WAEC SSCE - Further Mathematics - 2010

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To find the radians per second that the bicycle wheel turns, we need to find the angle in radians that the wheel turns in 1 second. The distance covered by the bicycle wheel in 1 second can be found by dividing the distance covered in 2 seconds by 2: \[\frac{350 \, \text{cm}}{2 \, \text{s}} = 175 \, \text{cm/s}\] To convert this to radians per second, we need to know the circumference of the wheel in centimeters. The circumference of the wheel is given by: \[\pi \times \text{diameter} = \pi \times 70 \, \text{cm} = 220 \pi \, \text{cm}\] So, in one revolution, the wheel covers a distance of 220π cm. The angle in radians that the wheel turns in 1 second can be found by dividing the distance covered in 1 second by the circumference of the wheel: \[\frac{175 \, \text{cm/s}}{220 \pi \, \text{cm/rev}} = \frac{5}{2\pi} \, \text{rev/s}\] To convert revolutions per second to radians per second, we multiply by 2π: \[\frac{5}{2\pi} \times 2\pi = 5 \, \text{rad/s}\] Therefore, the bicycle wheel turns at a rate of 5 radians per second. So the answer is (A) 5.

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Given, distance of a particle from a fixed point at time t seconds is given by, $$s = 7 + pt^{3} + t^{2}$$ We can find the acceleration of the particle by differentiating the distance equation twice with respect to time. $$\frac{d}{dt} s = \frac{d}{dt} (7 + pt^{3} + t^{2})$$ $$\Rightarrow \frac{ds}{dt} = 3pt^2 + 2t$$ $$\frac{d^{2}}{dt^{2}} s = \frac{d}{dt} (3pt^2 + 2t)$$ $$\Rightarrow \frac{d^{2}s}{dt^{2}} = 6pt + 2$$ We are given that the acceleration at t = 3 sec is \(8 ms^{-2}\), so we can substitute the values in the second derivative of the distance equation to get, $$6p(3) + 2 = 8$$ $$\Rightarrow 18p = 6$$ $$\Rightarrow p = \frac{6}{18} = \frac{1}{3}$$ Hence, the value of p is \(\frac{1}{3}\). Therefore, is correct.

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We can use the formula: P(A and B) = P(A) + P(B) - P(A or B) where A is the event of liking Science, B is the event of liking History, P(A and B) is the probability of selecting a pupil who likes both Science and History, and P(A or B) is the probability of selecting a pupil who likes either Science or History or both. We know that there are 50 pupils in the class, 35 like Science and 30 like History. This means that the maximum number of pupils who could like both Science and History is 30 (since 30 already like History). Therefore, we have: P(A and B) <= 30/50 = 0.6 Also, we know that P(A or B) = P(A) + P(B) - P(A and B) (as shown above). Therefore: P(A and B) = P(A) + P(B) - P(A or B) = 35/50 + 30/50 - P(A or B) = 65/50 - P(A or B) We also know that P(A or B) <= 1, since it is the probability of selecting a pupil who likes either Science or History or both. Therefore: P(A and B) >= 65/50 - 1 = 15/50 = 0.3 Thus, the probability of selecting a pupil who likes both Science and History is between 0.3 and 0.6. The only option that falls within this range is 0.30, so the answer is: - 0.30

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We can start by computing the values of \(h(a)\) and \(h(\frac{1}{a})\), and then find their difference. First, we have \(h(a) = a^{3} - \frac{1}{a^{3}}\). Next, we have \(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\). Therefore, \begin{align*} h(a) - h(\frac{1}{a}) &= \left(a^{3} - \frac{1}{a^{3}}\right) - \left(\frac{1}{a^{3}} - a^{3}\right) \\ &= a^{3} - \frac{1}{a^{3}} - \frac{1}{a^{3}} + a^{3} \\ &= 2a^{3} - \frac{2}{a^{3}}. \end{align*} So the answer is \(2a^{3} - \frac{2}{a^{3}}\). Therefore, is the correct answer.

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The final velocity of an object can be found using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we have: v = (-5, 3) + 3(3, -4) v = (-5, 3) + (9, -12) v = (4, -9) Therefore, the final velocity of the object is \(\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\). The correct option is (C).

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Direction cosines are the cosines of the angles that a vector makes with the positive x, y and z axes. Let's denote the vector by \(\mathbf{v} = 4\mathbf{i} - 3\mathbf{j} + 0\mathbf{k}\). Then the direction cosines of \(\mathbf{v}\) are: \begin{align*} \cos\alpha &= \frac{4}{5} \\ \cos\beta &= \frac{-3}{5} \\ \cos\gamma &= 0 \end{align*} Therefore, the correct option is: - \(\frac{4}{5}, -\frac{3}{5}\)

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There are 4 seats to be labeled, so Yomi can label the first seat with any of the 4 letters. After labeling the first seat, there are 3 letters left to choose from for the second seat, 2 letters for the third seat, and only 1 letter left for the last seat. Thus, there are a total of 4 x 3 x 2 x 1 = 24 possible ways to label the seats. Out of these 24 possible ways, only one arrangement is in alphabetical order (i.e., S, P, Q, R). Therefore, the probability that Yomi labeled the seats in alphabetical order is 1/24. Therefore, the correct option is (a) \(\frac{1}{24}\).

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We can find the midpoint M of AB by using the midpoint formula: \[\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}\] Substituting the values we have: \[\overrightarrow{OM} = \frac{(3i + 4j) + (5i - 6j)}{2} = \frac{8i - 2j}{2} = 4i - j\] Therefore, the answer is 4i - j.

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The function \(f(x) = \frac{x}{3-x}\) is defined for all real numbers except when the denominator is zero, i.e., when \(3-x=0\). Therefore, the domain of the function is all real numbers except \(x=3\), or in interval notation: \({x : x \in R, x \neq 3}\). So the correct option is: \({x : x \in R, x \neq 3}\).

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Given that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j, and PS = xi + yj. Since PQ + QR + RS = PS, we have: (8i - 5j) + (5i + 7j) + (7i + 3j) = xi + yj Simplifying the left-hand side, we get: 20i + 5j = xi + yj Equating the corresponding components, we have: 20 = x and 5 = y Therefore, the coordinates of point S are (20, 5). Hence, the correct answer is (20, 5).

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To find the axis of symmetry of a quadratic equation in the form of \(y = ax^2 + bx + c\), we need to use the formula: $$x = \frac{-b}{2a}$$ Comparing the given equation \(y = x^2 - 4x - 12\) with the standard form of the quadratic equation, we have: $$a = 1, b = -4, c = -12$$ Substituting the values in the formula, we get: $$x = \frac{-(-4)}{2(1)} = 2$$ Therefore, the axis of symmetry of the curve is the vertical line passing through the point \((2,0)\) and the equation of this line is: $$x = 2$$ Hence, the correct answer is "x = 2".

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To find the magnitude and direction of \(F_{1} - F_{2}\), we need to subtract the components of \(F_{2}\) from \(F_{1}\). \begin{align*} F_{1} - F_{2} &= (7i + 8j) - (3i + 4j)\\ &= 4i + 4j\\ \end{align*} The magnitude of \(F_{1} - F_{2}\) is given by: \begin{align*} |F_{1} - F_{2}| &= \sqrt{(4)^2 + (4)^2}\\ &= 4\sqrt{2} N\\ \end{align*} The direction of \(F_{1} - F_{2}\) is given by: \begin{align*} \theta &= \tan^{-1}\left(\frac{4j}{4i}\right)\\ &= \tan^{-1}(1)\\ &= 45°\\ \end{align*} Therefore, the magnitude and direction of \(F_{1} - F_{2}\) are \((4\sqrt{2} N, 045°)\). Answer is correct.

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The probability that only one of the husband and wife will be alive in 15 years can happen in two ways: either the husband is alive and the wife is not, or the wife is alive and the husband is not. The probability of the first case is (m)(1-n) = m - mn, and the probability of the second case is (n)(1-m) = n - mn. Therefore, the total probability that only one of them will be alive is (m - mn) + (n - mn) = m + n - 2mn. Hence, the correct answer is: m + n - 2mn.

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The height of the stone at any time t seconds is given by \(h = 45t - 9t^{2}\). To find the maximum height reached, we need to find the vertex of the parabolic function \(h\), since the vertex represents the highest point of the parabola. The vertex of the parabola is given by the formula \(\frac{-b}{2a}\), where \(a\) and \(b\) are the coefficients of the quadratic function. In this case, \(a = -9\) and \(b = 45\). Substituting these values into the formula for the vertex, we get: \[\frac{-b}{2a} = \frac{-45}{2(-9)} = \frac{45}{18} = 2.5.\] Therefore, the stone reaches its maximum height after 2.5 seconds. To find the maximum height, we substitute \(t = 2.5\) into the equation for \(h\): \[h = 45(2.5) - 9(2.5)^2 = 56.25\text{ m}.\] Therefore, the maximum height reached by the stone is 56.25 meters. The answer is (D) 56.25 m.

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To differentiate the given function with respect to x, we will use the chain rule and the power rule of differentiation. Let's start by using the power rule of differentiation to find the derivative of \((2x + \sqrt{x})^{2}\) with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\] Next, we use the chain rule to differentiate \(\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\) with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x}) = 2 + \frac{1}{2\sqrt{x}}\] Substituting this into the previous equation, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \left(2 + \frac{1}{2\sqrt{x}}\right)\] Expanding this expression, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 4(2x + \sqrt{x}) + 2\sqrt{x}(2x + \sqrt{x})\] Simplifying further, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 8x + 6\sqrt{x}\] Finally, using the constant multiple rule of differentiation, we can find the derivative of the given function: \[\frac{\mathrm d y}{\mathrm d x} = 2 \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(8x + 6\sqrt{x}) = 16x + 12\sqrt{x}\] Therefore, the correct answer is: \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\).

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We can use the formula for the nth term of a geometric progression to solve this problem. Let a be the first term and r be the common ratio, then the third term is ar^2 and the sixth term is ar^5. We are given that ar^2 = 10 and ar^5 = 80. Dividing the second equation by the first, we get: (ar^5)/(ar^2) = 80/10 Simplifying and canceling a, we get: r^3 = 8 Taking the cube root of both sides, we get: r = 2 Therefore, the common ratio is 2, and the answer is (a) 2.

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We can use the properties of logarithms to simplify this expression. Recall that \(\log_{a} b^{c} = c\log_{a} b\) and \(\log_{a} \sqrt{b} = \frac{1}{2}\log_{a} b\). So, \[\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8} = \frac{2\log_{5} 8}{\log_{5} 8} = \boxed{2}.\] Therefore, the answer is 2.

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The value of \(\sin(\theta + 25)\) varies between -1 and 1. Therefore, the maximum value of \(2 + \sin(\theta + 25)\) occurs when \(\sin(\theta + 25) = 1\). This occurs when \(\theta + 25 = \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer. Thus, \(\theta = \frac{\pi}{2} - 25 + 2n\pi\). Substituting this value of \(\theta\) into the expression for \(2 + \sin(\theta + 25)\), we get: \[2 + \sin\left(\frac{\pi}{2} + 2n\pi\right) = 2 + 1 = 3\] Therefore, the maximum value of \(2 + \sin(\theta + 25)\) is 3.

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The roots of a quadratic equation of the form \(ax^{2} + bx + c = 0\) are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\), then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\) is \(x^{2} - 6x + 6 = 0\). Therefore, the answer is.

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To find the magnitude of acceleration, we need to use Newton's Second Law, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, we have two forces: (2i - 5j)N and (-3i + 4j)N. To find the net force, we add these two vectors by adding their corresponding components: (2i - 5j)N + (-3i + 4j)N = -i - j N So the net force is (-i - j)N. Now, using Newton's Second Law, we can find the acceleration: F = ma where F is the net force, m is the mass, and a is the acceleration. In our case, the net force is (-i - j)N and the mass is 5kg, so we have: (-i - j)N = 5kg * a Solving for a, we get: a = (-i - j)N / 5kg To find the magnitude of a, we take the square root of the sum of the squares of its components: |a| = sqrt((-1)^2 + (-1)^2) N/kg = sqrt(2) N/kg Converting to \(ms^{-2}\) by dividing by 5, we get: |a| = sqrt(2)/5 \(ms^{-2}\) Therefore, the answer is \(\frac{\sqrt{2}}{5}\).

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If (x-3) is a factor of the quadratic expression, then the expression can be factored into the form: $$(x - 3)(ax + b)$$ where a and b are constants. Multiplying out the brackets gives: $$2x^{2} - 2x + p = (x - 3)(ax + b) = ax^{2} + (b - 3a)x - 3b$$ Since the coefficients of the quadratic expression are equal to those of the factored expression, we can equate the corresponding coefficients to get a system of equations: $$a = 2$$ $$b - 3a = -2$$ $$-3b = p$$ Solving these equations simultaneously, we get: $$a = 2, b = 3a - 2 = 4, p = -3b = -12$$ Therefore, the constant p is -12.

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Using the sum to product formula, we have: \begin{align*} \cos(60° + 45°) &= \cos 60° \cos 45° - \sin 60° \sin 45° \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} - \sqrt{6}}{4} \end{align*} Therefore, the correct answer is option C: \(\frac{\sqrt{2} - \sqrt{6}}{4}\).

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The expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we can find the values of x that make the denominator zero by setting it equal to zero and solving for x: $$x^{2} + 2x - 35 = 0$$ We can factor this quadratic equation as: $$(x + 7)(x - 5) = 0$$ So, the denominator is equal to zero when x = -7 or x = 5. Therefore, the expression is undefined for these values of x. So the correct answer is: - -7 or 5

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Let's use the formula for the mean (average) of a set of numbers: \[\text{mean} = \frac{\text{sum of all numbers}}{\text{number of numbers}}\] Let the age of the new pupil be x. Before the new pupil joined the class, the sum of ages of the 15 pupils would be: \[15 \times 14.2 = 213\] After the new pupil joined the class, the sum of ages of the 16 pupils would be: \[15 \times 14.2 + x = 213 + x\] We know that the new mean is 14.1, so we can set up an equation: \[\frac{15 \times 14.2 + x}{16} = 14.1\] Multiplying both sides by 16, we get: \[15 \times 14.2 + x = 16 \times 14.1\] Simplifying the left side, we get: \[213 + x = 225.6\] Subtracting 213 from both sides, we get: \[x = 12.6\] Therefore, the age of the new pupil is 12.6 years. Answer: (b) 12.6 years.

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(a) Histogram

Convert the class limits to continuous class boundaries. Since every class has width 5, the bar heights are the frequencies.

Marks	Class boundaries	Frequency	Cumulative frequency
40–44	39.5–44.5	4	4
45–49	44.5–49.5	9	13
50–54	49.5–54.5	18	31
55–59	54.5–59.5	23	54
60–64	59.5–64.5	10	64
65–69	64.5–69.5	6	70

The required histogram is:

(b) Median

The total frequency is

\[N=70,\qquad \frac{N}{2}=35.\]

The 35th observation lies in the class \(55\text{–}59\), since the cumulative frequency before this class is 31 and the cumulative frequency at the end is 54.

Using \(L=54.5\), \(c_f=31\), \(f=23\) and class width \(c=5\),

\[\begin{aligned}\text{Median} &=L+\frac{\left(\frac{N}{2}-c_f\right)}{f}\times c\\&=54.5+\frac{35-31}{23}\times5\\&=54.5+0.87\\&\approx55.4.\end{aligned}\]

Therefore, the estimated median mark is \(55.4\).

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The plank \(PQ\) is \(8\ \text{m}\) long, mass \(10\ \text{kg}\), supported at \(P\) (left end) and at \(R\), which is \(3\ \text{m}\) from \(Q\), i.e. \(5\ \text{m}\) from \(P\). The plank weight \(10g\) acts at the midpoint, \(4\ \text{m}\) from \(P\). A boy of weight \(20g\) stands at distance \(d\) from \(P\). (Take \(g = 10\ \text{m s}^{-2}\), so weights are \(100\ \text{N}\) and \(200\ \text{N}\).)

(a) When \(d = 1.5\ \text{m}\). Taking moments about \(P\) (so \(R_P\) has no moment):

\[R_R \times 5 = 10g(4) + 20g(1.5) = 40g + 30g = 70g \Rightarrow R_R = 14g = 140\ \text{N}.\]

Vertical equilibrium: \(R_P + R_R = 30g = 300\ \text{N} \Rightarrow R_P = 160\ \text{N}\).

So \(R_P = 160\ \text{N},\ R_R = 140\ \text{N}\).

(b) When the reactions are equal. Each reaction \(= \tfrac12(30g) = 15g = 150\ \text{N}\). Moments about \(P\):

\[15g \times 5 = 40g + 20g\,d \Rightarrow 75g = 40g + 20g\,d \Rightarrow d = \frac{35}{20} = 1.75\ \text{m}.\]

(c) When the plank tips. The plank tips about \(R\) when \(R_P = 0\). Taking moments about \(R\), the boy (on the \(Q\) side, \(d - 5\) from \(R\)) balances the plank weight (\(5 - 4 = 1\ \text{m}\) on the \(P\) side):

\[20g(d - 5) = 10g(1) \Rightarrow d - 5 = 0.5 \Rightarrow d = 5.5\ \text{m}.\]

The boy can walk \(5.5\ \text{m}\) from \(P\) before the plank tips over.

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Integrate the acceleration to get velocity, then position, applying the initial conditions.

\[v = \int (4 - 2t)\,dt = 4t - t^2 + C.\]

At \(t = 0,\ v = 5\), so \(C = 5\) and \(v = 4t - t^2 + 5\).

(a) Velocity is greatest when \(\dfrac{dv}{dt} = a = 0\):

\[4 - 2t = 0 \Rightarrow t = 2\ \text{s}.\]

(b) The particle is momentarily at rest when \(v = 0\):

\[4t - t^2 + 5 = 0 \Rightarrow t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0 \Rightarrow t = 5\ \text{s}\ (t>0).\]

(i) Time \(= 5\ \text{s}\).

Now integrate for displacement: \(s = \int(4t - t^2 + 5)\,dt = 2t^2 - \dfrac{t^3}{3} + 5t + D\). At \(t = 0,\ s = 18\), so \(D = 18\).

(ii) At \(t = 5\):

\[s = 2(25) - \frac{125}{3} + 25 + 18 = 93 - \frac{125}{3} = \frac{154}{3} \approx 51.3\ \text{m}.\]

The particle is \(\dfrac{154}{3}\ \text{m} \approx 51.3\ \text{m}\) from \(O\).

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(a) Let \(T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Using the images \((3,2)\to(-1,4)\) and \((-1,4)\to(7,11)\):

From the \(x\)-outputs: \(3a + 2b = -1\) and \(-a + 4b = 7\). Solving gives \(b = \tfrac{10}{7},\ a = -\tfrac{9}{7}\).

From the \(y\)-outputs: \(3c + 2d = 4\) and \(-c + 4d = 11\). Solving gives \(d = \tfrac{37}{14},\ c = -\tfrac{3}{7}\).

\[T = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}.\]

For \(P:(x,y)\to(x+y,\,x+2y)\), \(P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\).

(b) \(\displaystyle TP = \frac{1}{14}\begin{pmatrix} -18 & 20 \\ -6 & 37 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}.\)

(c) Image of \(X(4,3)\) under \(TP\):

\[\frac{1}{14}\begin{pmatrix} 2 & 22 \\ 31 & 68 \end{pmatrix}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac{1}{14}\begin{pmatrix} 74 \\ 328 \end{pmatrix} = \left(\frac{37}{7},\ \frac{164}{7}\right).\]

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None

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(a) The recurrence is \(T_n = 4T_{n-1} - 3\). Express the terms through \(T_1\):

\[T_2 = 4T_1 - 3,\qquad T_3 = 4T_2 - 3 = 4(4T_1 - 3) - 3 = 16T_1 - 15.\]

Given \(2T_3 = 5T_2\):

\[2(16T_1 - 15) = 5(4T_1 - 3) \Rightarrow 32T_1 - 30 = 20T_1 - 15 \Rightarrow 12T_1 = 15 \Rightarrow T_1 = \tfrac{5}{4}.\]

Then \(T_2 = 4(\tfrac54) - 3 = 2\) and \(T_3 = 4(2) - 3 = 5\).

The first three terms are \(\tfrac{5}{4},\ 2,\ 5\).

(b) Multiplying the matrices row by row:

Row 1: \(2(1) + 0\cdot m + 1\cdot r = 2 + r = k\).

Row 2: \(5(1) - 3m + r = 2 \Rightarrow -3m + r = -3\).

Row 3: \(0 + 4m + 6r = 26\).

From Row 2, \(r = 3m - 3\). Substitute into Row 3:

\[4m + 6(3m - 3) = 26 \Rightarrow 22m - 18 = 26 \Rightarrow m = 2,\]

so \(r = 3(2) - 3 = 3\) and \(k = 2 + r = 5\).

Answer: \(k = 5,\; m = 2,\; r = 3\).

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(a) With \(10\%\) faulty, \(P(\text{a bulb is good}) = 0.9\). For a sample of 5, \(P(\text{no faulty}) = (0.9)^5 = 0.59049\).

The batch is accepted if the first sample has no faulty, OR the first sample has at least one faulty but the second sample has none:

\[P(\text{accept}) = (0.9)^5 + \big[1 - (0.9)^5\big](0.9)^5 = 0.59049 + (0.40951)(0.59049).\]

\[= 0.59049 + 0.241813 = 0.832303 \approx 0.832.\]

(b) Bag of 15 marbles, 3 black, so \(P(\text{black}) = \tfrac{3}{15} = 0.2\), \(P(\text{not black}) = 0.8\), with replacement, \(n = 10\).

(i) Did not pick a black at all: \((0.8)^{10} = 0.1073742 \approx 0.107\).

(ii) Picked black at most three times \(= P(0)+P(1)+P(2)+P(3)\):

\[P(0)=0.107374,\ P(1)=\binom{10}{1}(0.2)(0.8)^9=0.268435,\]

\[P(2)=\binom{10}{2}(0.2)^2(0.8)^8=0.301990,\ P(3)=\binom{10}{3}(0.2)^3(0.8)^7=0.201327.\]

Sum \(= 0.879\) (to three decimal places).

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(a) Simplify the surds: \(\sqrt{48} = 4\sqrt{3},\ \sqrt{8} = 2\sqrt{2},\ \sqrt{27} = 3\sqrt{3}\).

So the expression is \(\dfrac{2\sqrt{2}}{\sqrt{3} - 2\sqrt{2}}\). Rationalise using \((\sqrt{3} + 2\sqrt{2})\):

This is of the form \(p + q\sqrt{r}\) with \(p = -\tfrac{8}{5},\ q = -\tfrac{2}{5},\ r = 6\).

(b) From \(V = A\log_{10}(M + N)\):

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(a) Expand the equation \((2x - 1) - p(x^2 + 2) = 0\) into standard quadratic form:

\[-px^2 + 2x - 1 - 2p = 0 \;\Rightarrow\; px^2 - 2x + (1 + 2p) = 0.\]

Here \(a = p,\ b = -2,\ c = 1 + 2p\). For real roots the discriminant must be non-negative:

\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-2)^2 - 4p(1 + 2p) \ge 0 \;\Rightarrow\; 4 - 4p - 8p^2 \ge 0.\]

Divide through by \(4\): \(1 - p - 2p^2 \ge 0\). Multiplying by \(-1\) (reversing the inequality):

\[2p^2 + p - 1 \le 0.\]

(For two distinct real roots the inequality is strict, \(2p^2 + p - 1 < 0\), as stated.)

(b) Solve \(2p^2 + p - 1 = 0\): factorising, \((2p - 1)(p + 1) = 0\), so \(p = \tfrac12\) or \(p = -1\).

Since the coefficient of \(p^2\) is positive, \(2p^2 + p - 1 \le 0\) between the roots:

\[-1 \le p \le \tfrac{1}{2}.\]

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(a) With \(\mathbf{x} = -2\mathbf{i} + 5\mathbf{j}\) and \(\mathbf{y} = \mathbf{i} - 7\mathbf{j}\):

(i) \(3\mathbf{x} + 2\mathbf{y} = (-6\mathbf{i} + 15\mathbf{j}) + (2\mathbf{i} - 14\mathbf{j}) = -4\mathbf{i} + \mathbf{j}\).

(ii) \(\mathbf{y} - 2\mathbf{x} = (\mathbf{i} - 7\mathbf{j}) - (-4\mathbf{i} + 10\mathbf{j}) = 5\mathbf{i} - 17\mathbf{j}\), so \(|\mathbf{y} - 2\mathbf{x}| = \sqrt{25 + 289} = \sqrt{314} \approx 17.7\).

(iii) \(\cos\theta = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}||\mathbf{y}|}\), where \(\mathbf{x}\cdot\mathbf{y} = (-2)(1) + (5)(-7) = -37\), \(|\mathbf{x}| = \sqrt{29}\), \(|\mathbf{y}| = \sqrt{50}\):

\[\cos\theta = \frac{-37}{\sqrt{29}\,\sqrt{50}} = \frac{-37}{38.08} = -0.9717 \Rightarrow \theta \approx 166.3^\circ.\]

(iv) \(\mathbf{x} + \mathbf{y} = -\mathbf{i} - 2\mathbf{j}\), \(|\mathbf{x} + \mathbf{y}| = \sqrt{5}\). Unit vector \(= \dfrac{-\mathbf{i} - 2\mathbf{j}}{\sqrt{5}} \approx -0.45\mathbf{i} - 0.89\mathbf{j}\).

(b) Conservation of momentum (block initially at rest). Let \(u\) be the bullet's initial speed:

\[0.084\,u = (0.084 + 20)(0.24) \Rightarrow u = \frac{20.084 \times 0.24}{0.084} = \frac{4.82016}{0.084} \approx 57.38\ \text{m s}^{-1}.\]

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Selecting \(5\) students, \(P(\text{boy}) = 0.6\) and \(P(\text{girl}) = 0.4\). Use the binomial distribution.

(a) Exactly 3 boys (so 2 girls):

\[\binom{5}{3}(0.6)^3(0.4)^2 = 10(0.216)(0.16) = 0.3456 \approx 0.346.\]

(b) At least 3 girls means \(3\), \(4\) or \(5\) girls, with \(P(\text{girl}) = 0.4\):

\[P(3) = \binom{5}{3}(0.4)^3(0.6)^2 = 10(0.064)(0.36) = 0.2304,\]

\[P(4) = \binom{5}{4}(0.4)^4(0.6) = 5(0.0256)(0.6) = 0.0768,\]

\[P(5) = (0.4)^5 = 0.01024.\]

Sum: \(0.2304 + 0.0768 + 0.01024 = 0.31744 \approx 0.317\).

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(a) Eight coins, \(n = 8\), \(p = 0.5\) for a head, so each outcome count is \(\binom{8}{r}/256\).

(i) Exactly 8 heads: \(\left(\tfrac12\right)^8 = \dfrac{1}{256} \approx 0.004\).

(ii) At least 5 heads \(= \dfrac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{256} = \dfrac{56+28+8+1}{256} = \dfrac{93}{256} \approx 0.363\).

(iii) At most 1 head \(= \dfrac{\binom{8}{0}+\binom{8}{1}}{256} = \dfrac{1+8}{256} = \dfrac{9}{256} \approx 0.035\).

(b) The word SHEEP has letters S, H, E, E, P (the E repeats). Choose and arrange 4 letters.

(i) Without restriction, split by how many E's are used:

• Both E's plus 2 of \(\{S,H,P\}\): \(\binom{3}{2}\times\dfrac{4!}{2!} = 3\times12 = 36\).
• At most one E, i.e. the four distinct letters \(S,H,E,P\): \(4! = 24\).

Total \(= 36 + 24 = 60\) arrangements.

(ii) With only one E, the other three must be \(S, H, P\), giving four distinct letters: \(4! = 24\) arrangements.

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Differentiate \(3x^2 + 2y^2 + xy + x - 7 = 0\) implicitly with respect to \(x\). Use the product rule on \(xy\):

\[6x + 4y\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) + 1 = 0.\]

Collect the \(\dfrac{dy}{dx}\) terms:

\[\frac{dy}{dx}(4y + x) = -(6x + y + 1) \;\Rightarrow\; \frac{dy}{dx} = -\frac{6x + y + 1}{4y + x}.\]

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

\[\frac{dy}{dx} = -\frac{6(-2) + 1 + 1}{4(1) + (-2)} = -\frac{-12 + 2}{4 - 2} = -\frac{-10}{2} = 5.\]

Answer: \(\dfrac{dy}{dx} = 5\) at \((-2, 1)\).

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The force is found from Newton's second law \(\mathbf{F} = m\mathbf{a}\), where \(\mathbf{a}\) is the second derivative of the position vector.

Given \(\mathbf{r} = (4t^3 - t^2)\mathbf{i} - (2t^2 - t)\mathbf{j}\), differentiate once for velocity:

\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = (12t^2 - 2t)\mathbf{i} - (4t - 1)\mathbf{j}.\]

Differentiate again for acceleration:

\[\mathbf{a} = \frac{d\mathbf{v}}{dt} = (24t - 2)\mathbf{i} - 4\mathbf{j}.\]

At \(t = 2\): \(\mathbf{a} = (24(2) - 2)\mathbf{i} - 4\mathbf{j} = 46\mathbf{i} - 4\mathbf{j}\).

With mass \(m = 3\ \text{kg}\):

\[\mathbf{F} = 3(46\mathbf{i} - 4\mathbf{j}) = 138\mathbf{i} - 12\mathbf{j}\ \text{N}.\]

Its magnitude is \(|\mathbf{F}| = \sqrt{138^2 + 12^2} = \sqrt{19188} \approx 138.5\ \text{N}\).

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(a) Cumulative frequency table

Marks	Class boundaries	Frequency	Cumulative frequency
23–25	22.5–25.5	3	3
26–28	25.5–28.5	7	10
29–31	28.5–31.5	15	25
32–34	31.5–34.5	21	46
35–37	34.5–37.5	10	56
38–40	37.5–40.5	4	60

Plot the upper class boundaries against the cumulative frequencies, including the starting point (22.5,0). The resulting less-than ogive is:

(b) Estimates from the ogive

(i) 80th percentile

The 80th percentile is the 48th value:

\[P_{80}=34.5+\frac{48-46}{10}\times 3=35.1\]

Hence, \(\boxed{P_{80}\approx35.1\text{ marks}}\).

(ii) Median

The median is the 30th value:

\[\text{Median}=31.5+\frac{30-25}{21}\times 3=32.2\]

Therefore, \(\boxed{\text{Median}\approx32.2\text{ marks}}\).

(iii) Semi-interquartile range

\[Q_1=28.5+\frac{15-10}{15}\times3=29.5\]

\[Q_3=31.5+\frac{45-25}{21}\times3=34.4\]

\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{34.4-29.5}{2}=2.45\]

Thus, \(\boxed{\text{semi-interquartile range}\approx2.4\text{ marks}}\).

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(a) For each value of \(x\), calculate \(y=2\cos x+3\sin x\), correct to two decimal places.

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)	\(210^\circ\)
\(y\)	2.00	3.23	3.60	3.00	1.60	-0.23	-2.00	-3.23

For example,

\[y(60^\circ)=2\cos60^\circ+3\sin60^\circ=2(0.5)+3(0.8660)=3.60.\]

(b) The required graph is shown below. The plotted curve is smooth, with the horizontal scale \(2\text{ cm}\) to \(60^\circ\) and vertical scale \(2\text{ cm}\) to 1 unit.

(c)(i) The maximum ordinate of the curve is

\[\boxed{3.61}\]

correct to two decimal places.

(c)(ii) Since

\[\frac{2}{3}\cos x+\sin x=\frac56,\]

multiplying by 3 gives

\[2\cos x+3\sin x=2.50.\]

Thus, draw the line \(y=2.50\) on the graph and read the abscissae of its intersections with the curve:

\[\boxed{x\approx10^\circ\text{ or }102^\circ}.\]

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Expand the curve: \(y = x(3 - x^2) = 3x - x^3\).

Differentiate to get the gradient of the tangent:

\[\frac{dy}{dx} = 3 - 3x^2.\]

At \(x = 2\): gradient of tangent \(= 3 - 3(4) = -9\).

The \(y\)-coordinate there is \(y = 2(3 - 4) = -2\), giving the point \((2, -2)\).

The normal is perpendicular to the tangent, so its gradient is

\[m_{\text{normal}} = -\frac{1}{-9} = \frac{1}{9}.\]

Equation of the normal through \((2, -2)\):

\[y - (-2) = \frac{1}{9}(x - 2) \;\Rightarrow\; 9y + 18 = x - 2 \;\Rightarrow\; x - 9y - 20 = 0.\]