Correct Answer: B. a = -5, b = 14

Explanation

The terms of the sequence can be written as : \(u_{r} = ar + b\) in this case, being that they have a regular common difference for each of the r terms.

We can rewrite the sequence as \(a + b, 2a + b, 3a + b,...\) where a is the common difference of the sequence and b is a given constant gotten by solving

\(a + b = 9\) or \(2a + b = 4\) or any other one.

The common difference here is 4 - 9 = -1 - 4 = -5.

\(-5 + b = 9 \implies b = 9 + 5 = 14\)

\(\therefore\) The equation can be written as \(u_{r} = -5r + 14\).

Correct Answer: B. 128

Explanation

\(S_{\infty} = \frac{a}{1 - r}\) (for an exponential series)

\(r = \frac{24}{96} = \frac{6}{24} = \frac{1}{4}\)

\(S_{\infty} = \frac{96}{1 - \frac{1}{4}} = \frac{96}{\frac{3}{4}}\)

= \(\frac{96 \times 4}{3} = 128\)

Correct Answer: A. \(\frac{3}{2}\) or \(1\)

Explanation

\((2t - 3s)(t - s) = 0 \implies (2t - 3s) = \text{0 or} (t - s) = 0\)

\(2t - 3s = 0 \implies 2t = 3s \therefore \frac{t}{s} = \frac{3}{2}\)

\(t - s = 0 \implies t = s \therefore \frac{t}{s} = 1\)

\(\frac{t}{s} = \frac{3}{2} or 1\)

Explanation

(a)(i) \((2 - \frac{1}{2}x)^{5} = ^{5}C_{5}(2)^{5}(-\frac{1}{2}x)^{0} + ^{5}C_{4}(2)^{4}(-\frac{1}{2}x)^{1} + ^{5}C_{3}(2)^{3}(-\frac{1}{2}x)^{2} + ^{5}C_{2}(2)^{2}(-\frac{1}{2}x)^{3} + ^{5}C_{1} (2)^{1}(-\frac{1}{2}x)^{4} + ^{5}C_{0} (2)^{0}(-\frac{1}{2}x)^{5}\)

= \(32 - 5(8x) + 10(2x^{2}) - 10(\frac{x^{3}}{2}) + 5(\frac{x^{4}}{8}) - \frac{x^{5}}{32}\)

= \(32 - 40x + 20x^{2} - 5x^{3} + \frac{5}{8}x^{4} - \frac{1}{32}x^{5}\)

(ii) \((1.99)^{5} = (2 - \frac{1}{2}x)^{5}\)

\(2 - \frac{1}{2}x = 1.99 \implies \frac{1}{2}x = 2 - 1.99 = 0.01\)

\(x = 0.01 \times 2 = 0.02\)

Put x = 0.02 in the equation, we have

\(32 - 40(0.02) + 20(0.02)^{2} - 5(0.02)^{3} + \frac{5}{8}(0.02)^{4} - \frac{1}{32}(0.02)^{5}\)

\(32 - 0.8 + 0.008 - ...\)

= \(31.208 \approxeq 31.21\)

(b) \(f(x) = x^{3} + qx^{2} + rx + 9\)

(x + 1) is a factor, hence, f(-1) = 0.

\(f(-1) = (-1)^{3} + q(-1)^{2} + r(-1) + 9\)

\(0 = -1 + q - r + 9\)

\(-8 = q - r ... (1)\)

(x + 2) has remainder -17, therefore f(-2) = -17.

\(-17 = (-2)^{3} + q(-2)^{2} + r(-2) + 9\)

\(-17 = -8 + 4q - 2r + 9\)

\(-18 = 4q - 2r \implies -9 = 2q - r ... (2)\)

From (1), r = q + 8.

\(\therefore 2q - q - 8 = -9\)

\(q = -1\)

\(r = -1 + 8 = 7\)

\((q, r) = (-1, 7)\)

Correct Answer: D. \(x: x \in R\)

Explanation

The domain of a function refers to the regions where the function is defined or has a value on a particular region.

\(\frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\) has a domain defined on all set of real numbers because the function is defined on the set of real numbers when the denominator \(\sqrt{9x^{2} + 1} \geq 0\).

\(\sqrt{9x^{2} + 1} \geq 0 \implies 9x^{2} + 1 \geq 0\) which because of the square sign has a value for all values of x, be it negative or positive.

Correct Answer: C. 0

Explanation

\(\frac{x + P}{(x-1)(x-3)} = \frac{Q}{x-1} + \frac{2}{x-3}\)

\(\frac{x + P}{(x-1)(x-3)} = \frac{Q(x-3) + 2(x-1)}{(x-1)(x-3)}\)

Comparing LHS and RHS of the equation, we have

\(x + P = Qx - 3Q + 2x -2\)

\(P = -3Q - 2\)

\(Q + 2 = 1 \implies Q = 1 - 2 = -1\)

\(P = -3(-1) - 2 = 3 - 2 = 1\)

\(P + Q = 1 + (-1) = 0\)

Correct Answer: B. 3.0 s

Explanation

\(F = ma \)

\(10 = 2.5a \implies a = 4 ms^{-2}\)

Since it is a retarding movement, then \(a = -4 ms^{-2}\).

\(v = u + at; v = 0 ms^{-1}, u = 12 ms^{-1}\)

\(0 = 12 + (-4t) \implies 0 = 12 - 4t\)

\(4t = 12 \implies t = 3 s\)

Correct Answer: B. \(\sqrt{3}\)

Explanation

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding this, we have \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the given equation, \(3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0\) (making the coefficients of \(x^{2}\) and \(y^{2}\) = 1 , we get that

\(-2a = 2 \implies a = -1\)

\(2b = 4 \implies b = 2\)

\(r^{2} - a^{2} - b^{2} = -2\)

\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3\)

\(r = \sqrt{3}\)

Explanation

(a)P(Bad Quality) = \(\frac{30}{10}\) = 0.3 and P(Good Quality) = 1 - 0.3 = 0.7

(a)(i) P(exactly 6) = \(^{12}C_6\) x (0.3)\(^6\) x (0.7)\(^6\) = 0.0792

(a)(ii) P(at least 4) = 1 - [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)]

Where P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = \(^{12}C_0\)(0.3)\(^6\)(0.7)\(^{12}\) + \(^{12}C_1\)(0.3)\(^1\)(0.7)\(^{11}\) + \(^{12}C_2\)(0.3)\(^2\)(0.7)\(^{10}\) + \(^{12}C_3\)(0.3)\(^3\)(0.7)\(^{9}\)

= 0.01384 + 0.071184 + 0.16779 + 0.2397

Therefore; P(at least 4) = 1 - 0.4925 = 0.5075

(a)(iii) P(no bad one) = \(^{12}C_0\)(0.3)\(^o\)(0.7)\(^{12}\) = 0.0138

(b) To find P(4 boys and 3 girls) = \(\frac{^8C_4 \times ^5C_3}{^{13}C_7}\)

= \(\frac{70 \times 10}{1716}\)

= \(\frac{175}{429}\)

= 0.4079

Explanation

(a) The position vector of L, \(\overrightarrow{OL} = 5i + 6j\)

The coordinates of L = (5, 6)

The position vector of M, \(\overrightarrow{OM} = 13i + 4j\)

The coordinates of M = (13, 4).

Let \(\overrightarrow{OK} = xi + yj\) be the position vector of K.

\(\overrightarrow{LK} = (x - 5)i + (y - 6)j\), \(\overrightarrow{KM} = (13 - x)i + (4 - y)j\)

\(LK : KM = 2 : 3 ; \frac{LK}{KM} = \frac{2}{3}\)

\(3[(x - 5)i + (y - 6)j] = 2[(13 - x)i + (4 - y)j]\)

Equating by components,

\(i : 3x - 15 = 26 - 2x \implies 5x = 41\)

\(x = \frac{41}{5}\)

\(j : 3y - 18 = 8 - 2y \implies 5y = 26\)

\(y = \frac{26}{5}\)

The position vector of K is \(\overrightarrow{OK} = \frac{41}{5} i + \frac{26}{5} j\)

(b) \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\)

= \(\begin{pmatrix} 8 \cos 60° \\ 8 \sin 60° \end{pmatrix} + \begin{pmatrix} 12 \cos 130° \\ 12 \sin 130° \end{pmatrix}\)

= \(\begin{pmatrix} 8 \times \frac{1}{2} \\ 8 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatric} -12 \times 0.6428 \\ 12 \times 0.7660 \end{pmatrix}\)

= \(\begin{pmatrix} 4 \\ 4\sqrt{3} \end{pmatrix} + \begin{pmatrix} -7.7136 \\ 9.9192 \end{pmatrix} = \begin{pmatrix} -3.7136 \\ 16.12 \end{pmatrix}\)

\(|AC| = \sqrt{(-3.7136)^{2} + (16.12)^{2}} = \sqrt{13.794 + 259.9}\)

= \(\sqrt{273.694} = 16.544 \approxeq 16.5 km\)

\(AB = (8 \cos 60 i + 8 \sin 60 j) ; AC = (-3.7136 i + 16.12 j)\)

(ii)

Let the angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) be \(\theta\).

Using sine rule,

\(\frac{\sin B}{AC} = \frac{\sin A}{BC}\)

\(\frac{\sin 110°}{16.5} = \frac{\sin \theta}{12}\)

\(\sin \theta = \frac{12 \sin 110°}{16.5}\)

\(\sin \theta = 0.6834\)

\(\theta = \sin^{-1} (0.6834) = 43.11°\)

Bearing of C from A = 60° + 43.11° = 103.11° \(\approxeq\) 103° (nearest degree)