WAEC SSCE - Further Mathematics - 2015 (Essay)
Frage 1 Bericht
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 \(22\sqrt2\).
Antwortdetails
\(a\,\Delta\,b=a^{3}-b^{3}\); 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 \(22\sqrt2\).
Frage 2 Bericht
(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 \(QM:MR=3:7\)):
\[\overrightarrow{PM}=\frac{7\,\overrightarrow{PQ}+3\,\overrightarrow{PR}}{10}=\frac{7(2i+8j)+3(11i-12j)}{10}\]\[=\frac{(14i+56j)+(33i-36j)}{10}=\frac{47i+20j}{10}=4.7i+2j\]Antwortdetails
(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 \(QM:MR=3:7\)):
\[\overrightarrow{PM}=\frac{7\,\overrightarrow{PQ}+3\,\overrightarrow{PR}}{10}=\frac{7(2i+8j)+3(11i-12j)}{10}\]\[=\frac{(14i+56j)+(33i-36j)}{10}=\frac{47i+20j}{10}=4.7i+2j\]Frage 3 Bericht
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
\[f(x)=x^{2}-2x-8\]Antwortdetails
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
\[f(x)=x^{2}-2x-8\]Frage 4 Bericht
(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:
\[D_x=\begin{vmatrix}-5&-1&2\\1&3&4\\-2&2&1\end{vmatrix}=-5(3-8)+1(1+8)+2(2+6)=25+9+16=50\]\[x=\frac{D_x}{D}=\frac{50}{-15}=-\frac{10}{3}\approx -3.33\]Antwortdetails
(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:
\[D_x=\begin{vmatrix}-5&-1&2\\1&3&4\\-2&2&1\end{vmatrix}=-5(3-8)+1(1+8)+2(2+6)=25+9+16=50\]\[x=\frac{D_x}{D}=\frac{50}{-15}=-\frac{10}{3}\approx -3.33\]Frage 5 Bericht
(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 \(d=14\).
Antwortdetails
(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 \(d=14\).
Frage 6 Bericht
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.
Antwortdetails
(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.
Frage 7 Bericht
(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 \(35\,\text{m/s}\):
\[35^{2}=15^{2}+2(4)s\;\Rightarrow\;1225=225+8s\;\Rightarrow\;8s=1000\]\[s=125\,\text{m}\]Antwortdetails
(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 \(35\,\text{m/s}\):
\[35^{2}=15^{2}+2(4)s\;\Rightarrow\;1225=225+8s\;\Rightarrow\;8s=1000\]\[s=125\,\text{m}\]Frage 8 Bericht
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.
Antwortdetails
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.
Frage 9 Bericht
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 \(x=30^{\circ},\ 120^{\circ},\ 210^{\circ},\ 300^{\circ}\).
Antwortdetails
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 \(x=30^{\circ},\ 120^{\circ},\ 210^{\circ},\ 300^{\circ}\).
Frage 10 Bericht
(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 \(P(B)=\tfrac59\):
\[P(0)=\left(\tfrac49\right)^{5}=\frac{1024}{59049}\]\[P(1)={}^{5}C_1\left(\tfrac59\right)\left(\tfrac49\right)^{4}=\frac{6400}{59049}\]\[P(2)={}^{5}C_2\left(\tfrac59\right)^{2}\left(\tfrac49\right)^{3}=\frac{16000}{59049}\]\[P(\le 2)=\frac{1024+6400+16000}{59049}=\frac{23424}{59049}\approx 0.3967\]Antwortdetails
(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 \(P(B)=\tfrac59\):
\[P(0)=\left(\tfrac49\right)^{5}=\frac{1024}{59049}\]\[P(1)={}^{5}C_1\left(\tfrac59\right)\left(\tfrac49\right)^{4}=\frac{6400}{59049}\]\[P(2)={}^{5}C_2\left(\tfrac59\right)^{2}\left(\tfrac49\right)^{3}=\frac{16000}{59049}\]\[P(\le 2)=\frac{1024+6400+16000}{59049}=\frac{23424}{59049}\approx 0.3967\]Frage 11 Bericht
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: \(\dfrac{43}{91}+\dfrac{48}{91}=1\).)
Antwortdetails
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: \(\dfrac{43}{91}+\dfrac{48}{91}=1\).)
Frage 12 Bericht
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
\[(x-4)^{2}+(y-4)^{2}=13\]Antwortdetails
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
\[(x-4)^{2}+(y-4)^{2}=13\]Frage 13 Bericht
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.
Forces \(3\,\text{N},\ 4\,\text{N},\ 2\,\text{N}\) along \(j,\ -i+j,\ i+j\) respectively.
Convert each to components using the unit vector of its direction.
\[3\,\text{N along }j=(0,1):\quad (0,3)\]\[4\,\text{N along }-i+j=\frac{(-1,1)}{\sqrt2}:\quad \left(-\frac{4}{\sqrt2},\frac{4}{\sqrt2}\right)=(-2\sqrt2,\,2\sqrt2)\]\[2\,\text{N along }i+j=\frac{(1,1)}{\sqrt2}:\quad \left(\frac{2}{\sqrt2},\frac{2}{\sqrt2}\right)=(\sqrt2,\,\sqrt2)\]Resultant components:
\[R_x=0-2\sqrt2+\sqrt2=-\sqrt2\approx-1.414\]\[R_y=3+2\sqrt2+\sqrt2=3+3\sqrt2\approx 7.243\]Magnitude:
\[|R|=\sqrt{(-\sqrt2)^{2}+(3+3\sqrt2)^{2}}=\sqrt{2+(27+18\sqrt2)}=\sqrt{54.46}\approx 7.4\,\text{N}\]The magnitude of the resultant is about \(7.4\,\text{N}\).
Antwortdetails
Forces \(3\,\text{N},\ 4\,\text{N},\ 2\,\text{N}\) along \(j,\ -i+j,\ i+j\) respectively.
Convert each to components using the unit vector of its direction.
\[3\,\text{N along }j=(0,1):\quad (0,3)\]\[4\,\text{N along }-i+j=\frac{(-1,1)}{\sqrt2}:\quad \left(-\frac{4}{\sqrt2},\frac{4}{\sqrt2}\right)=(-2\sqrt2,\,2\sqrt2)\]\[2\,\text{N along }i+j=\frac{(1,1)}{\sqrt2}:\quad \left(\frac{2}{\sqrt2},\frac{2}{\sqrt2}\right)=(\sqrt2,\,\sqrt2)\]Resultant components:
\[R_x=0-2\sqrt2+\sqrt2=-\sqrt2\approx-1.414\]\[R_y=3+2\sqrt2+\sqrt2=3+3\sqrt2\approx 7.243\]Magnitude:
\[|R|=\sqrt{(-\sqrt2)^{2}+(3+3\sqrt2)^{2}}=\sqrt{2+(27+18\sqrt2)}=\sqrt{54.46}\approx 7.4\,\text{N}\]The magnitude of the resultant is about \(7.4\,\text{N}\).
Frage 14 Bericht
(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: \(x^{2}+6x+9=(x+3)^{2}\).
\[\int\frac{x+3}{(x+3)^{2}}\,dx=\int\frac{1}{x+3}\,dx=\ln|x+3|+C\]Antwortdetails
(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: \(x^{2}+6x+9=(x+3)^{2}\).
\[\int\frac{x+3}{(x+3)^{2}}\,dx=\int\frac{1}{x+3}\,dx=\ln|x+3|+C\]Frage 15 Bericht
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 \(144\,\text{m}\).
Antwortdetails
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 \(144\,\text{m}\).
