WAEC SSCE - Further Mathematics - 2011
Frage 1 Bericht
The mean of 2, 5, (x + 2), 7 and 9 is 6. Find the median.
Antwortdetails
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!
Frage 2 Bericht
Two forces 10N and 15N act on an object at an angle of 120° to each other. Find the magnitude of the resultant.
Antwortdetails
Frage 3 Bericht
The equation of a circle is \(3x^{2} + 3y^{2} + 6x - 12y + 6 = 0\). Find its radius
Antwortdetails
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}}.
Frage 4 Bericht
Find the unit vector in the direction of (-5i + 12j).
Antwortdetails
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 \(\frac{1}{13}(-5i + 12j)\).
Frage 5 Bericht
A binary operation * is defined on the set of real numbers R, by a* b = -1. Find the identity element under the operation *.
Antwortdetails
Frage 6 Bericht
A basket contains 3 red and 1 white identical balls. A ball is drawn from the basket at random. Calculate the probability that it is either white or red.
Antwortdetails
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.
Frage 7 Bericht
Find, correct to two decimal places, the acute angle between \(p = \begin{pmatrix} 13 \\ 14 \end{pmatrix}\) and \(q = \begin{pmatrix} 12 \\ 5 \end{pmatrix}\).
Antwortdetails
Frage 8 Bericht
What percentage increase in the radius of a sphere will cause its volume to increase by 45%?
Antwortdetails
Frage 9 Bericht
If \(^{3x}C_{2} = 15\), find the value of x?
Antwortdetails
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.
Frage 10 Bericht
A body of mass 25kg changes its speed from 15m/s to 35m/s in 5 seconds by the action of an applied force F. Find the value of F.
Antwortdetails
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.
Frage 11 Bericht
The probability that Kofi and Ama hit a target in a shooting competition are \(\frac{1}{6}\) and \(\frac{1}{9}\) respectively. What is the probability that only one of them hit the target?
Antwortdetails
Frage 12 Bericht
Find the remainder when \(5x^{3} + 2x^{2} - 7x - 5\) is divided by (x - 2).
Antwortdetails
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).
Frage 13 Bericht
Find the standard deviation of the numbers 3,6,2,1,7 and 5.
Antwortdetails
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.
Frage 14 Bericht
If \(y^{2} + xy - x = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\).
Antwortdetails
Frage 18 Bericht
A function is defined by \(f(x) = \frac{3x + 1}{x^{2} - 1}, x \neq \pm 1\). Find f(-3).
Antwortdetails
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) \(-1\).
Frage 19 Bericht
| Marks | 5-7 | 8-10 | 11-13 | 14-16 | 17-19 | 20-22 |
| No of students | 4 | 7 | 26 | 41 | 14 | 8 |
The table above shows the distribution of marks of students in a class. Find the upper class boundary of the modal class.
Antwortdetails
Frage 20 Bericht
The functions f and g are defined on the set, R, of real numbers by \(f : x \to x^{2} - x - 6\) and \(g : x \to x - 1\). Find \(f \circ g(3)\).
Antwortdetails
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
Frage 21 Bericht
Determine the coefficient of \(x^{2}\) in the expansion of \((a + 3x)^{6}\).
Antwortdetails
Frage 22 Bericht
Given that \(q = 9i + 6j\) and \(r = 4i - 6j\), which of the following statements is true?
Antwortdetails
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.
Frage 23 Bericht
A force of 200N acting on a body of mass 20kg initially at rest causes it to move a distance of 320m along a straight line for t secs. Find the value of t.
Antwortdetails
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.
Frage 24 Bericht
The sum and product of the roots of a quadratic equation are \(\frac{4}{7}\) and \(\frac{5}{7}\) respectively. Find its equation.
Antwortdetails
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 \(7x^2 - 4x + 5 = 0\).
Frage 25 Bericht
Find the equation of a circle with centre (-3, -8) and radius \(4\sqrt{6}\).
Antwortdetails
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) \(x^{2} + y^{2} + 6x + 16y - 23 = 0\).
Frage 26 Bericht
A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Determine the acceleration when t = 2 secs.
Antwortdetails
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: \(\mathbf{10 ms^{-2}}\).
Frage 27 Bericht
A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Calculate the distance covered in the first 2 seconds.
Antwortdetails
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.
Frage 28 Bericht
Evaluate \(\frac{1}{1 - \sin 60°}\), leaving your answer in surd form.
Antwortdetails
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}}.
Frage 29 Bericht
Evaluate \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\)
Antwortdetails
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 \(\frac{-\sqrt{3}}{2}\)
Frage 30 Bericht
\(f(x) = p + qx\), where p and q are constants. If f(1) = 7 and f(5) = 19, find f(3).
Antwortdetails
Frage 31 Bericht
The inverse of a function is given by \(f^{-1} : x \to \frac{x + 1}{4}\).
Antwortdetails
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.
Frage 32 Bericht
The fourth term of a geometric sequence is 2 and the sixth term is 8. Find the common ratio.
Antwortdetails
Frage 33 Bericht
In how many ways can 3 prefects be chosen out of 8 prefects?
Antwortdetails
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.
Frage 34 Bericht
Four doctors and two nurses are to sit round a circular table. In how many ways can this be done if the nurses are to sit together?
Frage 35 Bericht
If \(\begin{pmatrix} 3 & 2 \\ 7 & x \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix} \), find x.
Antwortdetails
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.
Frage 37 Bericht
\(f(x) = (x^{2} + 3)^{2}\) is defines on the set of real numbers, R. Find the gradient of f(x) at x = \(\frac{1}{2}\).
Antwortdetails
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.
Frage 38 Bericht
Express 75° in radians, leaving your answer in terms of \(\pi\).
Antwortdetails
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) \(\frac{5\pi}{12}\).
Frage 39 Bericht
A line is perpendicular to \(3x - y + 11 = 0\) and passes through the point (1, -5). Find its equation.
Antwortdetails
Frage 40 Bericht
If \(\log_{9} 3 + 2x = 1\), find x.
Frage 41 Bericht
A tyre manufacturing company researched into the life span of one type of their motorcycle tyres. The results were as follows :
| Distance (100km) |
10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 |
| Number of tyres | 30 | 69 | 93 | 57 | 36 | 15 |
(a) Draw a histogram for the distribution.
(b) Use the histogram to estimate the mode.
(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:
| Distance (100 km) | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 30 |
| 20–29 | 19.5–29.5 | 69 |
| 30–39 | 29.5–39.5 | 93 |
| 40–49 | 39.5–49.5 | 57 |
| 50–59 | 49.5–59.5 | 36 |
| 60–69 | 59.5–69.5 | 15 |
(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 \(\boxed{33.5\times100=3350\text{ km}}\).
Antwortdetails
(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:
| Distance (100 km) | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 30 |
| 20–29 | 19.5–29.5 | 69 |
| 30–39 | 29.5–39.5 | 93 |
| 40–49 | 39.5–49.5 | 57 |
| 50–59 | 49.5–59.5 | 36 |
| 60–69 | 59.5–69.5 | 15 |
(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 \(\boxed{33.5\times100=3350\text{ km}}\).
Frage 42 Bericht
Solve \(2^{(2y + 2)} - 9(2^{y}) = -2\).
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 \(u=2^{y}\):
- \(2^{y}=2\Rightarrow y=1\)
- \(2^{y}=\tfrac14=2^{-2}\Rightarrow y=-2\)
\[\boxed{y=1\ \text{or}\ y=-2}\]
Antwortdetails
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 \(u=2^{y}\):
- \(2^{y}=2\Rightarrow y=1\)
- \(2^{y}=\tfrac14=2^{-2}\Rightarrow y=-2\)
\[\boxed{y=1\ \text{or}\ y=-2}\]
Frage 43 Bericht
The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the :
(a) matrix A ; (b) inverse of A ; (c) point whose image is (-1, 1).
Frage 44 Bericht
A bag contains 4 red, 6 blue and 8 green identical marbles.
(a) If three marbles are drawn at random, without replacement, calculate the probability that :
(i) all will be green ; (ii) all will have the same colour.
(b) If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.
Frage 45 Bericht
A survey conducted revealed that four out of every twenty taxi drivers do not have a valid driving license. If 6 drivers are selected at random, calculate, correct to three decimal places, the probability that
(a) exactly 2 ;
(b) more than 3 ;
(c) at least 5; have valid driving license.
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:
\[\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.015.\](b) more than 3 valid \(=P(4)+P(5)+P(6)\):
\[P(4)=\binom{6}{4}(0.8)^4(0.2)^2=15(0.4096)(0.04)=0.24576,\] \[P(5)=\binom{6}{5}(0.8)^5(0.2)=6(0.32768)(0.2)=0.393216,\quad P(6)=(0.8)^6=0.262144.\] \[\text{Sum}=0.24576+0.393216+0.262144=0.901.\](c) at least 5 valid \(=P(5)+P(6)=0.393216+0.262144=0.655\).
Antwortdetails
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:
\[\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.015.\](b) more than 3 valid \(=P(4)+P(5)+P(6)\):
\[P(4)=\binom{6}{4}(0.8)^4(0.2)^2=15(0.4096)(0.04)=0.24576,\] \[P(5)=\binom{6}{5}(0.8)^5(0.2)=6(0.32768)(0.2)=0.393216,\quad P(6)=(0.8)^6=0.262144.\] \[\text{Sum}=0.24576+0.393216+0.262144=0.901.\](c) at least 5 valid \(=P(5)+P(6)=0.393216+0.262144=0.655\).
Frage 46 Bericht
(a) If \(A = \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}\), find the values of x and y such that \(BA = 2\begin{pmatrix} 3 & 7 \\ -2 & x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix}\).
(b) Two functions, f and g are defined by \(f : x \to \frac{1}{2}x + 1\) and \(g : x \to \frac{5x - 1}{3}\). Find :
(i) \(g^{-1}\) ; (ii) \(g^{-1} \circ f\).
(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) \(g^{-1}\circ f(x)=g^{-1}\!\left(\tfrac12 x+1\right)=\dfrac{3\left(\tfrac12 x+1\right)+1}{5}=\dfrac{\tfrac32 x+4}{5}=\dfrac{3x+8}{10}\)
Antwortdetails
(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) \(g^{-1}\circ f(x)=g^{-1}\!\left(\tfrac12 x+1\right)=\dfrac{3\left(\tfrac12 x+1\right)+1}{5}=\dfrac{\tfrac32 x+4}{5}=\dfrac{3x+8}{10}\)
Frage 47 Bericht
(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and BP. The strings make angles 50° and 60° respectively with the downward vertical.
(i) Express the forces acting on P in component form; (ii) If P is at rest, write down the vector equation connecting all the forces; (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5 kg moves with initial velocity \(\frac{1}{2} m/s\) and final velocity \(\frac{3}{4} m/s \). Find the magnitude of its change in momentum.
(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:
\[\vec{T_1}=\begin{pmatrix}-T_1\sin 50^\circ\\ T_1\cos 50^\circ\end{pmatrix},\quad \vec{T_2}=\begin{pmatrix}T_2\sin 60^\circ\\ T_2\cos 60^\circ\end{pmatrix},\quad \vec{W}=\begin{pmatrix}0\\-65\end{pmatrix}.\](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 \(110^\circ,120^\circ,130^\circ\)):
\[T_1=\frac{65\sin 120^\circ}{\sin 110^\circ}=\frac{65(0.8660)}{0.9397}=59.9\ \text{N},\quad T_2=\frac{65\sin 130^\circ}{\sin 110^\circ}=\frac{65(0.7660)}{0.9397}=53.0\ \text{N}.\]Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).
(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).
Antwortdetails
(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:
\[\vec{T_1}=\begin{pmatrix}-T_1\sin 50^\circ\\ T_1\cos 50^\circ\end{pmatrix},\quad \vec{T_2}=\begin{pmatrix}T_2\sin 60^\circ\\ T_2\cos 60^\circ\end{pmatrix},\quad \vec{W}=\begin{pmatrix}0\\-65\end{pmatrix}.\](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 \(110^\circ,120^\circ,130^\circ\)):
\[T_1=\frac{65\sin 120^\circ}{\sin 110^\circ}=\frac{65(0.8660)}{0.9397}=59.9\ \text{N},\quad T_2=\frac{65\sin 130^\circ}{\sin 110^\circ}=\frac{65(0.7660)}{0.9397}=53.0\ \text{N}.\]Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).
(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).
Frage 48 Bericht
A particle of mass 400g is moving under the action of two forces \(F_{1} = (35N, 210°), F_{2} = (35\sqrt{3} N, 300°)\) and a resistance of 40N. Find the magnitude of the
(a) resultant of \(F_{1}\) and \(F_{2}\).
(b) resultant force acting on the particle.
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 \(\boxed{30\text{N}}\).
Antwortdetails
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 \(\boxed{30\text{N}}\).
Frage 49 Bericht
If \(f(x) = 6x^{3} + 13x^{2} + 2x - 5\) and \(f(-1) = 0\), find the factors of f(x).
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
\[f(x) = (x + 1)(2x - 1)(3x + 5).\]
Antwortdetails
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
\[f(x) = (x + 1)(2x - 1)(3x + 5).\]
Frage 50 Bericht
(a) Write the following as column vectors: \(r = (10N, 090°) ; q = (8N, 135°)\).
(b) Use your answer in (a) to find \((r + q)\).
(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:
\[r+q=\begin{pmatrix}0-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}\approx\begin{pmatrix}-5.66\\ 15.66\end{pmatrix}\text{N}\]
Antwortdetails
(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:
\[r+q=\begin{pmatrix}0-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}\approx\begin{pmatrix}-5.66\\ 15.66\end{pmatrix}\text{N}\]
Frage 51 Bericht
If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 7x + 4 = 0\), find the equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\).
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 \(x^2 - (\text{sum})x + (\text{product}) = 0\):
\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]
Antwortdetails
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 \(x^2 - (\text{sum})x + (\text{product}) = 0\):
\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]
Frage 52 Bericht
(a) Write down the binomial expansion of \((2 - x)^{5}\) in ascending powers of x.
(b) Use your expansion in (a) to evaluate \((1.98)^{5}\) correct to four decimal places.
(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: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).
\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]
Antwortdetails
(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: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).
\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]
Frage 53 Bericht
The table shows the frequency distribution of marks scored by some candidates in an examination.
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Freq | 2 | 5 | 8 | 18 | 20 | 15 | 5 | 4 | 2 | 1 |
(a) Draw the cumulative frequency curve of the distribution.
(b) Use your graph to estimate the :
(i) semi-interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.
(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 \(74\).
\[\text{Number scoring at least }72=80-74=6\]
\[\text{Percentage with distinction}=\frac{6}{80}\times100\%=\boxed{7.5\%}\]
Antwortdetails
(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 \(74\).
\[\text{Number scoring at least }72=80-74=6\]
\[\text{Percentage with distinction}=\frac{6}{80}\times100\%=\boxed{7.5\%}\]
Frage 54 Bericht
(a) Evaluate \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\) correct to two decimal places.
(b) A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done if one particular tutor must be on the committee and two particular pupils must not be on the committee?
(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:
\[5\times56=280\ \text{ways}\]
Antwortdetails
(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:
\[5\times56=280\ \text{ways}\]
Frage 55 Bericht
(a) \(m \begin{pmatrix} 2 \\ 1 \end{pmatrix} + n \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}\) where m and n are scalars. Find the value of (m + n).
(b) A(-1, 3), B(2, -1) and C(5, 3) are the vertices of \(\Delta\) ABC.
(i) Express in column notation, the unit vectors parallel to AB and AC.
(ii) Use a dot product to calculate \(\stackrel \frown{BAC}\), correct to the nearest degree.
(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):
\[2m-n=5,\qquad m+2n=-4.\]From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).
\[m+n=\frac{6}{5}-\frac{13}{5}=-\frac{7}{5}=-1.4.\](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
\[\cos(\widehat{BAC})=\frac{18}{5\times 6}=\frac{18}{30}=0.6\Rightarrow \widehat{BAC}=53^\circ.\]Antwortdetails
(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):
\[2m-n=5,\qquad m+2n=-4.\]From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).
\[m+n=\frac{6}{5}-\frac{13}{5}=-\frac{7}{5}=-1.4.\](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
\[\cos(\widehat{BAC})=\frac{18}{5\times 6}=\frac{18}{30}=0.6\Rightarrow \widehat{BAC}=53^\circ.\]Frage 56 Bericht
(a) Find, from first principles, the derivative of \(f(x) = (2x + 3)^{2}\).
(b) Evaluate : \(\int_{1} ^{2} \frac{(x + 1)(x^{2} - 2x + 2)}{x^{2}} \mathrm {d} x\)
(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 \(x=1\): \(\tfrac12-1-2=-2.5\).
\[=-1-(-2.5)=\tfrac32\]
Antwortdetails
(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 \(x=1\): \(\tfrac12-1-2=-2.5\).
\[=-1-(-2.5)=\tfrac32\]
Frage 57 Bericht
(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the :
(i) acceleration of the particle ; (ii) velocity of the particle after 3 seconds.
(b) \(F_{1} = (2i + 3j)N, F_{2} = (-5j)N \) and \(F_{3} = (6i - 4j)N\) act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.
(a) Resolve the polar forces into components.
\[F_1=(3,210^\circ):\ (3\cos210^\circ,3\sin210^\circ)=(-2.598,-1.5).\] \[F_2=(4,120^\circ):\ (4\cos120^\circ,4\sin120^\circ)=(-2,3.464).\]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 \(F_4=-(8,-6)=(-8,6)\).
\[|F_4|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10\ \text{N},\quad \theta=180^\circ-\tan^{-1}\!\tfrac{6}{8}=143.1^\circ.\]The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.
Antwortdetails
(a) Resolve the polar forces into components.
\[F_1=(3,210^\circ):\ (3\cos210^\circ,3\sin210^\circ)=(-2.598,-1.5).\] \[F_2=(4,120^\circ):\ (4\cos120^\circ,4\sin120^\circ)=(-2,3.464).\]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 \(F_4=-(8,-6)=(-8,6)\).
\[|F_4|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10\ \text{N},\quad \theta=180^\circ-\tan^{-1}\!\tfrac{6}{8}=143.1^\circ.\]The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.
Frage 58 Bericht
(a) Find the equation of the tangent to curve \(\frac{x^{2}}{4} + y^{2} = 1\) at point \(1, \frac{\sqrt{3}}{2}\).
(b) Express \(\frac{3x + 2}{x^{2} + x - 2}\) in partial fractions.
(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 \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).
\[\frac{3x+2}{x^2+x-2}=\frac{4}{3(x+2)}+\frac{5}{3(x-1)}\]
Antwortdetails
(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 \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).
\[\frac{3x+2}{x^2+x-2}=\frac{4}{3(x+2)}+\frac{5}{3(x-1)}\]
