WAEC SSCE - Further Mathematics - 2015
Question 1 Report
Find \(\int \frac{x^{3} + 5x + 1}{x^{3}} \mathrm {d} x\)
Answer Details
We can write the given expression as: $$\int \frac{x^{3} + 5x + 1}{x^{3}} \mathrm {d} x = \int \left(1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}\right) \mathrm {d} x$$ Now, we integrate each term separately: $$\int \mathrm {d} x + 5 \int \frac{1}{x^{2}} \mathrm {d} x + \int \frac{1}{x^{3}} \mathrm {d} x = x - \frac{5}{x} - \frac{1}{2x^{2}} + c$$ Therefore, the correct option is: $$\boxed{x - \frac{5}{x} - \frac{1}{2x^{2}} + c}$$
Question 2 Report
Given that \(r = 2i - j\), \(s = 3i + 5j\) and \(t = 6i - 2j\), find the magnitude of \(2r + s - t\).
Answer Details
We are given that: \begin{align*} r &= 2i - j\\ s &= 3i + 5j\\ t &= 6i - 2j \end{align*} We can now substitute these expressions into \(2r+s-t\) and simplify: \begin{align*} 2r + s - t &= 2(2i-j) + (3i+5j) - (6i - 2j)\\ &= 4i - 2j + 3i + 5j - 6i + 2j\\ &= i + 5j \end{align*} The magnitude of a vector with components \(a\) and \(b\) is given by \(\sqrt{a^2 + b^2}\). So, the magnitude of \(i+5j\) is: \begin{align*} \sqrt{i^2 + 5j^2} &= \sqrt{1^2 + 5^2} \\ &= \sqrt{26} \end{align*} Therefore, the answer is \(\sqrt{26}\).
Question 3 Report
Given that \(P = \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix}\) and \(Q = \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix}\), find PQ - QP.
Question 4 Report
A line passes through the origin and the point \((1\frac{1}{4}, 2\frac{1}{2})\), what is the gradient of the line?
Answer Details
Question 5 Report
Each of the 90 students in a class speak at least Igbo or Hausa. If 56 students speak Igbo and 50 speak Hausa, find the probability that a student selected at random from the class speaks Igbo only.
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Question 6 Report
If \(36, p, \frac{9}{4}, q\) are consecutive terms of an exponential sequence (G.P.). Find the sum of p and q.
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Question 7 Report
The gradient of a curve at the point (-2, 0) is \(3x^{2} - 4x\). Find the equation of the curve.
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Question 8 Report
In how many ways can a committee of 5 be selected from 8 students if 2 particular students are to be included?
Answer Details
Since 2 particular students must be on the committee, we only need to select 3 more students from the remaining 6 students. The number of ways to do this is the number of combinations of 3 students from a set of 6, which is given by: \(\binom{6}{3}=\frac{6!}{3!3!}=20\) Therefore, the answer is 20.
Question 10 Report
A line passes through the origin and the point \((1\frac{1}{4}, 2\frac{1}{2})\). Find the y-coordinate of the line when x = 4.
Answer Details
Question 11 Report
Find the 3rd term of \((\frac{x}{2} - 1)^{8}\) in descending order of x.
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Question 12 Report
Given the statements:
p : the subject is difficult
q : I will do my best
Which of the following is equivalent to 'Although the subject is difficult, I will do my best'?
Answer Details
The given sentence can be rewritten as "Despite the subject being difficult, I will do my best." We can now use the logical operators to find the equivalent statement. "Despite the subject being difficult" can be represented as \(\sim p\) (not p), and "I will do my best" is equivalent to q. Then, the statement "Although the subject is difficult, I will do my best" is equivalent to \(\sim p \wedge q\). Therefore, the correct option is \( \sim p \wedge q \).
Question 13 Report
A particle starts from rest and moves in a straight line such that its acceleration after t secs is given by \(a = (3t - 2) ms^{-2}\). Find the distance covered after 3 secs.
Answer Details
To solve this problem, we need to use the kinematic equation: $$s = ut + \frac{1}{2}at^2$$ where: - s is the distance covered - u is the initial velocity (which is zero in this case) - a is the acceleration - t is the time taken We are given that the acceleration is given by \(a = (3t-2)ms^{-2}\). Integrating this with respect to time, we get: $$v = \int a dt = \int (3t-2) dt = \frac{3}{2}t^2 - 2t + C$$ where v is the velocity of the particle at time t, and C is a constant of integration. Since the particle starts from rest, we have v = 0 when t = 0, so: $$0 = \frac{3}{2}(0)^2 - 2(0) + C$$ which gives C = 0. Therefore, the velocity of the particle at time t is given by: $$v = \frac{3}{2}t^2 - 2t$$ Now, we can use the equation for distance covered to find the distance covered by the particle after 3 seconds: $$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3t-2)t^2 = \frac{3}{2}t^3 - t^2$$ So, when t = 3, we have: $$s = \frac{3}{2}(3)^3 - (3)^2 = \frac{27}{2} - 9 = \frac{9}{2}m$$ Therefore, the distance covered by the particle after 3 seconds is \(\frac{9}{2}m\). Thus, the correct option is: - \(\frac{9}{2}m\)
Question 14 Report
Given that \(y = 4 - 9x\) and \(\Delta x = 0.1\), calculate \(\Delta y\).
Answer Details
Question 15 Report
In calculating the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers. If he obtained 20 as the mean, find the correct mean.
Answer Details
Let's call the sum of the 8 numbers as S. Then, the boy's calculation of the mean is: $$ \text{Mean} = \frac{S+17}{8} $$ However, the correct calculation of the mean should be: $$ \text{Correct Mean} = \frac{S+25}{8} $$ We know that the boy obtained 20 as the mean, so we can set the two expressions equal to each other and solve for S: $$ \frac{S+17}{8} = 20 \implies S+17 = 160 \implies S = 143 $$ Therefore, the correct mean is: $$ \text{Correct Mean} = \frac{S+25}{8} = \frac{143+25}{8} = \frac{168}{8} = 21 $$ So the correct mean is 21. Therefore, the answer is (3) 21.
Question 16 Report
Find the coordinates of the point which divides the line joining P(-2, 3) and Q(4, 6) internally in the ratio 2 : 3.
Answer Details
Question 17 Report
Forces of magnitude 8N and 5N act on a body as shown above. Calculate, correct to 2 dp, the angle that the resultant makes with the horizontal.
Question 18 Report
Forces \(F_{1} = (8N, 030°)\) and \(F_{2} = (10N, 150°)\) act on a particle. Find the horizontal component of the resultant force.
Answer Details
To find the horizontal component of the resultant force, we need to resolve each force into its horizontal and vertical components. For \(F_{1} = (8N, 030°)\), the horizontal component is given by \(F_{1}\cos{30°} = 8\cos{30°} \approx 6.93N\). For \(F_{2} = (10N, 150°)\), the horizontal component is given by \(F_{2}\cos{150°} = 10\cos{150°} \approx -4.32N\). The negative sign in the horizontal component of \(F_2\) means that the force is acting in the opposite direction. To find the horizontal component of the resultant force, we sum up the horizontal components of the two forces: \[\text{Horizontal component of resultant force} = 6.93N - 4.32N = 2.61N \approx 1.7N\] Therefore, the horizontal component of the resultant force is approximately 1.7N. The option that matches this value is (a) 1.7N.
Question 19 Report
The sum, \(S_{n}\), of a sequence is given by \(S_{n} = 2n^{2} - 5\). Find the 6th term.
Answer Details
To find the 6th term of the sequence, we need to find the difference between the sum of the first 6 terms and the sum of the first 5 terms. The sum of the first 5 terms is: $S_5 = 2(5)^2 - 5 = 45$ The sum of the first 6 terms is: $S_6 = 2(6)^2 - 5 = 67$ Therefore, the 6th term of the sequence is: $S_6 - S_5 = 67 - 45 = 22$ So the answer is 22.
Question 20 Report
Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r.
Answer Details
We know that the left side of the equation can be expanded to give: $$x^{2} + 4x + k = x^{2} + 2xr + r^{2} + 1$$ By comparing the coefficients of x and x^2, we get: $$2xr + 4x = 0 \implies x(2r+4) = 0$$ Since x cannot be equal to zero, we have: $$2r + 4 = 0 \implies r = -2$$ Substituting r = -2 in the original equation, we get: $$x^{2} + 4x + k = (x-2)^{2} + 1 = x^{2} - 4x + 5$$ Comparing the coefficients of x^2 and x, we get: $$k = 5$$ Therefore, the solution is k = 5, r = -2. Note that this matches with.
Question 22 Report
A 24N force acts on a body such that it changes its velocity from 5m/s to 9m/s in 2 secs.If the body is travelling in a straight line, calculate the distance covered in the period.
Answer Details
The problem involves finding the distance covered by a body when a force is applied to it and it changes its velocity from an initial value to a final value over a period of time. The force acting on the body is 24N and the change in velocity is (9 - 5)m/s = 4m/s. The time taken for this change is 2 seconds. Using the formula, distance = (initial velocity + final velocity) * time / 2, we can calculate the distance covered by the body. Substituting the given values, we get distance = (5m/s + 9m/s) * 2s / 2 = 14m Therefore, the distance covered by the body is 14m. Answer: 14m
Question 23 Report
Forces of magnitude 8N and 5N act on a body as shown above. Calculate, correct to 2 d.p., the resultant force acting at O.
Answer Details
Question 24 Report
Four fair coins are tossed once. Calculate the probability of having equal heads and tails.
Answer Details
When tossing a coin, there are two possible outcomes: heads or tails. Since we are tossing four coins, there are \(2^{4} = 16\) possible outcomes. We want to find the probability of having an equal number of heads and tails. The possible outcomes where we have an equal number of heads and tails are as follows: HHTT, HTHT, THHT, THTH, HTTH, TTHH. There are six such outcomes. Therefore, the probability of having an equal number of heads and tails is \(\frac{6}{16} = \frac{3}{8}\). So the answer is.
Question 25 Report
Solve: \(2\cos x - 1 = 0\).
Answer Details
To solve the equation \(2\cos x - 1 = 0\), we first isolate the cosine term by adding 1 to both sides, giving: $$2\cos x = 1$$ Next, we divide both sides by 2 to obtain: $$\cos x = \frac{1}{2}$$ From the unit circle or trigonometric ratios, we know that the solutions to this equation are the angles whose cosine is equal to 1/2, which are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\) (or their reference angles in the interval [0, 2π)). Therefore, the answer is \((\frac{\pi}{3}, \frac{5\pi}{3})\).
Question 26 Report
A mass of 75kg is placed on a lift. Find the force exerted by the floor of the lift on the mass when the lift is moving up with constant velocity. \([g = 9.8ms^{-2}]\)
Answer Details
When the lift is moving up with a constant velocity, the net force acting on the mass must be zero since the mass is not accelerating. Therefore, the force exerted by the floor of the lift on the mass is equal in magnitude and opposite in direction to the force of gravity acting on the mass. The force of gravity acting on the mass is given by: F = m*g = 75kg * 9.8m/s^2 = 735N Therefore, the force exerted by the floor of the lift on the mass when the lift is moving up with constant velocity is 735N. Hence, the correct option is: 735N.
Question 27 Report
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of candidates | 6 | 4 | 8 | 10 | 9 | 3 |
The table above shows the distribution of marks scored by students in a test. Find the interquartile range of the distribution.
Question 28 Report
If \(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \), find the values of x.
Question 29 Report
A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by \(a = (3t - 2) ms^{-2}\). Find the other time when the velocity would be zero.
Answer Details
The acceleration of the particle is given as \(a = (3t - 2) ms^{-2}\). We can integrate this to get the velocity function, \(v(t)\), of the particle: \[\int a dt = \int (3t - 2) dt\] \[v(t) = \frac{3}{2}t^2 - 2t + C\] where C is the constant of integration. Since the particle starts from rest, we have \(v(0) = 0\), which implies that C = 0. Therefore, the velocity function of the particle is given as: \[v(t) = \frac{3}{2}t^2 - 2t\] To find the other time when the velocity would be zero, we need to solve the equation \(v(t) = 0\): \[\frac{3}{2}t^2 - 2t = 0\] \[t(3t - 4) = 0\] This equation has two roots, \(t = 0\) and \(t = \frac{4}{3}\). Since the particle is starting from rest, we discard the root \(t = 0\). Therefore, the other time when the velocity would be zero is \(t = \frac{4}{3}\) seconds. Hence, the answer is \(\frac{4}{3} seconds\).
Question 30 Report
Given that \(\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}\), find P and Q.
Answer Details
We can begin by multiplying both sides of the equation by the denominator \((x+6)(x+3)\). This will eliminate the denominators on the left side of the equation, leaving us with: $$2x = P(x+3) + Q(x+6)$$ Next, we can expand the right side of the equation: $$2x = Px + 3P + Qx + 6Q$$ We can then simplify by collecting like terms: $$(P + Q)x + (3P + 6Q) = 2x$$ Since this equation must be true for all values of x, we can equate the coefficients of x and the constants on both sides of the equation: $$\begin{aligned} P + Q &= 0 \\ 3P + 6Q &= 2 \end{aligned}$$ Solving the first equation for P in terms of Q, we get: $$P = -Q$$ Substituting this expression for P into the second equation, we get: $$3(-Q) + 6Q = 2$$ Simplifying this equation, we get: $$3Q = 2$$ Thus, we have: $$Q = \frac{2}{3}$$ Substituting this value for Q into the equation P = -Q, we get: $$P = -\frac{2}{3}$$ Therefore, we have: $$P = -\frac{2}{3}\text{ and }Q = \frac{2}{3}$$ So the answer is, P = 4 and Q = -2.
Question 31 Report
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of candidates | 6 | 4 | 8 | 10 | 9 | 3 |
The table above shows the distribution of marks scored by students in a test. How many candidates scored above the median score?
Answer Details
To find the median score, we need to arrange the scores in ascending or descending order. From the table, we can see that there are 40 candidates in total. To determine the median score, we need to find the middle value. Since there are an even number of candidates (40), the median will be the average of the two middle values. The two middle values are the 20th and 21st scores, which are both 3. Therefore, the median score is 3. To find how many candidates scored above the median score, we can add up the number of candidates who scored 4 and 5, which are the scores above 3. From the table, we can see that 9 candidates scored 4 and 3 candidates scored 5, making a total of 12 candidates who scored above the median score of 3. Therefore, the answer is 12.
Question 32 Report
Given that \(f : x \to x^{2}\) and \(g : x \to x + 3\), where \(x \in R\), find \(f o g(2)\).
Answer Details
To find \(f \circ g(2)\), we first need to apply the function g to 2, which gives us \(g(2) = 2 + 3 = 5\). Then we apply the function f to the result, which gives us \(f(5) = 5^2 = 25\). Therefore, the answer is 25.
Question 33 Report
If \(2\sin^{2} \theta = 1 + \cos \theta, 0° \leq \theta \leq 90°\), find the value of \(\theta\).
Answer Details
We are given an equation in terms of trigonometric functions, and we are asked to find the value of one of the angles involved. One possible way to approach this problem is to use trigonometric identities to simplify the equation and then solve for the unknown angle. First, we can use the identity \(\sin^{2}\theta + \cos^{2}\theta = 1\) to rewrite the left-hand side of the equation as \(\sin^{2}\theta = 1 - \cos^{2}\theta\). Substituting this into the given equation, we get: \[2(1-\cos^{2}\theta) = 1 + \cos\theta\] Expanding and rearranging terms, we obtain a quadratic equation in terms of \(\cos\theta\): \[2\cos^{2}\theta + \cos\theta - 1 = 0\] We can solve this quadratic equation using the quadratic formula: \[\cos\theta = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\] where \(a=2\), \(b=1\), and \(c=-1\). Plugging in these values, we get: \[\cos\theta = \frac{-1 \pm \sqrt{1+8}}{4} = \frac{-1 \pm 3}{4}\] Therefore, the two possible values of \(\cos\theta\) are \(\frac{1}{2}\) and \(-1\). However, we are given that \(0° \leq \theta \leq 90°\), which means that \(\cos\theta \geq 0\) in this range. Therefore, we must choose \(\cos\theta = \frac{1}{2}\), which gives: \[\cos\theta = \frac{1}{2} \implies \theta = \cos^{-1}\frac{1}{2} \approx 60°\] Therefore, the answer is (B) 60°.
Question 35 Report
Solve: \(4(2^{x^2}) = 8^{x}\)
Answer Details
We can rewrite \(8^x\) as \((2^3)^x = 2^{3x}\). Substituting this into the equation, we have: $$4(2^{x^2}) = 2^{3x}$$ Dividing both sides by \(2^{x^2}\) and simplifying, we get: $$2^{x^2 - 3x + 2} = 0$$ Using the zero-product property of multiplication, we can set the exponent equal to zero: $$x^2 - 3x + 2 = 0$$ Factoring the quadratic, we get: $$(x-1)(x-2) = 0$$ Therefore, the solutions are \(x=1\) and \(x=2\). Hence, the answer is (1, 2).
Question 36 Report
Which of the following is a factor of the polynomial \(6x^{4} + 2x^{3} + 15x + 5\)?
Answer Details
To determine if a given expression is a factor of a polynomial, we can use the factor theorem which states that if a polynomial f(a) is divided by x - a, then the remainder is zero if and only if x - a is a factor of the polynomial. To use the factor theorem in this case, we can try to divide the polynomial \(6x^{4} + 2x^{3} + 15x + 5\) by each of the given expressions to see if the remainder is zero. We can use long division or synthetic division to perform the division. Using synthetic division, we can test each option as follows: - For 3x + 1: -1/3 | 6 2 0 15 5 | -2 2 -5 -------------- 6 0 2 10 0 The remainder is zero, so 3x + 1 is a factor of the polynomial. - For x + 1: -1 | 6 2 0 15 5 | -6 -8 -7 ------------ 6 -4 -8 8 The remainder is not zero, so x + 1 is not a factor of the polynomial. - For 2x + 1: -1/2 | 6 2 0 15 5 | -3 -2.5 ------------ 6 0 3 12.5 The remainder is not zero, so 2x + 1 is not a factor of the polynomial. - For x + 2: -2 | 6 2 0 15 5 | -12 -20 -30 ------------- 6 -10 -20 -15 The remainder is not zero, so x + 2 is not a factor of the polynomial. Therefore, only 3x + 1, is a factor of the polynomial \(6x^{4} + 2x^{3} + 15x + 5\). The answer is: 3x + 1.
Question 37 Report
Find the minimum value of \(y = x^{2} + 6x - 12\).
Answer Details
To find the minimum value of the quadratic function, we need to determine its vertex. We can do this by completing the square or using the formula for the x-coordinate of the vertex, which is given by: $$x = -\frac{b}{2a}$$ where a and b are the coefficients of the quadratic function in standard form: \(y = ax^{2} + bx + c\). In this case, we have: $$y = x^{2} + 6x - 12$$ where a = 1 and b = 6. Therefore, the x-coordinate of the vertex is: $$x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3$$ To find the minimum value of y, we substitute x = -3 into the quadratic function: $$y = (-3)^{2} + 6(-3) - 12 = -21$$ Therefore, the minimum value of y is -21. So, the correct option is (a) -21.
Question 38 Report
If \(\alpha\) and \(\beta\) are the roots of \(x^{2} + x - 2 = 0\), find the value of \(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}\).
Answer Details
To find the value of \(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}\), we first need to find the values of \(\alpha\) and \(\beta\). Given that \(x^{2} + x - 2 = 0\), we can factorize the quadratic equation as \((x-1)(x+2)=0\). Thus, the roots are \(\alpha = 1\) and \(\beta = -2\). Now, we can substitute these values into the expression to get: \[\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{1}{1^{2}} + \frac{1}{(-2)^{2}} = 1 + \frac{1}{4} = \frac{5}{4}\] Therefore, the answer is \(\frac{5}{4}\).
Question 39 Report
Given that \(f : x \to \frac{2x - 1}{x + 2}, x \neq -2\), find \(f^{-1}\), the inverse of f.
Answer Details
Question 40 Report
If \(x = i - 3j\) and \(y = 6i + j\), calculate the angle between x and y.
Question 41 Report
A binary operation A is defined on the set of real numbers, R, by \(a \Delta b = a^{3} - b^{3}\). Without using calculator, find the value of \((\sqrt{3} + \sqrt{2}) \Delta (\sqrt{3} - \sqrt{2})\) leaving the answer in surd form.
\(a\,\Delta\,b=a^{3}-b^{3}\); evaluate \((\sqrt3+\sqrt2)\,\Delta\,(\sqrt3-\sqrt2)\).
By the definition,
\[(\sqrt3+\sqrt2)\,\Delta\,(\sqrt3-\sqrt2)=(\sqrt3+\sqrt2)^{3}-(\sqrt3-\sqrt2)^{3}\]Expand each cube:
\[(\sqrt3+\sqrt2)^{3}=3\sqrt3+3(3)\sqrt2+3\sqrt3(2)+2\sqrt2=9\sqrt3+11\sqrt2\]\[(\sqrt3-\sqrt2)^{3}=3\sqrt3-3(3)\sqrt2+3\sqrt3(2)-2\sqrt2=9\sqrt3-11\sqrt2\]Subtract:
\[(9\sqrt3+11\sqrt2)-(9\sqrt3-11\sqrt2)=22\sqrt2\]The value is \(22\sqrt2\).
Answer Details
\(a\,\Delta\,b=a^{3}-b^{3}\); evaluate \((\sqrt3+\sqrt2)\,\Delta\,(\sqrt3-\sqrt2)\).
By the definition,
\[(\sqrt3+\sqrt2)\,\Delta\,(\sqrt3-\sqrt2)=(\sqrt3+\sqrt2)^{3}-(\sqrt3-\sqrt2)^{3}\]Expand each cube:
\[(\sqrt3+\sqrt2)^{3}=3\sqrt3+3(3)\sqrt2+3\sqrt3(2)+2\sqrt2=9\sqrt3+11\sqrt2\]\[(\sqrt3-\sqrt2)^{3}=3\sqrt3-3(3)\sqrt2+3\sqrt3(2)-2\sqrt2=9\sqrt3-11\sqrt2\]Subtract:
\[(9\sqrt3+11\sqrt2)-(9\sqrt3-11\sqrt2)=22\sqrt2\]The value is \(22\sqrt2\).
Question 42 Report
Solve : \(\tan (2x - 15)° - 1 = 0\), for values of x such that \(0° \leq x \leq 360°\).
Solve \(\tan(2x-15)^{\circ}-1=0\) for \(0^{\circ}\le x\le 360^{\circ}\).
\[\tan(2x-15)^{\circ}=1\]The tangent is \(1\) at \(45^{\circ}\) and every \(180^{\circ}\) thereafter, so
\[2x-15=45+180n\quad(n\in\mathbb{Z})\]\[2x=60+180n\;\Rightarrow\;x=30+90n\]Take values of \(x\) in \([0^{\circ},360^{\circ}]\):
- \(n=0:\ x=30^{\circ}\)
- \(n=1:\ x=120^{\circ}\)
- \(n=2:\ x=210^{\circ}\)
- \(n=3:\ x=300^{\circ}\)
Hence \(x=30^{\circ},\ 120^{\circ},\ 210^{\circ},\ 300^{\circ}\).
Answer Details
Solve \(\tan(2x-15)^{\circ}-1=0\) for \(0^{\circ}\le x\le 360^{\circ}\).
\[\tan(2x-15)^{\circ}=1\]The tangent is \(1\) at \(45^{\circ}\) and every \(180^{\circ}\) thereafter, so
\[2x-15=45+180n\quad(n\in\mathbb{Z})\]\[2x=60+180n\;\Rightarrow\;x=30+90n\]Take values of \(x\) in \([0^{\circ},360^{\circ}]\):
- \(n=0:\ x=30^{\circ}\)
- \(n=1:\ x=120^{\circ}\)
- \(n=2:\ x=210^{\circ}\)
- \(n=3:\ x=300^{\circ}\)
Hence \(x=30^{\circ},\ 120^{\circ},\ 210^{\circ},\ 300^{\circ}\).
Question 43 Report
Points (2, 1) and (6, 7) are opposite vertices of a square which is inscribed in a circle. Find the :
(a) centre of the circle ; (b) equation of the circle.
Square inscribed in a circle with opposite vertices \((2,1)\) and \((6,7)\).
Opposite vertices of the square are the ends of a diagonal, which is also a diameter of the circle.
(a) Centre = midpoint of the diagonal:
\[\left(\frac{2+6}{2},\frac{1+7}{2}\right)=(4,4)\](b) Equation. The diameter length is
\[d=\sqrt{(6-2)^{2}+(7-1)^{2}}=\sqrt{16+36}=\sqrt{52}\]so the radius is \(r=\dfrac{\sqrt{52}}{2}=\sqrt{13}\) and \(r^{2}=13\). The equation of the circle is
\[(x-4)^{2}+(y-4)^{2}=13\]Answer Details
Square inscribed in a circle with opposite vertices \((2,1)\) and \((6,7)\).
Opposite vertices of the square are the ends of a diagonal, which is also a diameter of the circle.
(a) Centre = midpoint of the diagonal:
\[\left(\frac{2+6}{2},\frac{1+7}{2}\right)=(4,4)\](b) Equation. The diameter length is
\[d=\sqrt{(6-2)^{2}+(7-1)^{2}}=\sqrt{16+36}=\sqrt{52}\]so the radius is \(r=\dfrac{\sqrt{52}}{2}=\sqrt{13}\) and \(r^{2}=13\). The equation of the circle is
\[(x-4)^{2}+(y-4)^{2}=13\]Question 44 Report
There are 8 boys and 6 girls in a class. If two students are selected at random from the class, find the probability that they are of
(a) the same sex ;
(b) different sex.
8 boys and 6 girls; two students chosen at random.
Total ways to choose 2 from 14:
\[{}^{14}C_2=\frac{14\times 13}{2}=91\](a) Same sex.
\[{}^{8}C_2+{}^{6}C_2=\frac{8\times7}{2}+\frac{6\times5}{2}=28+15=43\]\[P(\text{same sex})=\frac{43}{91}\](b) Different sex.
\[{}^{8}C_1\times{}^{6}C_1=8\times6=48\]\[P(\text{different sex})=\frac{48}{91}\](Check: \(\dfrac{43}{91}+\dfrac{48}{91}=1\).)
Answer Details
8 boys and 6 girls; two students chosen at random.
Total ways to choose 2 from 14:
\[{}^{14}C_2=\frac{14\times 13}{2}=91\](a) Same sex.
\[{}^{8}C_2+{}^{6}C_2=\frac{8\times7}{2}+\frac{6\times5}{2}=28+15=43\]\[P(\text{same sex})=\frac{43}{91}\](b) Different sex.
\[{}^{8}C_1\times{}^{6}C_1=8\times6=48\]\[P(\text{different sex})=\frac{48}{91}\](Check: \(\dfrac{43}{91}+\dfrac{48}{91}=1\).)
Question 45 Report
Forces of magnitude 3N, 4N and 2N act along the vectors \(j ; -i + j\) and \(i + j\) respectively. Calculate, correct to one decimal place, the magnitude of the resultant of the forces.
Question 46 Report
A car moving with an initial velocity, u, travels in a straight line with a constant acceleration of 3ms\(^{-2}\) until it attains a velocity of 33ms\(^{-1}\) after 6 seconds. Calculate the distance travelled by the car.
Car: initial velocity \(u\), \(a=3\,\text{ms}^{-2}\), reaches \(v=33\,\text{ms}^{-1}\) after \(t=6\,\text{s}\).
First find \(u\) using \(v=u+at\):
\[33=u+3(6)\;\Rightarrow\;u=33-18=15\,\text{ms}^{-1}\]Distance travelled, using \(s=ut+\tfrac12 at^{2}\):
\[s=15(6)+\tfrac12(3)(6)^{2}=90+54=144\,\text{m}\](Check with \(s=\tfrac{u+v}{2}t=\tfrac{15+33}{2}\times 6=24\times 6=144\,\text{m}\).)
The car travels \(144\,\text{m}\).
Answer Details
Car: initial velocity \(u\), \(a=3\,\text{ms}^{-2}\), reaches \(v=33\,\text{ms}^{-1}\) after \(t=6\,\text{s}\).
First find \(u\) using \(v=u+at\):
\[33=u+3(6)\;\Rightarrow\;u=33-18=15\,\text{ms}^{-1}\]Distance travelled, using \(s=ut+\tfrac12 at^{2}\):
\[s=15(6)+\tfrac12(3)(6)^{2}=90+54=144\,\text{m}\](Check with \(s=\tfrac{u+v}{2}t=\tfrac{15+33}{2}\times 6=24\times 6=144\,\text{m}\).)
The car travels \(144\,\text{m}\).
Question 47 Report
(a) Given that \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\); find the :
(i) angle between the vectors AB and AC ; (ii) unit vector along \(\overrightarrow{AB} - \overrightarrow{BC}\).
(b) P, Q, R and M are points in the \(O_{XY}\) plane. If \(\overrightarrow{PQ} = 2i + 8j , \overrightarrow{PR} = 11i - 12j\) and M divides QR internally in the ratio 3 : 7, find \(\overrightarrow{PM}\).
(a) \(\overrightarrow{AB}=\binom{4}{5},\ \overrightarrow{BC}=\binom{-3}{5}\).
First, \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\binom{1}{10}\).
(i) Angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):
\[\cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}|\,|\overrightarrow{AC}|}=\frac{(4)(1)+(5)(10)}{\sqrt{41}\,\sqrt{101}}=\frac{54}{64.35}=0.8392\]\[\theta=\cos^{-1}(0.8392)\approx 33.0^{\circ}\](ii) Unit vector along \(\overrightarrow{AB}-\overrightarrow{BC}\):
\[\overrightarrow{AB}-\overrightarrow{BC}=\binom{4-(-3)}{5-5}=\binom{7}{0},\quad|\cdot|=7\]\[\text{unit vector}=\frac{1}{7}\binom{7}{0}=\binom{1}{0}=i\](b) \(\overrightarrow{PQ}=2i+8j,\ \overrightarrow{PR}=11i-12j\); M divides QR in the ratio \(3:7\).
Using the section formula from P (with \(QM:MR=3:7\)):
\[\overrightarrow{PM}=\frac{7\,\overrightarrow{PQ}+3\,\overrightarrow{PR}}{10}=\frac{7(2i+8j)+3(11i-12j)}{10}\]\[=\frac{(14i+56j)+(33i-36j)}{10}=\frac{47i+20j}{10}=4.7i+2j\]Answer Details
(a) \(\overrightarrow{AB}=\binom{4}{5},\ \overrightarrow{BC}=\binom{-3}{5}\).
First, \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\binom{1}{10}\).
(i) Angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):
\[\cos\theta=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}|\,|\overrightarrow{AC}|}=\frac{(4)(1)+(5)(10)}{\sqrt{41}\,\sqrt{101}}=\frac{54}{64.35}=0.8392\]\[\theta=\cos^{-1}(0.8392)\approx 33.0^{\circ}\](ii) Unit vector along \(\overrightarrow{AB}-\overrightarrow{BC}\):
\[\overrightarrow{AB}-\overrightarrow{BC}=\binom{4-(-3)}{5-5}=\binom{7}{0},\quad|\cdot|=7\]\[\text{unit vector}=\frac{1}{7}\binom{7}{0}=\binom{1}{0}=i\](b) \(\overrightarrow{PQ}=2i+8j,\ \overrightarrow{PR}=11i-12j\); M divides QR in the ratio \(3:7\).
Using the section formula from P (with \(QM:MR=3:7\)):
\[\overrightarrow{PM}=\frac{7\,\overrightarrow{PQ}+3\,\overrightarrow{PR}}{10}=\frac{7(2i+8j)+3(11i-12j)}{10}\]\[=\frac{(14i+56j)+(33i-36j)}{10}=\frac{47i+20j}{10}=4.7i+2j\]Question 48 Report
(a) Edem and his wife were invited to a dinner by a family of 5. They all sat in such a way in such a way that Edem sat next to his wife. Find the number of ways of seating them in a row.
(b) A bag contains 4 red and 5 black identical balls. If 5 balls are selected at random, one after the other with replacement, find the probability that :
(i) a red ball was picked 3 times ; (ii) a black ball was picked at most 2 times.
(a) Edem and his wife join a family of 5, so 7 people sit in a row with Edem next to his wife.
Tie Edem and his wife into a single block. Then there are \(6\) units to arrange in a row:
\[6!=720\]Within the block, Edem and his wife can swap in \(2!=2\) ways. Total:
\[720\times 2=1440\ \text{ways}\](b) Bag of 4 red and 5 black (total 9), 5 draws with replacement, so \(P(R)=\dfrac49,\ P(B)=\dfrac59\).
(i) Red picked exactly 3 times:
\[{}^{5}C_3\left(\tfrac49\right)^{3}\left(\tfrac59\right)^{2}=10\cdot\frac{64}{729}\cdot\frac{25}{81}=\frac{16000}{59049}\approx 0.2710\](ii) Black picked at most 2 times (0, 1 or 2), with \(P(B)=\tfrac59\):
\[P(0)=\left(\tfrac49\right)^{5}=\frac{1024}{59049}\]\[P(1)={}^{5}C_1\left(\tfrac59\right)\left(\tfrac49\right)^{4}=\frac{6400}{59049}\]\[P(2)={}^{5}C_2\left(\tfrac59\right)^{2}\left(\tfrac49\right)^{3}=\frac{16000}{59049}\]\[P(\le 2)=\frac{1024+6400+16000}{59049}=\frac{23424}{59049}\approx 0.3967\]Answer Details
(a) Edem and his wife join a family of 5, so 7 people sit in a row with Edem next to his wife.
Tie Edem and his wife into a single block. Then there are \(6\) units to arrange in a row:
\[6!=720\]Within the block, Edem and his wife can swap in \(2!=2\) ways. Total:
\[720\times 2=1440\ \text{ways}\](b) Bag of 4 red and 5 black (total 9), 5 draws with replacement, so \(P(R)=\dfrac49,\ P(B)=\dfrac59\).
(i) Red picked exactly 3 times:
\[{}^{5}C_3\left(\tfrac49\right)^{3}\left(\tfrac59\right)^{2}=10\cdot\frac{64}{729}\cdot\frac{25}{81}=\frac{16000}{59049}\approx 0.2710\](ii) Black picked at most 2 times (0, 1 or 2), with \(P(B)=\tfrac59\):
\[P(0)=\left(\tfrac49\right)^{5}=\frac{1024}{59049}\]\[P(1)={}^{5}C_1\left(\tfrac59\right)\left(\tfrac49\right)^{4}=\frac{6400}{59049}\]\[P(2)={}^{5}C_2\left(\tfrac59\right)^{2}\left(\tfrac49\right)^{3}=\frac{16000}{59049}\]\[P(\le 2)=\frac{1024+6400+16000}{59049}=\frac{23424}{59049}\approx 0.3967\]Question 49 Report
(a) The functions \(f : x \to x^{2} + 1\) and \(g : x \to 5 - 3x\) are defined on the set of the real numbers, R.
(i) State the domain of \(f^{-1}\), the inverse of f ; (ii) find \(g^{-1} (2)\).
(b) Evaluate : \(\int \frac{(x + 3)}{x^{2} + 6x + 9} \mathrm {d} x\)
(a) \(f:x\to x^{2}+1\) and \(g:x\to 5-3x\) on \(\mathbb{R}\).
(i) Domain of \(f^{-1}\). The domain of the inverse equals the range of \(f\). Since \(x^{2}\ge 0\), \(f(x)=x^{2}+1\ge 1\). Hence
\[\text{Domain of }f^{-1}=\{x:x\ge 1\}=[1,\infty)\](ii) \(g^{-1}(2)\). Let \(y=5-3x\Rightarrow x=\dfrac{5-y}{3}\), so \(g^{-1}(x)=\dfrac{5-x}{3}\).
\[g^{-1}(2)=\frac{5-2}{3}=1\](b) Evaluate \(\displaystyle\int\frac{x+3}{x^{2}+6x+9}\,dx\).
The denominator is a perfect square: \(x^{2}+6x+9=(x+3)^{2}\).
\[\int\frac{x+3}{(x+3)^{2}}\,dx=\int\frac{1}{x+3}\,dx=\ln|x+3|+C\]Answer Details
(a) \(f:x\to x^{2}+1\) and \(g:x\to 5-3x\) on \(\mathbb{R}\).
(i) Domain of \(f^{-1}\). The domain of the inverse equals the range of \(f\). Since \(x^{2}\ge 0\), \(f(x)=x^{2}+1\ge 1\). Hence
\[\text{Domain of }f^{-1}=\{x:x\ge 1\}=[1,\infty)\](ii) \(g^{-1}(2)\). Let \(y=5-3x\Rightarrow x=\dfrac{5-y}{3}\), so \(g^{-1}(x)=\dfrac{5-x}{3}\).
\[g^{-1}(2)=\frac{5-2}{3}=1\](b) Evaluate \(\displaystyle\int\frac{x+3}{x^{2}+6x+9}\,dx\).
The denominator is a perfect square: \(x^{2}+6x+9=(x+3)^{2}\).
\[\int\frac{x+3}{(x+3)^{2}}\,dx=\int\frac{1}{x+3}\,dx=\ln|x+3|+C\]Question 50 Report
(a)(i) Write down the binomial expansion of \((1 + x)^{4}\).
(ii) Use the result in (a)(i) to evaluate, correct to three decimal places \((\frac{5}{4})^{4}\).
(b) The first, second and fifth terms of a linear sequence (A.P) are three consecutive terms of an exponential sequence (G.P). If the first term of the linear sequence is 7, find the common difference.
(a)(i) Binomial expansion of \((1+x)^{4}\):
\[(1+x)^{4}=1+4x+6x^{2}+4x^{3}+x^{4}\](ii) Evaluate \(\left(\dfrac54\right)^{4}=\left(1+\dfrac14\right)^{4}\) with \(x=\dfrac14\):
\[=1+4\left(\tfrac14\right)+6\left(\tfrac14\right)^{2}+4\left(\tfrac14\right)^{3}+\left(\tfrac14\right)^{4}\]\[=1+1+0.375+0.0625+0.003906=2.441406\approx 2.441\](b) A.P. with first term 7; its 1st, 2nd and 5th terms form a G.P.
The three A.P. terms are \(7,\ 7+d,\ 7+4d\). For a G.P. the middle term squared equals the product of the outer terms:
\[(7+d)^{2}=7(7+4d)\]\[49+14d+d^{2}=49+28d\]\[d^{2}-14d=0\;\Rightarrow\;d(d-14)=0\]Taking the non-trivial value, the common difference is \(d=14\).
Answer Details
(a)(i) Binomial expansion of \((1+x)^{4}\):
\[(1+x)^{4}=1+4x+6x^{2}+4x^{3}+x^{4}\](ii) Evaluate \(\left(\dfrac54\right)^{4}=\left(1+\dfrac14\right)^{4}\) with \(x=\dfrac14\):
\[=1+4\left(\tfrac14\right)+6\left(\tfrac14\right)^{2}+4\left(\tfrac14\right)^{3}+\left(\tfrac14\right)^{4}\]\[=1+1+0.375+0.0625+0.003906=2.441406\approx 2.441\](b) A.P. with first term 7; its 1st, 2nd and 5th terms form a G.P.
The three A.P. terms are \(7,\ 7+d,\ 7+4d\). For a G.P. the middle term squared equals the product of the outer terms:
\[(7+d)^{2}=7(7+4d)\]\[49+14d+d^{2}=49+28d\]\[d^{2}-14d=0\;\Rightarrow\;d(d-14)=0\]Taking the non-trivial value, the common difference is \(d=14\).
Question 51 Report
(a) If \(\frac{\sqrt{5} + 4}{3 - 2\sqrt{5}} - \frac{2 + \sqrt{5}}{4 - 2\sqrt{5}} = a + b\sqrt{5}\), find the values of a and b.
(b)(i) Evaluate : \(\begin{vmatrix} 2 & -1 & 2 \\ 1 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix}\)
(ii) Using the result in b(i), find, correct to two decimal places, the value of x in the system of equations.
\(2x - y + 2z + 5 = 0\)
\(x + 3y + 4z - 1 = 0\)
\(x + 2y + z + 2 = 0\)
(a) Simplify \(\dfrac{\sqrt5+4}{3-2\sqrt5}-\dfrac{2+\sqrt5}{4-2\sqrt5}=a+b\sqrt5\).
Rationalise the first fraction by \(3+2\sqrt5\) (denominator \(9-20=-11\)):
\[\frac{(\sqrt5+4)(3+2\sqrt5)}{-11}=\frac{22+11\sqrt5}{-11}=-2-\sqrt5\]Rationalise the second by \(4+2\sqrt5\) (denominator \(16-20=-4\)):
\[\frac{(2+\sqrt5)(4+2\sqrt5)}{-4}=\frac{18+8\sqrt5}{-4}=-\frac{9}{2}-2\sqrt5\]Subtract:
\[(-2-\sqrt5)-\left(-\frac{9}{2}-2\sqrt5\right)=-2+\frac{9}{2}+(-\sqrt5+2\sqrt5)=\frac{5}{2}+\sqrt5\]Hence \(a=\dfrac{5}{2},\ b=1\).
(b)(i) Evaluate the determinant.
\[\begin{vmatrix}2&-1&2\\1&3&4\\1&2&1\end{vmatrix}=2(3-8)-(-1)(1-4)+2(2-3)=-10-3-2=-15\](ii) Writing the system as \(2x-y+2z=-5,\ x+3y+4z=1,\ x+2y+z=-2\), the coefficient determinant is \(-15\). By Cramer's rule, replace the x-column with the constants:
\[D_x=\begin{vmatrix}-5&-1&2\\1&3&4\\-2&2&1\end{vmatrix}=-5(3-8)+1(1+8)+2(2+6)=25+9+16=50\]\[x=\frac{D_x}{D}=\frac{50}{-15}=-\frac{10}{3}\approx -3.33\]Answer Details
(a) Simplify \(\dfrac{\sqrt5+4}{3-2\sqrt5}-\dfrac{2+\sqrt5}{4-2\sqrt5}=a+b\sqrt5\).
Rationalise the first fraction by \(3+2\sqrt5\) (denominator \(9-20=-11\)):
\[\frac{(\sqrt5+4)(3+2\sqrt5)}{-11}=\frac{22+11\sqrt5}{-11}=-2-\sqrt5\]Rationalise the second by \(4+2\sqrt5\) (denominator \(16-20=-4\)):
\[\frac{(2+\sqrt5)(4+2\sqrt5)}{-4}=\frac{18+8\sqrt5}{-4}=-\frac{9}{2}-2\sqrt5\]Subtract:
\[(-2-\sqrt5)-\left(-\frac{9}{2}-2\sqrt5\right)=-2+\frac{9}{2}+(-\sqrt5+2\sqrt5)=\frac{5}{2}+\sqrt5\]Hence \(a=\dfrac{5}{2},\ b=1\).
(b)(i) Evaluate the determinant.
\[\begin{vmatrix}2&-1&2\\1&3&4\\1&2&1\end{vmatrix}=2(3-8)-(-1)(1-4)+2(2-3)=-10-3-2=-15\](ii) Writing the system as \(2x-y+2z=-5,\ x+3y+4z=1,\ x+2y+z=-2\), the coefficient determinant is \(-15\). By Cramer's rule, replace the x-column with the constants:
\[D_x=\begin{vmatrix}-5&-1&2\\1&3&4\\-2&2&1\end{vmatrix}=-5(3-8)+1(1+8)+2(2+6)=25+9+16=50\]\[x=\frac{D_x}{D}=\frac{50}{-15}=-\frac{10}{3}\approx -3.33\]Question 52 Report
The table shows the distribution of the heights of a group of people.
| Height (m) |
0.4 - 0.5 | 0.6 - 0.9 | 1.0 - 1.2 | 1.3 - 1.4 | 1.5 - 1.7 |
| Number of people | 2 | 8 | 12 | 6 | 6 |
(a) Draw a histogram to illustrate the distribution.
(b) Using an assumed mean of 1.1m, find, correct to one decimal place, the mean height of the group.
(a) Histogram. The classes have unequal widths, so the bars must be drawn using frequency density, not raw frequency:
\[\text{frequency density}=\frac{\text{frequency}}{\text{class width}}\]First close the gaps between successive classes to obtain the true class boundaries, then compute the width and frequency density of each class.
| Height (m) | Class boundaries | Width | Frequency | Freq. density |
| 0.4 - 0.5 | 0.35 - 0.55 | 0.2 | 2 | 10 |
| 0.6 - 0.9 | 0.55 - 0.95 | 0.4 | 8 | 20 |
| 1.0 - 1.2 | 0.95 - 1.25 | 0.3 | 12 | 40 |
| 1.3 - 1.4 | 1.25 - 1.45 | 0.2 | 6 | 30 |
| 1.5 - 1.7 | 1.45 - 1.75 | 0.3 | 6 | 20 |
Plot frequency density on the vertical axis against height (class boundaries) on the horizontal axis, drawing the bars with no gaps between them:
(b) Mean height (assumed-mean method, \(A=1.1\) m). Use the class mid-values \(x\) and the deviation \(d=x-A\):
\[\bar{x}=A+\frac{\sum fd}{\sum f}\]| Height (m) | Mid-value \(x\) | Frequency \(f\) | \(d=x-1.1\) | \(fd\) |
| 0.4 - 0.5 | 0.45 | 2 | -0.65 | -1.30 |
| 0.6 - 0.9 | 0.75 | 8 | -0.35 | -2.80 |
| 1.0 - 1.2 | 1.10 | 12 | 0.00 | 0.00 |
| 1.3 - 1.4 | 1.35 | 6 | 0.25 | 1.50 |
| 1.5 - 1.7 | 1.60 | 6 | 0.50 | 3.00 |
| Total | 34 | 0.40 |
Therefore
\[\bar{x}=1.1+\frac{0.40}{34}=1.1+0.0118=1.1118\text{ m}\] \[\boxed{\bar{x}\approx 1.1\text{ m (to 1 decimal place)}}\]The mean height of the group is 1.1 m.
Answer Details
(a) Histogram. The classes have unequal widths, so the bars must be drawn using frequency density, not raw frequency:
\[\text{frequency density}=\frac{\text{frequency}}{\text{class width}}\]First close the gaps between successive classes to obtain the true class boundaries, then compute the width and frequency density of each class.
| Height (m) | Class boundaries | Width | Frequency | Freq. density |
| 0.4 - 0.5 | 0.35 - 0.55 | 0.2 | 2 | 10 |
| 0.6 - 0.9 | 0.55 - 0.95 | 0.4 | 8 | 20 |
| 1.0 - 1.2 | 0.95 - 1.25 | 0.3 | 12 | 40 |
| 1.3 - 1.4 | 1.25 - 1.45 | 0.2 | 6 | 30 |
| 1.5 - 1.7 | 1.45 - 1.75 | 0.3 | 6 | 20 |
Plot frequency density on the vertical axis against height (class boundaries) on the horizontal axis, drawing the bars with no gaps between them:
(b) Mean height (assumed-mean method, \(A=1.1\) m). Use the class mid-values \(x\) and the deviation \(d=x-A\):
\[\bar{x}=A+\frac{\sum fd}{\sum f}\]| Height (m) | Mid-value \(x\) | Frequency \(f\) | \(d=x-1.1\) | \(fd\) |
| 0.4 - 0.5 | 0.45 | 2 | -0.65 | -1.30 |
| 0.6 - 0.9 | 0.75 | 8 | -0.35 | -2.80 |
| 1.0 - 1.2 | 1.10 | 12 | 0.00 | 0.00 |
| 1.3 - 1.4 | 1.35 | 6 | 0.25 | 1.50 |
| 1.5 - 1.7 | 1.60 | 6 | 0.50 | 3.00 |
| Total | 34 | 0.40 |
Therefore
\[\bar{x}=1.1+\frac{0.40}{34}=1.1+0.0118=1.1118\text{ m}\] \[\boxed{\bar{x}\approx 1.1\text{ m (to 1 decimal place)}}\]The mean height of the group is 1.1 m.
Question 53 Report
Five finalists in a beauty pageant were ranked by two judges X and Y as shown in the table :
| Judges | Anne | Linda | Susan | Rose | Erica |
| X | 1 | 4 | 3 | 5 | 2 |
| Y | 3 | 2 | 4 | 5 | 1 |
Calculate the Spearman's rank correlation coefficient.
Spearman's rank correlation coefficient is \(r_s=1-\dfrac{6\sum d^2}{n(n^2-1)}\), where \(d\) is the difference between paired ranks and \(n\) is the number of individuals. Here \(n=5\).
| Finalist | Judge X | Judge Y | \(d=X-Y\) | \(d^2\) |
|---|---|---|---|---|
| Anne | 1 | 3 | -2 | 4 |
| Linda | 4 | 2 | 2 | 4 |
| Susan | 3 | 4 | -1 | 1 |
| Rose | 5 | 5 | 0 | 0 |
| Erica | 2 | 1 | 1 | 1 |
| \(\sum d^2\) | 10 | |||
\[r_s=1-\frac{6\times 10}{5(5^2-1)}=1-\frac{60}{5\times 24}=1-\frac{60}{120}=1-0.5=0.5\]
The Spearman's rank correlation coefficient is 0.5, indicating a moderate positive agreement between the two judges.
Answer Details
Spearman's rank correlation coefficient is \(r_s=1-\dfrac{6\sum d^2}{n(n^2-1)}\), where \(d\) is the difference between paired ranks and \(n\) is the number of individuals. Here \(n=5\).
| Finalist | Judge X | Judge Y | \(d=X-Y\) | \(d^2\) |
|---|---|---|---|---|
| Anne | 1 | 3 | -2 | 4 |
| Linda | 4 | 2 | 2 | 4 |
| Susan | 3 | 4 | -1 | 1 |
| Rose | 5 | 5 | 0 | 0 |
| Erica | 2 | 1 | 1 | 1 |
| \(\sum d^2\) | 10 | |||
\[r_s=1-\frac{6\times 10}{5(5^2-1)}=1-\frac{60}{5\times 24}=1-\frac{60}{120}=1-0.5=0.5\]
The Spearman's rank correlation coefficient is 0.5, indicating a moderate positive agreement between the two judges.
Question 54 Report
If \(f ' '(x) = 2\), \(f ' (1) = 0\) and \(f(0) = - 8\), find f(x).
Given \(f''(x)=2,\ f'(1)=0,\ f(0)=-8\), find \(f(x)\).
Integrate \(f''(x)=2\) once:
\[f'(x)=2x+C_1\]Use \(f'(1)=0\):
\[2(1)+C_1=0\;\Rightarrow\;C_1=-2,\qquad f'(x)=2x-2\]Integrate again:
\[f(x)=x^{2}-2x+C_2\]Use \(f(0)=-8\):
\[0-0+C_2=-8\;\Rightarrow\;C_2=-8\]Therefore
\[f(x)=x^{2}-2x-8\]Answer Details
Given \(f''(x)=2,\ f'(1)=0,\ f(0)=-8\), find \(f(x)\).
Integrate \(f''(x)=2\) once:
\[f'(x)=2x+C_1\]Use \(f'(1)=0\):
\[2(1)+C_1=0\;\Rightarrow\;C_1=-2,\qquad f'(x)=2x-2\]Integrate again:
\[f(x)=x^{2}-2x+C_2\]Use \(f(0)=-8\):
\[0-0+C_2=-8\;\Rightarrow\;C_2=-8\]Therefore
\[f(x)=x^{2}-2x-8\]Question 55 Report
(a) A bucket full of water with a mass of 8kg is pulled out of a well with a light inextensible rope. Find its acceleration when tha tension in the rope is 150N. [Take \(g = 10ms^{-2}\)].
(b) A mass of 12kg is acted upon by a force F, changing its speed from 15 m/s to 25 m/s after covering a distance of 50m. Find the :
(i) value of F ; (ii) distance covered when its speed is 35 m/s.
(a) Bucket of mass \(8\,\text{kg}\) pulled up, rope tension \(150\,\text{N}\), \(g=10\,\text{ms}^{-2}\).
Newton's second law (upward positive):
\[T-mg=ma\;\Rightarrow\;150-8(10)=8a\]\[70=8a\;\Rightarrow\;a=8.75\,\text{ms}^{-2}\ \text{(upward)}\](b) Mass \(12\,\text{kg}\), speed rises \(15\to 25\,\text{m/s}\) over \(50\,\text{m}\).
(i) Find the acceleration from \(v^{2}=u^{2}+2as\):
\[25^{2}=15^{2}+2a(50)\;\Rightarrow\;625=225+100a\;\Rightarrow\;a=4\,\text{ms}^{-2}\]\[F=ma=12\times 4=48\,\text{N}\](ii) Distance from the start (at \(15\,\text{m/s}\)) until the speed is \(35\,\text{m/s}\):
\[35^{2}=15^{2}+2(4)s\;\Rightarrow\;1225=225+8s\;\Rightarrow\;8s=1000\]\[s=125\,\text{m}\]Answer Details
(a) Bucket of mass \(8\,\text{kg}\) pulled up, rope tension \(150\,\text{N}\), \(g=10\,\text{ms}^{-2}\).
Newton's second law (upward positive):
\[T-mg=ma\;\Rightarrow\;150-8(10)=8a\]\[70=8a\;\Rightarrow\;a=8.75\,\text{ms}^{-2}\ \text{(upward)}\](b) Mass \(12\,\text{kg}\), speed rises \(15\to 25\,\text{m/s}\) over \(50\,\text{m}\).
(i) Find the acceleration from \(v^{2}=u^{2}+2as\):
\[25^{2}=15^{2}+2a(50)\;\Rightarrow\;625=225+100a\;\Rightarrow\;a=4\,\text{ms}^{-2}\]\[F=ma=12\times 4=48\,\text{N}\](ii) Distance from the start (at \(15\,\text{m/s}\)) until the speed is \(35\,\text{m/s}\):
\[35^{2}=15^{2}+2(4)s\;\Rightarrow\;1225=225+8s\;\Rightarrow\;8s=1000\]\[s=125\,\text{m}\]
