WAEC SSCE - General Mathematics - 1995 (Essay)
Question 1 Rapport
(a) Using mathematical tables, find ; (i) \(2 \sin 63.35°\) ; (ii) \(\log \cos 44.74°\);
(b) Find the value of K given that \(\log K - \log (K - 2) = \log 5\);
(c) Use logarithm tables to evaluate \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\)
(a)(i) From tables, \(\sin 63.35° = 0.8937\).
\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)
(ii) \(\cos 44.74° = 0.7103\).
\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)
(b) \(\log K - \log(K-2) = \log 5\)
\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)
\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)
(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.
| Expression | Log |
|---|---|
| \((3.68)^2\) | \(2 \times 0.5658 = 1.1316\) |
| \(6.705\) | \(0.8264\) |
| Numerator total | \(1.9580\) |
| \(\sqrt{0.3581}\) | \(\tfrac{1}{2}(\bar{1}.5540) = \bar{1}.7770\) |
\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)
Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).
Détails de la réponse
(a)(i) From tables, \(\sin 63.35° = 0.8937\).
\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)
(ii) \(\cos 44.74° = 0.7103\).
\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)
(b) \(\log K - \log(K-2) = \log 5\)
\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)
\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)
(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.
| Expression | Log |
|---|---|
| \((3.68)^2\) | \(2 \times 0.5658 = 1.1316\) |
| \(6.705\) | \(0.8264\) |
| Numerator total | \(1.9580\) |
| \(\sqrt{0.3581}\) | \(\tfrac{1}{2}(\bar{1}.5540) = \bar{1}.7770\) |
\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)
Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).
Question 2 Rapport
(a) Given that \(p = x + ym^{3}\), find m in terms of p, x and y.
(b) Using the method of completing the square, find the roots of the equation \(x^{2} - 6x + 7 = 0\), correct to 1 decimal place.
(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.
(a) \(p = x + ym^3\). Make \(m\) the subject.
\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)
(b) \(x^2 - 6x + 7 = 0\) by completing the square.
\(x^2 - 6x = -7\)
Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:
\((x-3)^2 = -7 + 9 = 2\)
\(x - 3 = \pm\sqrt{2} = \pm 1.414\)
\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)
(c) Let the consecutive odd numbers be \(n\) and \(n+2\).
\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)
\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)
Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).
Détails de la réponse
(a) \(p = x + ym^3\). Make \(m\) the subject.
\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)
(b) \(x^2 - 6x + 7 = 0\) by completing the square.
\(x^2 - 6x = -7\)
Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:
\((x-3)^2 = -7 + 9 = 2\)
\(x - 3 = \pm\sqrt{2} = \pm 1.414\)
\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)
(c) Let the consecutive odd numbers be \(n\) and \(n+2\).
\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)
\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)
Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).
Question 3 Rapport
(a) Factorise : \(px - 2px - 4qy + 2py\)
(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \((P \cup Q)' = P' \cap Q'\)
(a) Grouping the four terms \(px - 2qx - 4qy + 2py\) in pairs: \[(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q) = (p - 2q)(x + 2y).\] So the expression factorises as \(\mathbf{(p - 2q)(x + 2y)}.\)
(b) \(U = \{1,2,3,4,5,6,7,8,9,10\},\ P = \{1,2,4,6,10\},\ Q = \{2,3,6,9\}.\)
Left side: \(P \cup Q = \{1,2,3,4,6,9,10\}\), so \[(P \cup Q)' = \{5, 7, 8\}.\]
Right side: \(P' = \{3,5,7,8,9\}\) and \(Q' = \{1,4,5,7,8,10\}\), so \[P' \cap Q' = \{5, 7, 8\}.\]
Since \((P \cup Q)' = \{5,7,8\} = P' \cap Q'\), the identity is verified (De Morgan's law).
Détails de la réponse
(a) Grouping the four terms \(px - 2qx - 4qy + 2py\) in pairs: \[(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q) = (p - 2q)(x + 2y).\] So the expression factorises as \(\mathbf{(p - 2q)(x + 2y)}.\)
(b) \(U = \{1,2,3,4,5,6,7,8,9,10\},\ P = \{1,2,4,6,10\},\ Q = \{2,3,6,9\}.\)
Left side: \(P \cup Q = \{1,2,3,4,6,9,10\}\), so \[(P \cup Q)' = \{5, 7, 8\}.\]
Right side: \(P' = \{3,5,7,8,9\}\) and \(Q' = \{1,4,5,7,8,10\}\), so \[P' \cap Q' = \{5, 7, 8\}.\]
Since \((P \cup Q)' = \{5,7,8\} = P' \cap Q'\), the identity is verified (De Morgan's law).
Question 4 Rapport
A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining
(a) two red balls ;
(b) two blue balls or two black balls ;
(c) one black and one red ball in any order.
Total balls \(= 5 + 3 + 2 = 10\). Since each ball is replaced, the two draws are independent and the probabilities on each draw stay the same.
\(P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2},\quad P(\text{black}) = \tfrac{3}{10},\quad P(\text{red}) = \tfrac{2}{10} = \tfrac{1}{5}\)
(a) Two red balls:
\(P = \tfrac{1}{5}\times\tfrac{1}{5} = \dfrac{1}{25}\)
(b) Two blue OR two black (mutually exclusive, so add):
\(P = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{3}{10}\right)^2 = \tfrac{1}{4} + \tfrac{9}{100} = \tfrac{25}{100} + \tfrac{9}{100} = \dfrac{34}{100} = \dfrac{17}{50}\)
(c) One black and one red in any order (black-then-red or red-then-black):
\(P = 2\times\tfrac{3}{10}\times\tfrac{2}{10} = 2\times\tfrac{6}{100} = \dfrac{12}{100} = \dfrac{3}{25}\)
Détails de la réponse
Total balls \(= 5 + 3 + 2 = 10\). Since each ball is replaced, the two draws are independent and the probabilities on each draw stay the same.
\(P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2},\quad P(\text{black}) = \tfrac{3}{10},\quad P(\text{red}) = \tfrac{2}{10} = \tfrac{1}{5}\)
(a) Two red balls:
\(P = \tfrac{1}{5}\times\tfrac{1}{5} = \dfrac{1}{25}\)
(b) Two blue OR two black (mutually exclusive, so add):
\(P = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{3}{10}\right)^2 = \tfrac{1}{4} + \tfrac{9}{100} = \tfrac{25}{100} + \tfrac{9}{100} = \dfrac{34}{100} = \dfrac{17}{50}\)
(c) One black and one red in any order (black-then-red or red-then-black):
\(P = 2\times\tfrac{3}{10}\times\tfrac{2}{10} = 2\times\tfrac{6}{100} = \dfrac{12}{100} = \dfrac{3}{25}\)
Question 5 Rapport
The frequency table shows the marks scored by 32 students in a test.
| Marks scored | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No of students | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
Find the :
(a)(i) mean ; (ii) median ; (iii) mode of the marks;
(b) percentage of the students who scored at least 8 marks.
Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).
| Mark (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| f | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
| fx | 2 | 6 | 12 | 16 | 20 | 24 | 35 | 24 | 18 | 10 |
| Cum. f | 2 | 5 | 9 | 13 | 17 | 21 | 26 | 29 | 31 | 32 |
(a)(i) Mean. \(\sum fx = 167\).
\[\bar{x}=\frac{167}{32}\approx 5.22.\]
(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.
\[\text{Median}=\frac{5+5}{2}=5.\]
(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.
(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.
\[\frac{6}{32}\times100\% = 18.75\%.\]
Détails de la réponse
Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).
| Mark (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| f | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
| fx | 2 | 6 | 12 | 16 | 20 | 24 | 35 | 24 | 18 | 10 |
| Cum. f | 2 | 5 | 9 | 13 | 17 | 21 | 26 | 29 | 31 | 32 |
(a)(i) Mean. \(\sum fx = 167\).
\[\bar{x}=\frac{167}{32}\approx 5.22.\]
(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.
\[\text{Median}=\frac{5+5}{2}=5.\]
(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.
(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.
\[\frac{6}{32}\times100\% = 18.75\%.\]
Question 6 Rapport
(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/.
(b) P and Q are two towns on the earth's surface on latitude 56°N. Thei longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and \(\pi = \frac{22}{7}\)]
(a) Finding /PZ/
The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.
In a parallelogram opposite sides are equal, so:
- In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
- In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).
Since \(P, Q, Y, Z\) are collinear in that order:
\[ PZ = PQ + QY + YZ \]
\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]
/PZ/ = 8 cm.
(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N
Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:
\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]
The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:
\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]
Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):
\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]
\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]
\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]
Distance PQ \(\approx\) 4374 km (to the nearest km).
Détails de la réponse
(a) Finding /PZ/
The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.
In a parallelogram opposite sides are equal, so:
- In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
- In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).
Since \(P, Q, Y, Z\) are collinear in that order:
\[ PZ = PQ + QY + YZ \]
\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]
/PZ/ = 8 cm.
(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N
Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:
\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]
The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:
\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]
Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):
\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]
\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]
\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]
Distance PQ \(\approx\) 4374 km (to the nearest km).
Question 7 Rapport
(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.
[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]
(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.
(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.
\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)
(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).
- \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
- \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
- \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)
Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.
\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)
Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).
Détails de la réponse
(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.
\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)
(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).
- \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
- \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
- \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)
Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.
\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)
Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).
Question 8 Rapport
The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.
| % Mark | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 |
| No of pupils | 68 | 184 | 294 | 402 | 480 | 310 | 164 | 98 |
(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.
(a) Cumulative frequency table
| Marks (%) | Number of pupils, \(f\) | Cumulative frequency |
|---|---|---|
| 11 - 20 | 68 | 68 |
| 21 - 30 | 184 | 252 |
| 31 - 40 | 294 | 546 |
| 41 - 50 | 402 | 948 |
| 51 - 60 | 480 | 1428 |
| 61 - 70 | 310 | 1738 |
| 71 - 80 | 164 | 1902 |
| 81 - 90 | 98 | 2000 |
Using upper class boundaries, plot the cumulative frequencies and join the points with a smooth increasing curve.
(b)(i) Cut-off mark
The 300 candidates admitted are the highest scorers. Hence the number below the cut-off mark is
\[2000-300=1700.\]
From the ogive, at cumulative frequency \(1700\), the corresponding mark is approximately \(68\).
Therefore, the cut-off mark is 68 marks (68%).
(b)(ii) Probability of scoring at least \(45\%\)
From the ogive, the cumulative frequency at \(45\%\) is approximately \(720\).
\[\text{Number scoring at least }45\%=2000-720=1280.\]
\[P(\text{score at least }45\%)=\frac{1280}{2000}=0.64.\]
Détails de la réponse
(a) Cumulative frequency table
| Marks (%) | Number of pupils, \(f\) | Cumulative frequency |
|---|---|---|
| 11 - 20 | 68 | 68 |
| 21 - 30 | 184 | 252 |
| 31 - 40 | 294 | 546 |
| 41 - 50 | 402 | 948 |
| 51 - 60 | 480 | 1428 |
| 61 - 70 | 310 | 1738 |
| 71 - 80 | 164 | 1902 |
| 81 - 90 | 98 | 2000 |
Using upper class boundaries, plot the cumulative frequencies and join the points with a smooth increasing curve.
(b)(i) Cut-off mark
The 300 candidates admitted are the highest scorers. Hence the number below the cut-off mark is
\[2000-300=1700.\]
From the ogive, at cumulative frequency \(1700\), the corresponding mark is approximately \(68\).
Therefore, the cut-off mark is 68 marks (68%).
(b)(ii) Probability of scoring at least \(45\%\)
From the ogive, the cumulative frequency at \(45\%\) is approximately \(720\).
\[\text{Number scoring at least }45\%=2000-720=1280.\]
\[P(\text{score at least }45\%)=\frac{1280}{2000}=0.64.\]
Question 9 Rapport
The quantity y is partly constant and partly varies inversely as the square of x.
(a) Write down the relationship between x and y.
(b) When x = 1, y = 11 and when x = 2, y = 5, find the value of y when x = 4.
(a) "Partly constant and partly varies inversely as the square of \(x\)" means \[y = a + \frac{b}{x^2},\] where \(a\) and \(b\) are constants.
(b) Using the given values:
When \(x = 1, y = 11\): \(a + b = 11.\quad(1)\)
When \(x = 2, y = 5\): \(a + \dfrac{b}{4} = 5.\quad(2)\)
Subtract \((2)\) from \((1)\): \(b - \dfrac{b}{4} = 6 \Rightarrow \dfrac{3b}{4} = 6 \Rightarrow b = 8.\) Then \(a = 11 - 8 = 3.\)
So \(y = 3 + \dfrac{8}{x^2}.\) When \(x = 4\): \[y = 3 + \frac{8}{16} = 3 + 0.5 = \mathbf{3.5}.\]
Détails de la réponse
(a) "Partly constant and partly varies inversely as the square of \(x\)" means \[y = a + \frac{b}{x^2},\] where \(a\) and \(b\) are constants.
(b) Using the given values:
When \(x = 1, y = 11\): \(a + b = 11.\quad(1)\)
When \(x = 2, y = 5\): \(a + \dfrac{b}{4} = 5.\quad(2)\)
Subtract \((2)\) from \((1)\): \(b - \dfrac{b}{4} = 6 \Rightarrow \dfrac{3b}{4} = 6 \Rightarrow b = 8.\) Then \(a = 11 - 8 = 3.\)
So \(y = 3 + \dfrac{8}{x^2}.\) When \(x = 4\): \[y = 3 + \frac{8}{16} = 3 + 0.5 = \mathbf{3.5}.\]
Question 10 Rapport
(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| \(y = 2\cos 2x - 1\) | 1.0 | 0 | 1.0 |
(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \(y = 2 \cos 2x - 1\) for \(0° \leq x \leq 180°\).
(c) On the same axis, draw the graph of \(y = \frac{1}{180} (x - 360)\)
(d) Use your graphs to find the : (i) values of x for which \(2 \cos 2x + \frac{1}{2} = 0\); (ii) roots of the equation \(2 \cos 2x - \frac{x}{180} + 1 = 0\).
(a) For each value of x, evaluate \(y=2\cos 2x-1\).
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=2\cos2x-1\) | 1.0 | 0 | -2.0 | -3.0 | -2.0 | 0 | 1.0 |
(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.
For the straight line, the plotting values are:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=\dfrac{x-360}{180}\) | -2.00 | -1.83 | -1.67 | -1.50 | -1.33 | -1.17 | -1.00 |
(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]
(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]
Détails de la réponse
(a) For each value of x, evaluate \(y=2\cos 2x-1\).
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=2\cos2x-1\) | 1.0 | 0 | -2.0 | -3.0 | -2.0 | 0 | 1.0 |
(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.
For the straight line, the plotting values are:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=\dfrac{x-360}{180}\) | -2.00 | -1.83 | -1.67 | -1.50 | -1.33 | -1.17 | -1.00 |
(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]
(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]
Question 11 Rapport
(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :
(i) Y from Z ; (ii) Z from X .
(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E.
(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.
[Take \(\pi = \frac{22}{7}\), R = 6400km].
(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).
\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)
From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).
Z is due East of X, so its North component is \(0\):
\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)
(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).
East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).
(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).
(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).
(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)
\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).
(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).
Détails de la réponse
(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).
\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)
From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).
Z is due East of X, so its North component is \(0\):
\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)
(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).
East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).
(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).
(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).
(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)
\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).
(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).
Question 12 Rapport
(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105°.
(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.
(c) Through D, construct a line I perpendicular to BC.
(d) If the line I meets AC at P, measure /BP/.
Construction and measurement
- Draw \(BC=8.1\text{ cm}\).
- At \(B\), construct an angle of \(105^\circ\). Construct \(90^\circ\) and \(120^\circ\), then bisect the angle between them to obtain \(105^\circ\).
- On the \(105^\circ\) ray, use compasses to mark \(BA=7.5\text{ cm}\). Join \(A\) to \(C\).
- Divide \(BC\) in the ratio \(3:2\). Since there are \(3+2=5\) equal parts, \[ BD=\frac{3}{5}\times 8.1=4.86\text{ cm}. \] Mark \(D\) \(4.86\text{ cm}\) from \(B\) along \(BC\).
- Construct the perpendicular to \(BC\) through \(D\). Where it meets \(AC\) is \(P\).
- Measure \(BP\).
The measurement from \(B\) to \(P\) is approximately
\[ \boxed{BP\approx 5.4\text{ cm}} \]Examination reminder: The ratio \(3:2\) means that \(BC\) is split into \(5\) equal parts altogether, with \(D\) located \(3\) of those parts from \(B\), not \(2\) parts from \(B\).
Détails de la réponse
Construction and measurement
- Draw \(BC=8.1\text{ cm}\).
- At \(B\), construct an angle of \(105^\circ\). Construct \(90^\circ\) and \(120^\circ\), then bisect the angle between them to obtain \(105^\circ\).
- On the \(105^\circ\) ray, use compasses to mark \(BA=7.5\text{ cm}\). Join \(A\) to \(C\).
- Divide \(BC\) in the ratio \(3:2\). Since there are \(3+2=5\) equal parts, \[ BD=\frac{3}{5}\times 8.1=4.86\text{ cm}. \] Mark \(D\) \(4.86\text{ cm}\) from \(B\) along \(BC\).
- Construct the perpendicular to \(BC\) through \(D\). Where it meets \(AC\) is \(P\).
- Measure \(BP\).
The measurement from \(B\) to \(P\) is approximately
\[ \boxed{BP\approx 5.4\text{ cm}} \]Examination reminder: The ratio \(3:2\) means that \(BC\) is split into \(5\) equal parts altogether, with \(D\) located \(3\) of those parts from \(B\), not \(2\) parts from \(B\).
