

\((-6i - j) . (-3i + 8j) = (\sqrt{37})(\sqrt{73}) \cos \theta\)

\(18 - 8 = \sqrt{2701} \cos \theta\)

\(\cos \theta = \frac{10}{\sqrt{2701}} = 0.1924\)

\(\theta = 78.907° \approxeq 78.9°\) (to 1 d.p)

(d) Area of PQRM = \((\sqrt{37})(\sqrt{73}) \sin 78.907°\)

= \(\sqrt{2701} \sin 78.907°\)

= \(51 \text{sq. units}\)

Correct Answer: B. \(\frac{4}{9}\)

Explanation

Let the number of candidates that speak both Igbo and Hausa = x

The number that speaks Igbo only = 56 - x

The number that speaks Hausa only = 50 - x

56 - x + x + 50 - x = 90

106 - x = 90

x = 16

P(Igbo only) = \(\frac{56 - 16}{90} = \frac{4}{9}\)

Correct Answer: B. \(125 m\)

Explanation

(b) \(\bar{x} = \frac{\sum X}{N} = \frac{228}{10} = 22.8\)

\(\bar{y} = \frac{\sum Y}{N} = \frac{115}{10} = 11.5\)

(c) See graph above.

Correct Answer: A. 9

Explanation

\(\int^3_0(px^2 + 16)\) = 129

\(\frac{px^2 + 1}{0 + 1} + 16x|^3_0 = 129\)

\(\frac{px^3}{3} + 16x|^3_0 = 129\)

(\(\frac{p(3)^3}{3} + 16(3)\)) - 0 = 129

9p + 48 = 129

9p = 129 - 48

\(\frac{9p}{9} = \frac{81}{9}\)

p = 9

Correct Answer: D. -3

Explanation

\(g(x) = \frac{x + 1}{x + 2}, x \neq 2\)

Let y = x, then \(g(y) = \frac{y + 1}{y + 2}\)

Let x = g(y), so that \(x = \frac{y + 1}{y + 2}\)

\(x(y + 2) = y + 1\)

\(xy + 2x = y + 1 \implies xy - y = 1 - 2x\)

\(y(x - 1) = 1 - 2x \implies y = \frac{1 - 2x}{x - 1}\)

\(y = g^{-1}(x) = \frac{1 - 2x}{x - 1}\)

\(g^{-1}(2) = \frac{1 - 2(2)}{2 - 1} = -3\)

Explanation

(a) \(\sqrt{4x - 3} - \sqrt{2x - 5} = 2\)

Squaring both sides, we have

\(4x - 3 - 2\sqrt{8x^{2} - 26x + 15} + 2x - 5 = 4\)

\(6x - 8 - 4 = 2\sqrt{8x^{2} - 26x + 15}\)

\(6x - 12 = 2\sqrt{8x^{2} - 26x + 15}\)

\(3x - 6 = \sqrt{8x^{2} - 26x + 15}\)

Squaring both sides, we have

\(9x^{2} - 36x + 36 = 8x^{2} - 26x + 15\)

\(9x^{2} - 8x^{2} - 36x + 26x + 36 - 15 = 0\)

\(x^{2} - 10x + 21 = 0\)

Factorising, we have

\((x - 7)(x - 3) = 0\)

\(\text{x = 7 or 3}\).

(b)

We integrate with respect to y.

\(y^{2} = 4x \implies x = \frac{y^{2}}{4}\)

\(x + y = 0 \implies x = -y\)

To get the boundary, equate the two values of x;

\(\frac{y^{2}}{4} = - y \implies y^{2} = -4y\)

\(y^{2} + 4y = 0 \implies y(y + 4) = 0\)

\(y = 0; y = -4\)

\(\int_{-4}^{0} (\frac{y^{2}}{4}) \mathrm {d} y = [\frac{y^{3}}{12}]|_{-4}^{0}\)

\(0 - \frac{-4^{3}}{12} = 0 - \frac{-64}{12} = \frac{64}{12} sq. units\)

\(\int_{-4}^{0} - y \mathrm {d} y = [-\frac{y^{2}}{2}]|_{-4}^{0}\)

\(0 - \frac{-(-4^{2})}{2} = \frac{16}{2} = 8 sq. units\)

Area enclosed = \(8 - \frac{64}{12} = \frac{32}{12} = 2\frac{2}{3} sq. units\)

Correct Answer: D. \(9\frac{9}{16}\)

Explanation

\(T_{n} = ar^{n-1}\) (for an exponential sequence)

\(T_{1} = 36 = a\)

\(T_{2} = ar = 36r = p\)

\(T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}\)

\(T_{4} = ar^{3} = 36r^{3} = q\)

\(36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}\)

\(r = \sqrt{\frac{1}{16}} = \frac{1}{4}\)

\( p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}\)

\(p + q = 9 + \frac{9}{16} = 9\frac{9}{16}\)

Explanation

Let the linear transformation T be represented by the following:

\(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} -1 & 7 \\ 4 & 11 \end{pmatrix}\).

\(3a + 2b = -1 ... (1)\)

\(-a + 4b = 7 ... (2)\)

\(3c + 2d = 4 ... (3)\)

\(-c + 4d = 11 ... (4)\)

\((2) \times 3 : -3a + 12b = 21... (5)\)

\((1) + (5) : 14b = 20 \implies b = \frac{10}{7}\)

Substitute for b in (1),

\(3a + 2(\frac{10}{7}) = -1 \implies 3a = \frac{-27}{7}\)

\(a = \frac{-9}{7}\)

Solving for c and d,

\((4) \times 3 : -3c + 12d = 33 ... (6)\)

\((3) + (6) : 14d = 37 \implies d = \frac{37}{14}\)

\(-c + 4(\frac{37}{14}) = 11 \implies -c = 11 - \frac{74}{7} = \frac{3}{7}\)

\(c = \frac{-3}{7}\)

\(\therefore T = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix}\)

\(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

From \(P : (x, y) \to (x + y, x + 2y)\)

\(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} x + y \\ x + 2y \end{pmatrix}\)

\(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

\(\therefore P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\end{pmatrix}\)

(b) \(TP = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{-9}{7} + \frac{10}{7} & \frac{-9}{7} + \frac{20}{7} \\ \frac{-3}{7} + \frac{37}{14} & \frac{-3}{7} + \frac{74}{14} \end{pmatrix}\)

= \(\begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{14} & \frac{34}{7} \end{pmatrix}\)

(c) The image of X(4, 3) under TP:

\(TP : \begin{pmatrix} 4 \\ 3 \end{pmatrix} \to \begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{17} & \frac{34}{7} \endd{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{4}{7} + \frac{33}{7} \\ \frac{62}{7} + \frac{102}{7} \end{pmatrix}\)

= \(\begin{pmatrix} \frac{37}{7} \\ \frac{164}{7} \end{pmatrix}\)

The image of X(4, 3) under TP is \((\frac{37}{7}, \frac{164}{7})\).

Explanation

(a) \(\int _{0} ^{1} \frac{3}{1 + x^{2}} \mathrm {d} x\)

h = 0.25 ;

\(y_{1} = 3, y_{2} = 2.82 , y_{3} = 2.40 , y_{4} = 1.92 , y_{5} = 1.5\)

\(y_{1} + y_{5} = 3 + 1.5 = 4.5\)

\(y_{2} + y_{3} + y_{4} = 2.82 + 2.40 + 1.92 = 7.14\)

\(2(y_{2} + y_{3} + y_{4}) = 14.28\)

= \(3[\frac{1}{2} h ( y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4})]\)

= \(3[\frac{1}{2} (0.25) (4.5 + 14.28)]\)

= \(3(0.125)(18.78) = 2.3475\)

\(\approxeq 2.348\) (4 sig. figures).

(b) \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\)

\(A^{2} + A + 2I = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

= \(\begin{pmatrix} 4 + 3 & 2 + 2 \\ 6 + 6 & 3 + 4 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} 7 & 4 \\ 12 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} 11 & 5 \\ 15 & 11\end{pmatrix}\)

\(T : \begin{pmatrix} 1 \\ 2 \end{pmatrix} \to \begin{pmatrix} 11 & 5 \\ 15 & 11 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)

=\(\begin{pmatrix} 11 + 10 \\ 15 + 22 \end{pmatrix} = \begin{pmatrix} 21 \\ 37 \end{pmatrix}\)

The image of (1, 2) = (21, 37).