WAEC SSCE - Further Mathematics - 2011

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To convert from degrees to radians, we use the conversion formula: radians = (pi/180) x degrees So to convert 75 degrees to radians, we have: radians = (pi/180) x 75 radians = (5pi/12) Therefore, the answer is (A) \(\frac{5\pi}{12}\).

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To evaluate \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\), we can use the formula for the cosine of the sum of two angles, which states that \[\cos(a+b) = \cos a \cos b - \sin a \sin b.\] Using this formula, we can simplify \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\) as follows: \[\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}\] Recall that \(\cos \frac{\pi}{2} = 0\) and \(\sin \frac{\pi}{2} = 1\), and we also know that \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\) from the unit circle. Substituting these values into the equation above, we get: \[\cos (\frac{\pi}{2} + \frac{\pi}{3}) = 0 \cdot \frac{1}{2} - 1 \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}\] Therefore, the value of \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\) is \(\frac{-\sqrt{3}}{2}\)

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To determine which of the statements is true, we can use the properties of vectors. We start by finding the magnitude of vector r using the formula: |magnitude of r| = sqrt(x^2 + y^2) where x and y are the coefficients of the i and j terms respectively. |magnitude of r| = sqrt(4^2 + (-6)^2) = sqrt(52) So the statement "The magnitude of r is 52 units" is true. Next, we can find the dot product of vectors q and r. If the dot product is equal to zero, then the vectors are perpendicular. If the dot product is non-zero, then the vectors are not perpendicular. q . r = (9 * 4) + (6 * -6) = 36 - 36 = 0 Since the dot product of q and r is zero, the statement "r and q are perpendicular" is true. Therefore, the correct answer is: - r and q are perpendicular.

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The formula for finding the number of combinations of r items out of n distinct items is given by the formula: $$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$ Where n! is the factorial of n, that is the product of all positive integers from 1 to n. In this question, we are given that $$^{3x}C_{2} = 15$$ Substituting the given values into the formula above, we have: $$^{3x}C_{2} = \frac{(3x)!}{2!(3x-2)!} = 15$$ Simplifying this equation, we have: $$\frac{(3x)(3x-1)(3x-2)!}{2!} = 15$$ Multiplying both sides by 2, we get: $$(3x)(3x-1)(3x-2)! = 30$$ We can observe that 3x-2! is the factorial of (3x-2) which is an integer, and 3x-1 and 3x are consecutive integers. Therefore, we can re-write the equation above as: $$(3x)(3x-1) = 10$$ Expanding the left-hand side, we have: $$9x^2 - 3x = 10$$ Bringing all the terms to one side, we get: $$9x^2 - 3x - 10 = 0$$ We can then factorize this quadratic equation as follows: $$(3x - 5)(3x + 2) = 0$$ Using the zero-product property, we get: $$3x - 5 = 0 \quad \text{or} \quad 3x + 2 = 0$$ Solving for x in each equation, we get: $$x = \frac{5}{3} \quad \text{or} \quad x = -\frac{2}{3}$$ Since x represents the number of items in the set from which we are selecting combinations, it must be a positive integer. Therefore, the only valid solution is: $$x = \frac{5}{3} = 1.67 \approx 2$$ Hence, the value of x is 2.

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To solve this problem, we can use the kinematic equation: s = ut + (1/2)at^2 where s is the distance travelled, u is the initial velocity (0 in this case), a is the acceleration, and t is the time taken. We can rearrange this equation to solve for t: t = sqrt((2s)/a) where sqrt() means "square root of". In this case, the force of 200N causes an acceleration of: a = F/m = 200/20 = 10 m/s^2 And the distance travelled is given as 320m. Plugging in the values, we get: t = sqrt((2s)/a) = sqrt((2 * 320)/10) = sqrt(64) = 8s Therefore, the value of t is 8s, which is.

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Let the quadratic equation be \(ax^2 + bx + c = 0\). According to the problem, we have: Sum of the roots: \(-\frac{b}{a} = \frac{4}{7}\) Product of the roots: \(\frac{c}{a} = \frac{5}{7}\) Using the formula for the sum and product of roots, we get: \(-\frac{b}{a} = \frac{4}{7} \implies b = -\frac{4}{7}a\) \(\frac{c}{a} = \frac{5}{7} \implies c = \frac{5}{7}a\) Substituting these values in the quadratic equation, we get: \(ax^2 - \frac{4}{7}ax + \frac{5}{7}a = 0\) Multiplying both sides by \(\frac{7}{a}\), we get: \(7x^2 - 4x + 5 = 0\) Therefore, the quadratic equation is \(7x^2 - 4x + 5 = 0\).

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To solve this problem, we just need to perform the matrix multiplication on the left-hand side and equate it with the right-hand side, and then solve for the unknown variable x. \(\begin{pmatrix} 3 & 2 \\ 7 & x \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\) Performing the matrix multiplication on the left-hand side gives: \(\begin{pmatrix} 3(2) + 2(3) \\ 7(2) + x(3) \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\) Simplifying the left-hand side gives: \(\begin{pmatrix} 12 \\ 14 + 3x \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\) We can see that the first element of both matrices is already equal to 12, so we only need to equate the second elements: \(14 + 3x = 29\) Solving for x gives: \(3x = 15\) \(x = 5\) Therefore, the answer is x = 5.

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The probability of an event happening is the number of ways that event can happen, divided by the total number of possible outcomes. In this case, the total number of balls in the basket is 4, so there are 4 possible outcomes. The number of ways to choose a red ball is 3, since there are 3 red balls in the basket. The number of ways to choose a white ball is 1, since there is only 1 white ball in the basket. Therefore, the probability of choosing a red ball or a white ball is: \[P(\text{red or white}) = \frac{\text{number of red or white balls}}{\text{total number of balls}} = \frac{3+1}{4} = \frac{4}{4} = 1\] So the answer is option D: 1, meaning that it is certain that the ball drawn will be either red or white.

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To find the distance covered in the first 2 seconds, we need to integrate the velocity function from 0 to 2 seconds: \begin{align*} \text{Distance} &= \int_{0}^{2} v\,dt \\ &= \int_{0}^{2} (3t^2 - 2t)\,dt \\ &= \left[\frac{3}{3}t^3 - \frac{2}{2}t^2\right]_{0}^{2} \\ &= \left(3\cdot 2^2 - 2\cdot 2^2\right) - \left(3\cdot 0^2 - 2\cdot 0^2\right) \\ &= 4 \text{ m}. \end{align*} Therefore, the distance covered in the first 2 seconds is 4 meters. Answer choice (b) is correct.

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To find \(f \circ g(3)\), we first need to apply the function g to the input 3. Since \(g : x \to x - 1\), we have: \[g(3) = 3 - 1 = 2.\] Now we can use the output of g(3) as the input to the function f. Since \(f : x \to x^{2} - x - 6\), we have: \[f(g(3)) = f(2) = 2^{2} - 2 - 6 = -4.\] Therefore, the value of \(f \circ g(3)\) is -4

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To find the unit vector in the direction of a given vector, we need to divide the vector by its magnitude. The magnitude of vector (-5i + 12j) can be found using the Pythagorean theorem as follows: $$\left|\begin{pmatrix}-5 \\ 12 \\\end{pmatrix}\right| = \sqrt{(-5)^2 + (12)^2} = 13$$ Therefore, the unit vector in the direction of (-5i + 12j) is: $$\frac{1}{13}\begin{pmatrix}-5 \\ 12 \\\end{pmatrix} = \frac{1}{13}(-5i + 12j)$$ So the correct option is \(\frac{1}{13}(-5i + 12j)\).

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To find the remainder when the polynomial \(5x^3 + 2x^2 - 7x - 5\) is divided by \((x-2)\), we can use the Remainder Theorem which states that the remainder of the polynomial division can be found by evaluating the polynomial at the root of the divisor. In this case, the root of the divisor \((x-2)\) is \(x=2\). So, we evaluate the polynomial at \(x=2\) as follows: \begin{align*} 5(2)^3 + 2(2)^2 - 7(2) - 5 &= 40 + 8 - 14 - 5 \\ &= 29 \end{align*} Therefore, the remainder is \(29\), and the correct option is (c).

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We know that \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\). Substituting this value in the given expression, we get: \[\frac{1}{1 - \sin 60^{\circ}} = \frac{1}{1 - \frac{\sqrt{3}}{2}}\] Rationalizing the denominator by multiplying both numerator and denominator by \(1 + \frac{\sqrt{3}}{2}\), we get: \begin{align*} \frac{1}{1 - \sin 60^{\circ}} &= \frac{1}{1 - \frac{\sqrt{3}}{2}} \cdot \frac{1 + \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}} \\ &= \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{3}{4}} \\ &= \frac{1 + \frac{\sqrt{3}}{2}}{\frac{1}{4}} \\ &= 4 + 2\sqrt{3} \end{align*} Therefore, the answer is \boxed{4 + 2\sqrt{3}}.

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To find the gradient of \(f(x) = (x^2+3)^2\) at \(x=\frac{1}{2}\), we need to differentiate f(x) with respect to x and then substitute x = \(\frac{1}{2}\). Using the chain rule, we have: \begin{align*} \frac{d}{dx}[(x^2+3)^2] &= 2(x^2+3) \cdot \frac{d}{dx}(x^2+3) \\ &= 2(x^2+3) \cdot 2x \\ &= 4x(x^2+3) \end{align*} So, the gradient of \(f(x)\) is \(4x(x^2+3)\). Substituting \(x=\frac{1}{2}\), we get: \begin{align*} \text{Gradient of } f(x) \text{ at } x = \frac{1}{2} &= 4\left(\frac{1}{2}\right)\left(\left(\frac{1}{2}\right)^2+3\right) \\ &= 4\left(\frac{1}{2}\right)\left(\frac{13}{4}\right) \\ &= \frac{13}{2} \\ &= 6.5 \end{align*} Therefore, the gradient of \(f(x)\) at \(x=\frac{1}{2}\) is 6.5.

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The acceleration of a particle is given by the derivative of its velocity with respect to time. So we differentiate the given equation for velocity to find the acceleration: $$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t) = 6t - 2$$ When t = 2 seconds, the acceleration is: $$a = 6(2) - 2 = 10 \text{ ms}^{-2}$$ Therefore, the correct option is: \(\mathbf{10 ms^{-2}}\).

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To find the standard deviation of a set of numbers, we need to follow these steps: 1. Find the mean (average) of the numbers. 2. For each number, subtract the mean and square the result. 3. Find the mean of the squared differences. 4. Take the square root of the mean to get the standard deviation. So, let's apply these steps to the given set of numbers: 3, 6, 2, 1, 7, 5. 1. The mean is (3 + 6 + 2 + 1 + 7 + 5) / 6 = 4. 2. For each number, subtract the mean and square the result: \begin{align*} (3 - 4)^2 &= 1 \\ (6 - 4)^2 &= 4 \\ (2 - 4)^2 &= 4 \\ (1 - 4)^2 &= 9 \\ (7 - 4)^2 &= 9 \\ (5 - 4)^2 &= 1 \end{align*} 3. Find the mean of the squared differences: \begin{align*} \frac{1 + 4 + 4 + 9 + 9 + 1}{6} &= \frac{28}{6} \\ &= 4.67 \end{align*} 4. Take the square root of the mean to get the standard deviation: \begin{align*} \sqrt{4.67} &\approx 2.16 \end{align*} Therefore, the standard deviation of the given set of numbers is approximately 2.16. So, the correct answer is (b) 2.16.

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To find the equation of a circle with center \((a, b)\) and radius \(r\), we use the formula: \[(x - a)^2 + (y - b)^2 = r^2\] In this case, the center is \((-3, -8)\) and the radius is \(4\sqrt{6}\). So the equation of the circle is: \[(x - (-3))^2 + (y - (-8))^2 = (4\sqrt{6})^2\] which simplifies to: \[(x + 3)^2 + (y + 8)^2 = 96\] Expanding the left-hand side gives: \[x^2 + 6x + 9 + y^2 + 16y + 64 = 96\] which simplifies to: \[x^2 + y^2 + 6x + 16y - 23 = 0\] So the answer is (2) \(x^{2} + y^{2} + 6x + 16y - 23 = 0\).

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To find the original function from its inverse, we can simply swap the roles of x and y in the equation for the inverse and then solve for y. Starting with the given inverse function: \[f^{-1} : x \to \frac{x + 1}{4}\] Swapping x and y: \[x = \frac{y + 1}{4}\] Solving for y: \[y+1 = 4x\] \[y = 4x - 1\] Therefore, the original function is: \[f : x \to 4x - 1\] So the correct answer is.

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To find the value of force F, we need to use Newton's second law of motion which states that force is equal to mass multiplied by acceleration (F = ma). In this case, we know the mass of the body is 25kg and we can find the acceleration by using the formula: acceleration = change in velocity / time The change in velocity is 35m/s - 15m/s = 20m/s and the time is 5 seconds, so: acceleration = 20m/s / 5s = 4m/s^2 Now we can use Newton's second law to find the force: F = ma F = 25kg x 4m/s^2 F = 100N Therefore, the value of force F is 100N.

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There are different ways to approach this problem, but one common method is to use the formula for combinations. In general, the number of ways to choose k items out of n distinct items (without repetition and without order) is given by the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where n! (read as "n factorial") means the product of all positive integers up to n, and 0! is defined as 1. In this specific problem, we want to choose 3 prefects out of 8, so we have: \(\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\) Therefore, there are 56 ways to choose 3 prefects out of 8 prefects. So the answer is: 56.

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To find f(-3), we substitute -3 in the given function: \(f(-3) = \frac{3(-3) + 1}{(-3)^2 - 1} = \frac{-8}{8} = -1\) Therefore, the value of f(-3) is -1. So, the answer is (B) \(-1\).

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To find the radius of the given circle, we need to rewrite the equation in standard form, which is of the form \((x - a)^2 + (y - b)^2 = r^2\), where \((a, b)\) is the center of the circle and \(r\) is its radius. Completing the square for the given equation, we have: \begin{align*} 3x^{2} + 3y^{2} + 6x - 12y + 6 &= 0\\ 3(x^{2} + 2x) + 3(y^{2} - 4y) &= -6\\ 3(x^{2} + 2x + 1) + 3(y^{2} - 4y + 4) &= -6 + 3 + 12\\ 3(x + 1)^{2} + 3(y - 2)^{2} &= 9 \end{align*} Dividing both sides by 3, we have: \[(x + 1)^{2} + (y - 2)^{2} = 3\] Comparing this with the standard form, we can see that the center of the circle is \((-1, 2)\), and the radius is \(\sqrt{3}\). Therefore, the answer is \boxed{\sqrt{3}}.

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Certainly, I can help with that! To find the median of a set of numbers, we need to first put them in order from smallest to largest. So, let's arrange the given numbers in ascending order: 2, 5, x+2, 7, 9 The next step is to find the middle value. If there are an odd number of values, the median is simply the middle value. If there are an even number of values, the median is the average of the two middle values. In this case, we have five numbers, which is an odd number. So, the median is simply the middle value of the ordered set. To find the middle value, we need to count from either end until we get to the middle. Since there are five numbers, the middle value will be the third one. 2, 5, x+2, 7, 9 The third number is x+2, so that is the median. Now, we are given that the mean of the five numbers is 6. To find the mean, we add up all the numbers and divide by the total number of numbers: (2 + 5 + x + 2 + 7 + 9)/5 = 6 (25 + x)/5 = 6 25 + x = 30 x = 5 So, the value of x that makes the mean of the five numbers equal to 6 is 5. Plugging this value of x back into the original set, we have: 2, 5, 7, 7, 9 The median of this set is the middle value, which is 7. Therefore, the median of the given set of numbers is 7. I hope this helps you understand how to find the median of a set of numbers, as well as how to use the mean to solve for missing values in the set!

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(a) Weight \(W=6.5\times 10=65\ \text{N}\). Let string \(AP\) (\(50^\circ\) to the downward vertical) carry tension \(T_1\) up to the left and \(BP\) (\(60^\circ\)) carry \(T_2\) up to the right.

(i) Component form:

(ii) Equilibrium of P: \(\vec{T_1}+\vec{T_2}+\vec{W}=\vec{0}\).

(iii) Horizontal: \(T_1\sin 50^\circ=T_2\sin 60^\circ\). Using Lami's theorem (angles \(110^\circ,120^\circ,130^\circ\)):

Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).

(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).

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Four out of twenty lack a valid licence, so \(P(\text{no licence})=\tfrac{4}{20}=0.2\) and \(P(\text{valid})=0.8\). With \(n=6\), \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\) (\(X\) = number with valid licences).

(a) exactly 2 valid:

(b) more than 3 valid \(=P(4)+P(5)+P(6)\):

(c) at least 5 valid \(=P(5)+P(6)=0.393216+0.262144=0.655\).

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(a) Using \((a+b)^5=\sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\) with \(a=2\) and \(b=-x\):

\[(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5\]

The coefficients come from \(\binom{5}{k}2^{5-k}(-1)^k\): \(32,\,-80,\,80,\,-40,\,10,\,-1\).

(b) Choose \(x\) so that \(2-x=1.98\), i.e. \(x=0.02\). Substitute:

• \(32\)
• \(-80(0.02)=-1.6\)
• \(+80(0.02)^2=+0.032\)
• \(-40(0.02)^3=-0.00032\)
• \(+10(0.02)^4=+0.0000016\)
• \(-(0.02)^5\approx 0\)

Adding: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).

\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]

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(a) Cumulative frequency table

Marks	Frequency	Upper class boundary	Cumulative frequency
0–9	2	9.5	2
10–19	5	19.5	7
20–29	8	29.5	15
30–39	18	39.5	33
40–49	20	49.5	53
50–59	15	59.5	68
60–69	5	69.5	73
70–79	4	79.5	77
80–89	2	89.5	79
90–99	1	99.5	80

The total number of candidates is \(N=80\). Plot cumulative frequency against the upper class boundaries and draw a smooth increasing ogive.

(b)(i) Semi-interquartile range

\[Q_1\text{ is the }\frac{N}{4}=\frac{80}{4}=20\text{th value.}\]
From the ogive, \(Q_1\approx 32.5\).

\[Q_3\text{ is the }\frac{3N}{4}=\frac{3(80)}{4}=60\text{th value.}\]
From the ogive, \(Q_3\approx 53.0\).

\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{53.0-32.5}{2}=\boxed{10.25\text{ marks}}\]

(b)(ii) Percentage passing with distinction

From the ogive, the cumulative frequency at a mark of \(72\) is approximately \(74\).

\[\text{Number scoring at least }72=80-74=6\]

\[\text{Percentage with distinction}=\frac{6}{80}\times100\%=\boxed{7.5\%}\]

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(a) Evaluate each term.

\({}^{9}P_{3}=9\times8\times7=504\), and \({}^{15}C_{3}=\dfrac{15\times14\times13}{3!}=455\), so \(\dfrac{{}^{9}P_{3}}{{}^{15}C_{3}}=\dfrac{504}{455}=1.1077\).

\({}^{5}C_{3}=10\), and \({}^{3}P_{2}=3\times2=6\), so \(\dfrac{{}^{5}C_{3}}{{}^{3}P_{2}}=\dfrac{10}{6}=1.6667\).

Sum \(=1.1077+1.6667=2.7744\approx 2.77\) (2 d.p.).

(b) We form a committee of 2 tutors and 5 pupils from 6 tutors and 10 pupils.

Tutors: one particular tutor is already chosen, so pick 1 more tutor from the remaining 5: \({}^{5}C_{1}=5\).

Pupils: two particular pupils are excluded, so choose 5 pupils from the remaining 8: \({}^{8}C_{5}=56\).

By the multiplication principle:

\[5\times56=280\ \text{ways}\]

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(a) Histogram

Since all the class intervals have the same width, namely 10 (in units of 100 km), the bar heights are proportional to the frequencies. Use continuous class boundaries as follows:

(b) Estimate of the mode

The modal class is \(30\text{–}39\), with boundaries \(29.5\text{–}39.5\). From the intersection of the diagonals drawn in the modal rectangle on the histogram, the modal value is approximately \(33.5\).

Equivalently,

\[\text{Mode}=29.5+\frac{93-69}{(93-69)+(93-57)}\times 10\]

\[=29.5+\frac{24}{60}\times10=\boxed{33.5}\]

Thus, the estimated modal tyre life span is \(\boxed{33.5\times100=3350\text{ km}}\).

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(a) First principles: \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\) with \(f(x)=(2x+3)^2\).

\[f(x+h)-f(x)=\big[(2x+3)+2h\big]^2-(2x+3)^2\]

\[=2(2x+3)(2h)+(2h)^2=4h(2x+3)+4h^2\]

Divide by \(h\): \(4(2x+3)+4h\). Taking the limit as \(h\to0\):

\[f'(x)=4(2x+3)=8x+12\]

(b) First expand the numerator: \((x+1)(x^2-2x+2)=x^3-x^2+2\). Then

\[\frac{x^3-x^2+2}{x^2}=x-1+\frac{2}{x^2}=x-1+2x^{-2}\]

\[\int_{1}^{2}\left(x-1+2x^{-2}\right)dx=\left[\frac{x^2}{2}-x-\frac{2}{x}\right]_{1}^{2}\]

At \(x=2\): \(2-2-1=-1\). At \(x=1\): \(\tfrac12-1-2=-2.5\).

\[=-1-(-2.5)=\tfrac32\]

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(a) A force \((F,\theta)\) resolves into the column vector \(\begin{pmatrix}F\cos\theta\\ F\sin\theta\end{pmatrix}\), where \(\theta\) is measured anticlockwise from the positive x-axis.

For \(r=(10\text{N},090^{\circ})\): \(\cos90^{\circ}=0,\ \sin90^{\circ}=1\).

\[r=\begin{pmatrix}0\\ 10\end{pmatrix}\]

For \(q=(8\text{N},135^{\circ})\): \(\cos135^{\circ}=-\frac{\sqrt2}{2},\ \sin135^{\circ}=\frac{\sqrt2}{2}\).

\[q=\begin{pmatrix}8(-\frac{\sqrt2}{2})\\ 8(\frac{\sqrt2}{2})\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 4\sqrt2\end{pmatrix}\]

(b) Add corresponding components:

\[r+q=\begin{pmatrix}0-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}\approx\begin{pmatrix}-5.66\\ 15.66\end{pmatrix}\text{N}\]

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(a) Resolve the polar forces into components.

Resultant \(R=(-4.598,\ 1.964)\), so \(|R|=\sqrt{(-4.598)^2+(1.964)^2}=\sqrt{25.0}=5\ \text{N}\), at direction \(\theta=180^\circ-\tan^{-1}\!\frac{1.964}{4.598}=156.9^\circ\).

(i) Acceleration: \(a=\dfrac{|R|}{m}=\dfrac{5}{7}=0.71\ \text{m/s}^2\) in the direction \(156.9^\circ\).

(ii) Velocity after 3 s (from rest): \(v=at=0.71\times 3=2.14\ \text{m/s}\), along \(156.9^\circ\).

(b) Sum of the three forces: \(F_1+F_2+F_3=(2+0+6,\ 3-5-4)=(8,-6)\). The fourth force for equilibrium is \(F_4=-(8,-6)=(-8,6)\).

The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.

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For \(2x^2 - 7x + 4 = 0\), the sum and product of the roots \(\alpha,\beta\) are

\[\alpha + \beta = \frac{7}{2},\qquad \alpha\beta = \frac{4}{2} = 2.\]

Sum of the new roots \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}\). Using \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\):

\[\alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2(2) = \frac{49}{4} - 4 = \frac{33}{4}.\]

\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{33/4}{2} = \frac{33}{8}.\]

Product of the new roots \(\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha} = 1\).

The required equation is \(x^2 - (\text{sum})x + (\text{product}) = 0\):

\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]

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Since \(f(-1) = 0\), \((x + 1)\) is a factor of \(f(x) = 6x^3 + 13x^2 + 2x - 5\).

Divide \(f(x)\) by \((x + 1)\):

\[6x^3 + 13x^2 + 2x - 5 = (x + 1)(6x^2 + 7x - 5).\]

Now factor the quadratic \(6x^2 + 7x - 5\). We need two numbers multiplying to \(6\times(-5) = -30\) and adding to \(7\): these are \(10\) and \(-3\):

\[6x^2 + 10x - 3x - 5 = 2x(3x + 5) - 1(3x + 5) = (3x + 5)(2x - 1).\]

Therefore

\[f(x) = (x + 1)(2x - 1)(3x + 5).\]

Antwoorddetails

Resolve each force into components \((F\cos\theta,\ F\sin\theta)\).

\(F_{1}=(35,210^{\circ})\): \(\cos210^{\circ}=-\frac{\sqrt3}{2},\ \sin210^{\circ}=-\frac12\), so \(F_{1}=\left(-\tfrac{35\sqrt3}{2},\,-\tfrac{35}{2}\right)\approx(-30.31,\,-17.5)\).

\(F_{2}=(35\sqrt3,300^{\circ})\): \(\cos300^{\circ}=\frac12,\ \sin300^{\circ}=-\frac{\sqrt3}{2}\), so \(F_{2}=\left(\tfrac{35\sqrt3}{2},\,-\tfrac{35\cdot3}{2}\right)\approx(30.31,\,-52.5)\).

(a) Resultant of \(F_{1}\) and \(F_{2}\):

\[R=\big(-30.31+30.31,\ -17.5-52.5\big)=(0,\,-70)\]

Magnitude \(=\sqrt{0^{2}+(-70)^{2}}=70\text{N}\), directed vertically downward (along \(270^{\circ}\)).

(b) The 40N resistance opposes the motion, i.e. it acts opposite to this 70N resultant. The net (resultant) force acting on the particle is therefore

\[70-40=30\text{N}\]

directed the same way as the \(70\)N resultant (\(270^{\circ}\)). Its magnitude is \(\boxed{30\text{N}}\).

Antwoorddetails

(a) Differentiate \(\dfrac{x^2}{4}+y^2=1\) implicitly:

\[\frac{2x}{4}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{4y}\]

At \(\left(1,\tfrac{\sqrt3}{2}\right)\): \(\dfrac{dy}{dx}=-\dfrac{1}{4\cdot\frac{\sqrt3}{2}}=-\dfrac{1}{2\sqrt3}=-\dfrac{\sqrt3}{6}\).

Tangent: \(y-\dfrac{\sqrt3}{2}=-\dfrac{\sqrt3}{6}(x-1)\). Multiplying through by 6 and simplifying:

\[\sqrt3\,x+6y=4\sqrt3\quad\text{or}\quad x+2\sqrt3\,y=4\]

(b) Factorise the denominator: \(x^2+x-2=(x+2)(x-1)\). Write

\[\frac{3x+2}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\Rightarrow 3x+2=A(x-1)+B(x+2)\]

Put \(x=1\): \(5=3B\Rightarrow B=\tfrac53\). Put \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).

\[\frac{3x+2}{x^2+x-2}=\frac{4}{3(x+2)}+\frac{5}{3(x-1)}\]

Antwoorddetails

None

Antwoorddetails

(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):

From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).

(b) \(A(-1,3),B(2,-1),C(5,3)\). \(\vec{AB}=\begin{pmatrix}3\\-4\end{pmatrix},\ |\vec{AB}|=5\); \(\vec{AC}=\begin{pmatrix}6\\0\end{pmatrix},\ |\vec{AC}|=6\).

(i) Unit vectors: \(\hat{AB}=\begin{pmatrix}3/5\\-4/5\end{pmatrix},\quad \hat{AC}=\begin{pmatrix}1\\0\end{pmatrix}.\)

(ii) \(\vec{AB}\cdot\vec{AC}=3(6)+(-4)(0)=18\), so

Antwoorddetails

(a) Compute \(BA\):

\[BA=\begin{pmatrix}3&1\\2&3\end{pmatrix}\begin{pmatrix}-2&5\\4&3\end{pmatrix}=\begin{pmatrix}-6+4&15+3\\-4+12&10+9\end{pmatrix}=\begin{pmatrix}-2&18\\8&19\end{pmatrix}\]

Now the right-hand side:

\[2\begin{pmatrix}3&7\\-2&x\end{pmatrix}+\begin{pmatrix}y&4\\12&-3\end{pmatrix}=\begin{pmatrix}6+y&18\\8&2x-3\end{pmatrix}\]

Equating entries: top-left \(-2=6+y\Rightarrow y=-8\); bottom-right \(19=2x-3\Rightarrow x=11\). (The other entries \(18=18,\ 8=8\) check out.)

\[\boxed{x=11,\ y=-8}\]

(b)(i) Let \(y=\dfrac{5x-1}{3}\). Then \(3y=5x-1\Rightarrow x=\dfrac{3y+1}{5}\), so

\[g^{-1}(x)=\frac{3x+1}{5}\]

(ii) \(g^{-1}\circ f(x)=g^{-1}\!\left(\tfrac12 x+1\right)=\dfrac{3\left(\tfrac12 x+1\right)+1}{5}=\dfrac{\tfrac32 x+4}{5}=\dfrac{3x+8}{10}\)

Antwoorddetails

Write everything in terms of \(u=2^{y}\). Note \(2^{2y+2}=2^{2}\cdot 2^{2y}=4(2^{y})^{2}=4u^{2}\).

The equation \(2^{2y+2}-9(2^{y})=-2\) becomes:

\[4u^{2}-9u+2=0\]

Factorising (or using the formula \(u=\frac{9\pm\sqrt{81-32}}{8}=\frac{9\pm 7}{8}\)):

\[(4u-1)(u-2)=0\Rightarrow u=\tfrac14\ \text{or}\ u=2\]

Since \(u=2^{y}\):

• \(2^{y}=2\Rightarrow y=1\)
• \(2^{y}=\tfrac14=2^{-2}\Rightarrow y=-2\)

\[\boxed{y=1\ \text{or}\ y=-2}\]