WAEC SSCE - Further Mathematics - 2013

Akọwa Nkọwa

Akọwa Nkọwa

We can find the sum of the vectors U and V by adding their corresponding components, i.e., $$ U + V = \begin{pmatrix} -1 \\ 5 \end{pmatrix} + \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 9 \end{pmatrix}. $$ Then, the magnitude (length) of the vector U + V is given by the formula $$ |U + V| = \sqrt{(-3)^2 + 9^2} = \sqrt{90} = 3\sqrt{10}. $$ Therefore, the answer is \(3\sqrt{10}\).

Akọwa Nkọwa

Akọwa Nkọwa

To find the radius of a circle given its equation, we need to complete the square for both the x and y terms. We can do this by rearranging the equation as follows: \begin{align*} x^2 - 8x + y^2 + 9y &= -15 \\ (x^2 - 8x + 16) + (y^2 + 9y + 20.25) &= -15 + 16 + 20.25 \\ (x - 4)^2 + (y + 4.5)^2 &= 21.25 \end{align*} We can now see that the equation of the circle is in the standard form: \begin{equation*} (x - a)^2 + (y - b)^2 = r^2 \end{equation*} where the center of the circle is at point (a, b) and the radius is r. From the completed square form, we can identify that the center of the circle is at point (4, -4.5) and the radius is the square root of 21.25, which simplifies to: \begin{equation*} r = \sqrt{21.25} = \frac{1}{2} \sqrt{85} \end{equation*} Therefore, the correct answer is option (C), \(\frac{1}{2}\sqrt{85}\).

Akọwa Nkọwa

To factorize the given expression, we can use synthetic division or long division to divide \(f(x)\) by \(x - 3\) which is one of its factors as given. The result of this division is a quadratic expression that can be factored easily. Using synthetic division, we get: \begin{array}{c|cccc} & 2 & -3 & -11 & 6 \\ \hline 3 & & 6 & 9 & -6 \\ & & & -6 & 27 \\ \hline & 2 & 3 & -2 & 21 \end{array} Thus, \(f(x) = (x - 3)(2x^{2} + 3x - 2)\). Now we need to factorize the quadratic expression \(2x^{2} + 3x - 2\). We can use the quadratic formula or factorization by grouping to factorize the quadratic expression. Factorization by grouping is simpler and can be done as follows: \begin{align*} 2x^{2} + 3x - 2 &= 2x^{2} + 4x - x - 2 \\ &= 2x(x + 2) - 1(x + 2) \\ &= (2x - 1)(x + 2) \end{align*} Therefore, the complete factorization of \(f(x)\) is: $$f(x) = (x - 3)(2x - 1)(x + 2)$$ Hence, the correct option is (C) (x - 3)(x + 2)(2x - 1).

Akọwa Nkọwa

Akọwa Nkọwa

To find the third term in the expansion of \((a-b)^6\) in ascending powers of b, we can use the binomial theorem, which states that: \[(a-b)^6 = \sum_{n=0}^{6} {6 \choose n} a^{6-n}(-b)^n\] The third term is the term where n=2, so we plug in n=2 into the above formula and simplify: \[{6 \choose 2} a^{6-2}(-b)^2 = 15a^4b^2\] Therefore, the third term in the expansion of \((a-b)^6\) in ascending powers of b is \(15a^4b^2\). So the correct answer is option B, "15a^4b^2".

Akọwa Nkọwa

To solve this problem, we will use the section formula. Let the point which divides the line segment PQ internally in the ratio 2:3 be R(x,y). Then we have: \begin{align*} x = \frac{3\times(-2) + 2\times4}{2+3} = \frac{6}{5} \\ y = \frac{3\times3 + 2\times9}{2+3} = \frac{24}{5} \end{align*} Therefore, the coordinates of R are \((\frac{6}{5},\frac{24}{5})\), which corresponds to option (B).

Akọwa Nkọwa

To evaluate \(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x\), we first need to find the antiderivative of the integrand with respect to x. \[\int (8x - 4x^{2}) \mathrm {d} x = 4x^{2} - \frac{4}{3} x^{3} + C\] where C is the constant of integration. Using the limits of integration, we can evaluate the definite integral as follows: \[\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x = \left[4x^{2} - \frac{4}{3} x^{3}\right]_{0}^{2}\] \[= \left[4(2)^{2} - \frac{4}{3}(2)^{3}\right] - \left[4(0)^{2} - \frac{4}{3}(0)^{3}\right]\] \[= \frac{16}{3}\] Therefore, the answer is \(\frac{16}{3}\). Option (c) is the correct answer.

Akọwa Nkọwa

To find the magnitude of the resultant force, we can use the Law of Cosines, which states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the included angle. In this case, let the two forces be a and b, with an included angle of 120°. Then, the magnitude of the resultant force, which we'll call R, is: $$R^2 = a^2 + b^2 - 2ab\cos{120^\circ}$$ Since cosine of 120° is -1/2, we can simplify this to: $$R^2 = a^2 + b^2 + ab$$ Substituting the values given in the problem, we get: $$R^2 = (6\text{ N})^2 + (3\text{ N})^2 + (6\text{ N})(3\text{ N}) = 54\text{ N}^2$$ Taking the square root of both sides, we get: $$R = \sqrt{54\text{ N}^2} = 3\sqrt{6}\text{ N} \approx 7.746\text{ N}$$ Therefore, the correct answer is (D) \(3\sqrt{3}\) N.

Akọwa Nkọwa

To calculate the mean deviation, we first need to find the mean of the given data set. Mean = (1 + 2 + 3 + 4 + 5 + 5 + 6 + 7 + 8 + 9) / 10 = 50 / 10 = 5 Next, we find the deviation of each value from the mean by subtracting the mean from each value: 1 - 5 = -4 2 - 5 = -3 3 - 5 = -2 4 - 5 = -1 5 - 5 = 0 5 - 5 = 0 6 - 5 = 1 7 - 5 = 2 8 - 5 = 3 9 - 5 = 4 To find the mean deviation, we need to find the average of the absolute values of these deviations. Mean Deviation = (|-4| + |-3| + |-2| + |-1| + |0| + |0| + |1| + |2| + |3| + |4|) / 10 = 20 / 10 = 2 Therefore, the mean deviation of the given data set is 2. The answer is (a) 2.

Akọwa Nkọwa

Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), we can integrate both sides with respect to x to obtain the original function y(x). $$ \frac{\mathrm d y}{\mathrm d x} = \sqrt{x} \\ \int \frac{\mathrm d y}{\mathrm d x} \mathrm d x= \int \sqrt{x} \mathrm d x\\ y= \int x^{\frac{1}{2}} \mathrm d x \\ y= \frac{2}{3}x^{\frac{3}{2}} + c $$ where c is an arbitrary constant of integration. Therefore, the correct option is (b) \(\frac{2}{3}x^{\frac{3}{2}} + c\).

Akọwa Nkọwa

We can use trigonometric identities to solve the equation. First, we know that $\cos^2 \theta + \sin^2 \theta = 1$, so we can write: \begin{align*} 2\sin^2 \theta &= 1 + \cos \theta \\ 2(1 - \cos^2 \theta) &= 1 + \cos \theta \\ 2 - 2\cos^2 \theta &= 1 + \cos \theta \\ 2\cos^2 \theta + \cos \theta - 1 &= 0 \\ (2\cos \theta - 1)(\cos \theta + 1) &= 0 \\ \end{align*} Therefore, either $2\cos \theta - 1 = 0$ or $\cos \theta + 1 = 0$. Solving for $\theta$ in each case, we get: \begin{align*} 2\cos \theta - 1 &= 0 \\ \cos \theta &= \frac{1}{2} \\ \theta &= 60^\circ \\ \\ \cos \theta + 1 &= 0 \\ \cos \theta &= -1 \\ \theta &= 180^\circ \\ \end{align*} However, we are given that $0^\circ \leq \theta \leq 90^\circ$, so the only valid solution is $\theta = 60^\circ$. Therefore, the answer is (C) 60°.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To find g o f, we need to substitute f(x) into g(x). Therefore, we have: g o f = g(f(x)) = g(x^2) = sin(x^2) So the answer is \(\sin x^{2}\). Option (B).

Akọwa Nkọwa

Let's assume that the first term of the exponential sequence is "a" and the common ratio is "r". Then, we can write the fourth and ninth terms as: Fourth term: ar^3 = 192 Ninth term: ar^8 = 6 We can then divide the two equations to eliminate "a" and obtain a relationship between the two powers of "r": (ar^8) / (ar^3) = 6/192 r^5 = 1/32 Taking the fifth root of both sides, we get: r = (1/32)^(1/5) = 1/2 Therefore, the common ratio of the exponential sequence is \(\frac{1}{2}\).

Akọwa Nkọwa

We can simplify the left side of the equation as follows: $$a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = a^{\frac{5}{6} - \frac{1}{n}}$$ Since this expression equals 1, we have: $$a^{\frac{5}{6} - \frac{1}{n}} = 1$$ We can rewrite this as: $$\frac{5}{6} - \frac{1}{n} = 0$$ Solving for n, we get: $$\frac{1}{n} = \frac{5}{6}$$ $$n = \frac{6}{5} = 1.2$$ Therefore, the value of n is 1.20, which is.

Akọwa Nkọwa

We know that: \[\log a + \log b = \log ab\] Using this property, we can rewrite the given expression as: \[\log \left(\frac{1}{8} \cdot \frac{1}{2}\right) = \log \frac{1}{16}\] Now, we use another property of logarithms: \[\log a^b = b\log a\] to rewrite \(\log \frac{1}{16}\) in terms of \(\log 2\): \[\log \frac{1}{16} = \log 2^{-4} = -4\log 2\] Therefore, \(\log \frac{1}{8} + \log \frac{1}{2} = -4\log 2\) and the answer is: \(-4 \log 2\).

Akọwa Nkọwa

To differentiate the equation, we need to find the derivative of each term in it, using the rules of differentiation. \begin{align*} \frac{d}{dx}(x^{2} + xy - 5) &= \frac{d}{dx}(0) \\ \frac{d}{dx}(x^{2}) + \frac{d}{dx}(xy) - \frac{d}{dx}(5) &= 0 \\ 2x + y\frac{dx}{dx} + x\frac{dy}{dx} - 0 &= 0 \\ 2x + y + x\frac{dy}{dx} &= 0 \\ x\frac{dy}{dx} &= -(2x + y) \\ \frac{dy}{dx} &= -\frac{2x+y}{x} \end{align*} Therefore, the correct answer is \(\frac{-(2x + y)}{x}\).

Akọwa Nkọwa

To find the upper quartile, we first need to arrange the numbers in order from smallest to largest: 2, 5, 12, 17, 18, 29, 33, 41, 43, 45 The upper quartile divides the data set into quarters, with 25% of the data falling above this point. To find the upper quartile, we need to find the median of the upper half of the data set. The upper half of the data set is: 29, 33, 41, 43, 45 The median of this data set is the middle number, which is 41. Therefore, the upper quartile is 41. So the correct answer is.

Akọwa Nkọwa

To calculate the standard deviation, we need to first find the deviations of each sale from the mean sale, square them, sum them up, divide by the total number of sales, and then take the square root of the result. The mean sale is GH¢ 32.00, so the deviations of each sale from the mean are: - GH¢ -6.00 - GH¢ 7.00 - GH¢ 1.00 - GH¢ -7.00 - GH¢ 5.00 Squaring these deviations, we get: - 36 - 49 - 1 - 49 - 25 Summing these up, we get: 36 + 49 + 1 + 49 + 25 = 160 Dividing by the total number of sales (which is 5), we get: 160/5 = 32 Taking the square root of this result, we get: sqrt(32) = 5.66 Therefore, the standard deviation is GH¢ 5.66. So the answer is.

Akọwa Nkọwa

Akọwa Nkọwa

The problem states that a sub-committee must be formed consisting of 3 boys and 2 girls. Also, Ozo must be on the sub-committee. This means that we already have one of the three boys chosen. We need to choose two more boys from the remaining four boys, and two girls from the three girls. The number of ways to choose two boys from the remaining four boys is given by the combination formula: C(4,2) = 6. (Alternatively, we can list all the possible combinations of two boys from the remaining four boys: Kofi and John, Kofi and Ojo, Kofi and James, John and Ojo, John and James, Ojo and James. This gives us a total of 6 combinations.) Similarly, the number of ways to choose two girls from the three girls is given by the combination formula: C(3,2) = 3. (Alternatively, we can list all the possible combinations of two girls from the three girls: Rose and Ugo, Rose and Ama, Ugo and Ama. This gives us a total of 3 combinations.) To find the total number of ways to choose the sub-committee, we multiply the number of ways to choose two boys and two girls: 6 x 3 = 18. Therefore, the answer is 18.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

There are ten tickets on sale for the raffle draw, and two of them are winning tickets. If a guest buys two tickets, there are a total of \(_{10}C_2\) ways to choose two tickets out of the ten. This is because we can choose any two tickets out of ten in \(_{10}C_2\) ways, and each combination is equally likely. The number of ways to choose two winning tickets out of the two available winning tickets is \(_2C_2\), which is equal to 1. The probability of selecting two winning tickets is therefore: $$ \frac{\text{number of ways to choose two winning tickets}}{\text{number of ways to choose any two tickets}} = \frac{\binom{2}{2}}{\binom{10}{2}} = \frac{1}{\frac{10\times9}{2\times1}} = \frac{1}{45} $$ Therefore, the probability that both tickets are winning tickets is \(\frac{1}{45}\).

Akọwa Nkọwa

Akọwa Nkọwa

We can use the trigonometric identity \(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\) to rewrite the given equation as \(\sin \theta = \dfrac{\sin \theta}{\cos \theta}\). Multiplying both sides by \(\cos \theta\) and simplifying, we get \(\cos \theta = 1\). Therefore, the possible solutions are those angles whose cosine is 1, which are multiples of 360 degrees. Among the given options, only 0 degrees (option D) is a multiple of 360 degrees. So, the answer is \(\theta = 0^{\circ}\).

Akọwa Nkọwa

Akọwa Nkọwa

First, let's find the value of the expressions inside the parentheses: \begin{align*} t - 3s &= 2i + 3j - 3(3i - j)\\ &= 2i + 3j - 9i + 3j\\ &= -7i + 6j\\ \\ t + 3s &= 2i + 3j + 3(3i - j)\\ &= 2i + 3j + 9i - 3j\\ &= 11i \end{align*} Now we can substitute these expressions into the dot product: \begin{align*} (t - 3s).(t + 3s) &= (-7i + 6j).(11i)\\ &= -77i + 66j\\ \end{align*} Therefore, the answer is -77.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The function f is defined piecewise as follows: For x > 4, f(x) = 3x + 2 For x = 4, f(x) = 3x - 2 For x < 4, f(x) = 5x - 3 To find f(4) - f(-3), we first need to find f(4) and f(-3) separately. Since f(4) is defined as 3x - 2 when x = 4, we have f(4) = 3(4) - 2 = 10. Since f(x) is defined as 5x - 3 when x < 4 and we are looking for f(-3), we substitute -3 for x to get f(-3) = 5(-3) - 3 = -18. Now we can calculate f(4) - f(-3) as: f(4) - f(-3) = 10 - (-18) = 28 Therefore, the answer is 28.

Akọwa Nkọwa

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To find x such that PQ = R, we need to find the matrix product of P and Q, and then compare it to R. The matrix product of P and Q is given by: $$ PQ = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} (3\times 1) + (4\times -2) & (3\times 3) + (4\times 4) \\ (2\times 1) + (x\times -2) & (2\times 3) + (x\times 4) \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2-2x & 22+4x \end{pmatrix} $$ We are given that PQ = R, which means we can equate the corresponding entries of the two matrices. In particular, the entry in the first row and first column of PQ must be equal to the entry in the first row and first column of R, i.e. -5. This gives us the equation: $$ -5 = -5 $$ Next, we can equate the corresponding entries in the second row and first column of PQ and R, i.e. 2-2x = -8. Solving for x, we get: $$ 2-2x = -8 \Rightarrow x = 5 $$ Therefore, the value of x that satisfies the equation PQ = R is 5.

Akọwa Nkọwa

Class width \(=3\); mid-values \(13,16,19,22,25\). Code with \(u=\dfrac{x-19}{3}\), \(A=19\).

(a) Mean age.\[\bar{x}=19+\frac{\sum fu}{\sum f}\times 3=19+\frac{-18}{22}\times 3=19-2.4545=16.5\text{ years}\]

(b) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=3\sqrt{\frac{40}{22}-\left(\frac{-18}{22}\right)^2}\]\[=3\sqrt{1.8182-0.6694}=3\sqrt{1.1488}=3(1.0718)=3.22\]

To three significant figures: mean age = 16.5 years and standard deviation = 3.22 years.

Akọwa Nkọwa

(a) Displacement–time graph

The displacement is \(S = t^{2} - 6t + 5\). Compute \(S\) for each whole second in the interval \(0 \le t \le 6\):

Plotting these points and joining them with a smooth curve gives the parabola below (scale on the time axis: 2 cm to 1 s; on the displacement axis: 2 cm to 1 m).

(b) Reading the graph

(i) Time at which the velocity is zero.
The velocity is the gradient of the displacement–time graph, \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t}\). The velocity is zero where the tangent to the curve is horizontal, i.e. at the lowest point (turning point) of the parabola. From the graph this occurs at

(Check: \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t} = 2t - 6 = 0 \Rightarrow t = 3\ \text{s}.\))

(ii) Average velocity over the interval \(0 \le t \le 4\).
Average velocity is the change in displacement divided by the time taken (the gradient of the chord joining the two end points). From the graph, \(S = 5\) m at \(t = 0\) and \(S = -3\) m at \(t = 4\):

The average velocity is \(-2\ \text{m/s}\); the negative sign shows the net motion is directed back towards, and past, \(O\). (For interest, the total path length in this interval is \(9 + 1 = 10\) m, so the average speed is \(10/4 = 2.5\) m/s.)

(iii) Total distance covered in the interval \(0 \le t \le 5\).
Distance is the actual length of path travelled, so the two legs of the journey are added as positive lengths. From the graph the particle first moves from \(S = 5\) m down to the turning point \(S = -4\) m at \(t = 3\) s, then rises back to \(S = 0\) m at \(t = 5\) s.

Akọwa Nkọwa

Rectangle with one side three times the other; perimeter rises by 2%.

Let the shorter side be \(x\), so the longer side is \(3x\).

The perimeter \(P=8x\) is directly proportional to \(x\), so a \(2\%\) rise in perimeter is a \(2\%\) rise in \(x\). Using differentials for the area:

Alternatively, the new side is \(1.02x\), so the new area is \(3(1.02x)^{2}=1.0404\times 3x^{2}\), an increase of \(4.04\%\), which rounds to about \(4\%\).

The area increases by approximately \(4\%\).

Akọwa Nkọwa

(a) Winning is binomial with \( p = 0.8,\ q = 0.2,\ n = 4 \).

(i) Wins all races: \( p^4 = 0.8^4 = 0.4096. \)

(ii) Wins no race: \( q^4 = 0.2^4 = 0.0016. \)

(iii) Wins at most 3 races \( = 1 - P(\text{all 4}) = 1 - 0.4096 = 0.5904. \)

(b) Committee of 5 chosen from 5 girls and 10 boys (15 people). Total \( = \binom{15}{5} = 3003 \).

(i) Exactly 3 boys (so 2 girls): \[ \frac{\binom{10}{3}\binom{5}{2}}{3003} = \frac{120 \times 10}{3003} = \frac{1200}{3003} = \frac{400}{1001} \approx 0.400. \]

(ii) At least one girl \( = 1 - P(\text{no girl}) = 1 - \dfrac{\binom{10}{5}}{3003} = 1 - \dfrac{252}{3003} = \dfrac{2751}{3003} = \dfrac{917}{1001} \approx 0.916. \)

Akọwa Nkọwa

Resolve each force into east (x) and north (y) components using its bearing, where east \(= F\sin\theta\), north \(= F\cos\theta\).

\( \mathbf{P} = (4\sin 30^\circ,\ 4\cos 30^\circ) = (2.000,\ 3.464) \).
\( \mathbf{R} = (10\sin 300^\circ,\ 10\cos 300^\circ) = (-8.660,\ 5.000) \).

Resultant: \( \mathbf{P} + \mathbf{R} = (-6.660,\ 8.464) \).

The equilibrant \( \mathbf{E} \) is equal and opposite: \( \mathbf{E} = (6.660,\ -8.464) \).

Magnitude: \[ |\mathbf{E}| = \sqrt{6.660^2 + 8.464^2} = \sqrt{44.36 + 71.64} = \sqrt{116} \approx 10.8\text{ N}. \]

Direction (bearing): the equilibrant points east and south, so \[ \text{bearing} = 180^\circ - \tan^{-1}\!\left(\frac{6.660}{8.464}\right) = 180^\circ - 38.2^\circ \approx 142^\circ. \]

The equilibrating force is about \( 10.8\text{ N} \) on a bearing of \( 142^\circ \).

Akọwa Nkọwa

Momentum \( = \text{mass} \times \text{velocity} \), and change in momentum \( = m(\vec{v}_f - \vec{v}_i) \).

Initial velocity: \( \vec{v}_i = 40\mathbf{j} \); final velocity: \( \vec{v}_f = 30\mathbf{i} \); mass \( m = 0.1\text{ kg} \).

\[ \Delta \vec{p} = 0.1(30\mathbf{i} - 40\mathbf{j}) = (3\mathbf{i} - 4\mathbf{j})\text{ kg ms}^{-1}. \]

Magnitude: \[ |\Delta \vec{p}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\text{ kg ms}^{-1}. \]

The change in momentum is \( 3\mathbf{i} - 4\mathbf{j} \) (magnitude \(5\text{ kg ms}^{-1}\)).

Akọwa Nkọwa

Gradient of \(x^{3}+y^{3}-2xy=11\) at \((2,-1)\).

Differentiate implicitly with respect to \(x\), using the product rule on \(2xy\):

Collect the derivative terms:

At \((2,-1)\):

The gradient of the curve at \((2,-1)\) is \(14\).

Akọwa Nkọwa

(a) Differentiate \( y = \dfrac{x^2 + 1}{(x+1)^2} \) using the quotient rule with \( u = x^2+1,\ v = (x+1)^2 \):

\( u' = 2x,\ v' = 2(x+1) \).

\[ \frac{dy}{dx} = \frac{2x(x+1)^2 - (x^2+1)\,2(x+1)}{(x+1)^4} = \frac{2(x+1)\big[x(x+1) - (x^2+1)\big]}{(x+1)^4}. \]

\[ = \frac{2\big[x^2 + x - x^2 - 1\big]}{(x+1)^3} = \frac{2(x-1)}{(x+1)^3}. \]

(b)(i) Expand the determinant along the first row:

\[ \Delta = 1(3\cdot3 - (-1)\cdot1) - 2(2\cdot3 - (-1)(-1)) + (-1)(2\cdot1 - 3(-1)). \]

\[ \Delta = 1(10) - 2(5) - 1(5) = 10 - 10 - 5 = -5. \]

(ii) The system \( x+2y-z=4,\ 2x+3y-z=2,\ -x+y+3z=-1 \) has coefficient determinant \(\Delta = -5 \neq 0\), so use Cramer's rule.

\( \Delta_x = \begin{vmatrix} 4 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix} = 40 - 10 - 5 = 25 \Rightarrow x = \dfrac{25}{-5} = -5. \)

\( \Delta_y = \begin{vmatrix} 1 & 4 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 3 \end{vmatrix} = 5 - 20 - 0 = -15 \Rightarrow y = \dfrac{-15}{-5} = 3. \)

\( \Delta_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & 3 & 2 \\ -1 & 1 & -1 \end{vmatrix} = -5 - 0 + 20 = 15 \Rightarrow z = \dfrac{15}{-5} = -3. \)

Solution: \( x = -5,\ y = 3,\ z = -3 \).

Akọwa Nkọwa

(a) Prepare a table of values for \(y = \sin 2x\) and \(y = \cos x\) at intervals of \(30^\circ\) for \(0^\circ \le x \le 210^\circ\). For example, at \(x = 30^\circ\): \(\sin 2x = \sin 60^\circ = 0.87\) and \(\cos x = \cos 30^\circ = 0.87\); at \(x = 120^\circ\): \(\sin 2x = \sin 240^\circ = -0.87\) and \(\cos x = \cos 120^\circ = -0.50\). The complete table is:

Plotting these points, using a scale of 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 0.2 units on the \(y\)-axis, gives the two smooth curves below:

(b)(i) The truth set of \(\sin 2x = 0\) is read where the curve \(y = \sin 2x\) crosses the \(x\)-axis (i.e. where \(y = 0\)). From the graph these crossings occur at \(x = 0^\circ,\ 90^\circ\) and \(180^\circ\):

(b)(ii) The truth set of \(\sin 2x - \cos x = 0\), i.e. \(\sin 2x = \cos x\), is read where the two curves intersect. From the graph the curves cross at \(x = 30^\circ,\ 90^\circ\) and \(150^\circ\).

This agrees with the algebra: \(\sin 2x = \cos x \Rightarrow 2\sin x\cos x = \cos x \Rightarrow \cos x\,(2\sin x - 1) = 0\), so \(\cos x = 0 \Rightarrow x = 90^\circ\), or \(\sin x = \tfrac{1}{2} \Rightarrow x = 30^\circ,\ 150^\circ\). Hence the truth set is:

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(a) & (c) Scatter diagram, mean point and line of best fit

Plotting the eight pairs of readings using a scale of 2 cm to 10 units on the x-axis and 2 cm to 0.5 units on the y-axis gives the scatter diagram below. The mean point \((\bar{x},\bar{y})=(45,\,1.55)\) is marked, and the line of best fit is drawn straight through \((10,\,1.0)\) and \((45,\,1.55)\).

(b) Mean of x and mean of y

Therefore the mean point is \((\bar{x},\bar{y})=(45,\,1.55)\), which is plotted on the diagram above.

(d)(i) Equation of the line of best fit

The line passes through \((\bar{x},\bar{y})=(45,\,1.55)\) and \((10,\,1.0)\). Its gradient is

Using the point \((10,\,1.0)\):

Check: at \(x=45\), \(y=0.0157(45)+0.843=0.707+0.843=1.55=\bar{y}\), so the line indeed passes through the mean point.

(d)(ii) Height of plant after 75 days

Substituting \(x=75\) into the equation of the line of best fit:

Reading off the graph at \(x=75\) confirms this value. The height of the plant after 75 days is approximately 2.0 m (2.02 m).

Akọwa Nkọwa

(a) Equal coefficients of \(x^{2}\) and \(x^{3}\) in \((p+qx)^{7}\).

The general term is \({}^{7}C_r\,p^{7-r}(qx)^{r}\).

Set them equal:

(b) Weekly contributions 180, 260, 340, ...

This is an A.P. with \(a=180\), common difference \(d=80\), and \(n=16\).

He would have contributed \(\text{N}12{,}480.00\) in 16 weeks.

Akọwa Nkọwa

Line \(2y=x+3\) meets circle \(x^{2}+y^{2}-2x+6y-15=0\).

From the line, \(x=2y-3\). Substitute into the circle:

So \(y=0\) or \(y=2\).

• \(y=0:\ x=2(0)-3=-3\), giving \((-3,0)\).
• \(y=2:\ x=2(2)-3=1\), giving \((1,2)\).

N is in the first quadrant, so \(N=(1,2)\) and \(M=(-3,0)\).

Akọwa Nkọwa

Take downward as positive with initial speed \(u = 20\text{ ms}^{-1}\), \(g = 10\text{ ms}^{-2}\), height \(s = 45\text{ m}\).

(a) Time to reach the ground. Using \( s = ut + \tfrac12 g t^2 \):

\[ 45 = 20t + 5t^2 \Rightarrow 5t^2 + 20t - 45 = 0 \Rightarrow t^2 + 4t - 9 = 0. \]

\[ t = \frac{-4 + \sqrt{16 + 36}}{2} = \frac{-4 + \sqrt{52}}{2} = \frac{-4 + 7.211}{2} \approx 1.61\text{ s}. \]

(b) Speed on hitting the ground. Using \( v^2 = u^2 + 2gs \):

\[ v^2 = 20^2 + 2(10)(45) = 400 + 900 = 1300 \Rightarrow v = \sqrt{1300} \approx 36.1\text{ ms}^{-1}. \]

Akọwa Nkọwa

(a) Let the three terms be \( \dfrac{a}{r},\ a,\ ar \).

Product: \( \dfrac{a}{r}\cdot a \cdot ar = a^3 = 8 \Rightarrow a = 2. \)

Sum: \( \dfrac{2}{r} + 2 + 2r = 7 \Rightarrow \dfrac{2}{r} + 2r = 5 \Rightarrow 2r^2 - 5r + 2 = 0. \)

\[ (2r - 1)(r - 2) = 0 \Rightarrow r = \tfrac12 \text{ or } r = 2. \]

(i) The sequence is decreasing, so \( r = \dfrac12 \).

(ii) The three terms are \( \dfrac{a}{r} = 4,\ a = 2,\ ar = 1 \), i.e. \( 4,\ 2,\ 1 \).

(b) With \( f(x) = x + \dfrac{2}{x^2} \) and ordinates at \( x = 1,2,3,4,5 \) (\(h = 1\)):

\( f(1)=3,\ f(2)=2.5,\ f(3)=3.2222,\ f(4)=4.125,\ f(5)=5.08. \)

Trapezium rule: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]

\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]

Akọwa Nkọwa

(a) Maximum and minimum points of \( y = 2x^{3} - 3x^{2} - 12x + 4 \).

At stationary points \( \dfrac{dy}{dx} = 0 \):

Setting \( \dfrac{dy}{dx} = 0 \):

The nature of each point is tested with the second derivative:

At \( x = -1 \):

Maximum point \( (-1,\ 11) \).

At \( x = 2 \):

Minimum point \( (2,\ -16) \).

(b) Sketch of the curve.

To locate the curve, note the following guide points:

Since the leading coefficient is positive, the curve rises from the bottom-left, climbs to the maximum \((-1,\ 11)\), falls through the \(y\)-intercept \((0,\ 4)\) to the minimum \((2,\ -16)\), then rises again to the top-right, as shown below.

Akọwa Nkọwa

Choosing 3 prefects from 4 girls and 5 boys (9 people). Total selections: \[ \binom{9}{3} = 84. \]

(a) Only boys chosen: \[ \binom{5}{3} = 10, \qquad P = \frac{10}{84} = \frac{5}{42}. \]

(b) More girls than boys means either 3 girls (0 boys) or 2 girls and 1 boy:

\( 3G: \binom{4}{3} = 4 \); \( 2G,1B: \binom{4}{2}\binom{5}{1} = 6 \times 5 = 30 \).

\[ \text{Favourable} = 4 + 30 = 34, \qquad P = \frac{34}{84} = \frac{17}{42}. \]

Akọwa Nkọwa

(a) Bag: 5 blue, 4 green, 3 yellow (12 balls), draw 3 without replacement. Total selections \(={}^{12}C_3=220\).

(i) All three the same colour.\[{}^{5}C_3+{}^{4}C_3+{}^{3}C_3=10+4+1=15\]\[P=\frac{15}{220}=\frac{3}{44}\approx 0.068\]

(ii) Exactly two of the same colour (two alike and one different):\[{}^{5}C_2(7)+{}^{4}C_2(8)+{}^{3}C_2(9)=10(7)+6(8)+3(9)=70+48+27=145\]\[P=\frac{145}{220}=\frac{29}{44}\approx 0.659\](Check: all-different \(=5\times4\times3=60\); \(15+145+60=220\).)

(b) Spearman's rank correlation (Physics vs Chemistry ranks). \(n=7\).

\[r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{6(16)}{7(48)}=1-\frac{96}{336}=1-0.286=0.714\]

Answers: (a)(i) \(\tfrac{3}{44}\); (a)(ii) \(\tfrac{29}{44}\); (b) \(r_s\approx 0.714\).

Akọwa Nkọwa

(a) First compute the direction vector \( \mathbf{a} + \mathbf{b} - 2\mathbf{c} \):

\[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} - 2\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 14 - 4 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \end{pmatrix}. \]

Its magnitude is \( \sqrt{10^2 + 4^2} = \sqrt{116} = 2\sqrt{29} \). The required vector \( \mathbf{d} \) has magnitude \( \sqrt{41} \) in this direction:

\[ \mathbf{d} = \sqrt{41}\cdot\frac{1}{\sqrt{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} = \sqrt{\frac{41}{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} \approx \begin{pmatrix} 5.94 \\ 2.38 \end{pmatrix}. \]

So \( \mathbf{d} \approx 5.94\mathbf{i} + 2.38\mathbf{j} \) (check: \( 5.94^2 + 2.38^2 \approx 41 \)).

(b) With \( O \) the origin, \( \vec{OA} = (3,4) \) so \( |OA| = 5 \), and \( \vec{OB} = (3,n) \) with \( |OB| = \sqrt{9+n^2} \). The angle \( AOB = 30^\circ \):

\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]

\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]

\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]

\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]

\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]