WAEC SSCE - Further Mathematics - 2007 (Essay)
Ajụjụ 1 Ripọtì
(a) Given that \(\begin{vmatrix} 5 & 2 & -3 \\ -1 & k & 6 \\ 3 & 9 & (k + 2) \end{vmatrix} = -207\), find the values of the constant k.
(b) The equation of a curve is \(x(y^{2} + 1) - y(x^{2} + 1) + 4 = 0\). Find the:
(i) gradient of the curve at any point (x, y).
(ii) equation of the tangent to the curve at the point (-1, -3).
(a) Find k given the determinant equals \(-207\).
Expand along the first row:
\[5\big(k(k+2)-54\big)-2\big(-(k+2)-18\big)+(-3)\big(-9-3k\big)\]\[=5(k^{2}+2k-54)-2(-k-20)-3(-9-3k)\]\[=5k^{2}+10k-270+2k+40+27+9k\]\[=5k^{2}+21k-203\]Set equal to \(-207\):
\[5k^{2}+21k-203=-207\;\Rightarrow\;5k^{2}+21k+4=0\]\[k=\frac{-21\pm\sqrt{441-80}}{10}=\frac{-21\pm 19}{10}\]So \(k=-\dfrac{1}{5}\) or \(k=-4\).
(b) Curve \(x(y^{2}+1)-y(x^{2}+1)+4=0\).
Expand: \(xy^{2}+x-x^{2}y-y+4=0\). Differentiate implicitly:
\[y^{2}+2xy\,y'+1-2xy-x^{2}y'-y'=0\]\[y'(2xy-x^{2}-1)=-(y^{2}-2xy+1)\](i) Gradient at any point:
\[\frac{dy}{dx}=\frac{y^{2}-2xy+1}{x^{2}-2xy+1}\](ii) At \((-1,-3)\):
\[\frac{dy}{dx}=\frac{9-2(-1)(-3)+1}{1-2(-1)(-3)+1}=\frac{9-6+1}{1-6+1}=\frac{4}{-4}=-1\]Tangent through \((-1,-3)\):
\[y+3=-1(x+1)\;\Rightarrow\;y=-x-4\quad\text{i.e. }x+y+4=0\]Akọwa Nkọwa
(a) Find k given the determinant equals \(-207\).
Expand along the first row:
\[5\big(k(k+2)-54\big)-2\big(-(k+2)-18\big)+(-3)\big(-9-3k\big)\]\[=5(k^{2}+2k-54)-2(-k-20)-3(-9-3k)\]\[=5k^{2}+10k-270+2k+40+27+9k\]\[=5k^{2}+21k-203\]Set equal to \(-207\):
\[5k^{2}+21k-203=-207\;\Rightarrow\;5k^{2}+21k+4=0\]\[k=\frac{-21\pm\sqrt{441-80}}{10}=\frac{-21\pm 19}{10}\]So \(k=-\dfrac{1}{5}\) or \(k=-4\).
(b) Curve \(x(y^{2}+1)-y(x^{2}+1)+4=0\).
Expand: \(xy^{2}+x-x^{2}y-y+4=0\). Differentiate implicitly:
\[y^{2}+2xy\,y'+1-2xy-x^{2}y'-y'=0\]\[y'(2xy-x^{2}-1)=-(y^{2}-2xy+1)\](i) Gradient at any point:
\[\frac{dy}{dx}=\frac{y^{2}-2xy+1}{x^{2}-2xy+1}\](ii) At \((-1,-3)\):
\[\frac{dy}{dx}=\frac{9-2(-1)(-3)+1}{1-2(-1)(-3)+1}=\frac{9-6+1}{1-6+1}=\frac{4}{-4}=-1\]Tangent through \((-1,-3)\):
\[y+3=-1(x+1)\;\Rightarrow\;y=-x-4\quad\text{i.e. }x+y+4=0\]Ajụjụ 2 Ripọtì
Forces \(F_{1} = (10 N, 090°), F_{2} = (20 N, 210°)\) and \(F_{3} = (4 N, 330°)\) act on a body at rest on a smooth table. Find, correct to one decimal place, the magnitude of the resultant force.
Resolve each force into components with bearings measured clockwise from north, using \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).
- \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
- \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
- \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)
Sum the components:
\[\sum x = 10 - 10 - 2 = -2,\qquad \sum y = 0 - 17.32 + 3.46 = -13.86\]Magnitude of the resultant:
\[R = \sqrt{(-2)^2 + (-13.86)^2} = \sqrt{4 + 192.0} = \sqrt{196.0} = 14.0\,\text{N}\]The magnitude of the resultant force is \(14.0\,\text{N}\) (to 1 decimal place).
Akọwa Nkọwa
Resolve each force into components with bearings measured clockwise from north, using \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).
- \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
- \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
- \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)
Sum the components:
\[\sum x = 10 - 10 - 2 = -2,\qquad \sum y = 0 - 17.32 + 3.46 = -13.86\]Magnitude of the resultant:
\[R = \sqrt{(-2)^2 + (-13.86)^2} = \sqrt{4 + 192.0} = \sqrt{196.0} = 14.0\,\text{N}\]The magnitude of the resultant force is \(14.0\,\text{N}\) (to 1 decimal place).
Ajụjụ 3 Ripọtì
The second term of a geometric progression is 3. If its sum to infinity is \(\frac{25}{2}\), find the value of its common ratio.
Let the first term be \(a\) and common ratio \(r\), with \(|r| < 1\).
Second term: \(ar = 3\), so \(a = \dfrac{3}{r}\).
Sum to infinity: \(\dfrac{a}{1 - r} = \dfrac{25}{2}\). Substitute \(a\):
\[\frac{3/r}{1 - r} = \frac{25}{2} \ \Rightarrow\ \frac{3}{r(1 - r)} = \frac{25}{2}\] \[6 = 25r(1 - r) \ \Rightarrow\ 25r^2 - 25r + 6 = 0\]Solve:
\[r = \frac{25 \pm \sqrt{625 - 600}}{50} = \frac{25 \pm 5}{50}\] \[r = \frac{3}{5} \quad\text{or}\quad r = \frac{2}{5}\]Both satisfy \(|r| < 1\), so the common ratio is \(r = \dfrac{3}{5}\) or \(r = \dfrac{2}{5}\).
Akọwa Nkọwa
Let the first term be \(a\) and common ratio \(r\), with \(|r| < 1\).
Second term: \(ar = 3\), so \(a = \dfrac{3}{r}\).
Sum to infinity: \(\dfrac{a}{1 - r} = \dfrac{25}{2}\). Substitute \(a\):
\[\frac{3/r}{1 - r} = \frac{25}{2} \ \Rightarrow\ \frac{3}{r(1 - r)} = \frac{25}{2}\] \[6 = 25r(1 - r) \ \Rightarrow\ 25r^2 - 25r + 6 = 0\]Solve:
\[r = \frac{25 \pm \sqrt{625 - 600}}{50} = \frac{25 \pm 5}{50}\] \[r = \frac{3}{5} \quad\text{or}\quad r = \frac{2}{5}\]Both satisfy \(|r| < 1\), so the common ratio is \(r = \dfrac{3}{5}\) or \(r = \dfrac{2}{5}\).
Ajụjụ 4 Ripọtì
(a) Write down the first four terms of the binomial expansion of \((2 - \frac{1}{2})^{5}\) in ascending powers of x.
(b) Use your expansion in (a) above to find, correct to two decimal places, the value of \((1.99)^{5}\).
(a) Expand \(\left(2 - \tfrac{x}{2}\right)^5\) in ascending powers of \(x\), using \(\binom{5}{k}2^{5-k}\left(-\tfrac{x}{2}\right)^k\):
- \(k = 0:\ 2^5 = 32\)
- \(k = 1:\ 5 \cdot 2^4 \cdot \left(-\tfrac{x}{2}\right) = -40x\)
- \(k = 2:\ 10 \cdot 2^3 \cdot \tfrac{x^2}{4} = 20x^2\)
- \(k = 3:\ 10 \cdot 2^2 \cdot \left(-\tfrac{x^3}{8}\right) = -5x^3\)
First four terms: \(32 - 40x + 20x^2 - 5x^3\).
(b) Since \(1.99 = 2 - 0.01\), set \(\tfrac{x}{2} = 0.01\), i.e. \(x = 0.02\):
\[(1.99)^5 \approx 32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3\] \[= 32 - 0.8 + 0.008 - 0.00004 = 31.20796 \approx 31.21\]Akọwa Nkọwa
(a) Expand \(\left(2 - \tfrac{x}{2}\right)^5\) in ascending powers of \(x\), using \(\binom{5}{k}2^{5-k}\left(-\tfrac{x}{2}\right)^k\):
- \(k = 0:\ 2^5 = 32\)
- \(k = 1:\ 5 \cdot 2^4 \cdot \left(-\tfrac{x}{2}\right) = -40x\)
- \(k = 2:\ 10 \cdot 2^3 \cdot \tfrac{x^2}{4} = 20x^2\)
- \(k = 3:\ 10 \cdot 2^2 \cdot \left(-\tfrac{x^3}{8}\right) = -5x^3\)
First four terms: \(32 - 40x + 20x^2 - 5x^3\).
(b) Since \(1.99 = 2 - 0.01\), set \(\tfrac{x}{2} = 0.01\), i.e. \(x = 0.02\):
\[(1.99)^5 \approx 32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3\] \[= 32 - 0.8 + 0.008 - 0.00004 = 31.20796 \approx 31.21\]Ajụjụ 5 Ripọtì
(a) Solve the equation : \(\sqrt{4x - 3} - \sqrt{2x - 5} = 2\).
(b) Find the finite area enclosed by the curve \(y^{2} = 4x\) and the line \(y + x = 0\).
(a) Solve \(\sqrt{4x-3}-\sqrt{2x-5}=2\).
Isolate one surd and square:
\[\sqrt{4x-3}=2+\sqrt{2x-5}\]\[4x-3=4+4\sqrt{2x-5}+(2x-5)\]\[4x-3=2x-1+4\sqrt{2x-5}\]\[2x-2=4\sqrt{2x-5}\;\Rightarrow\;x-1=2\sqrt{2x-5}\]Square again:
\[x^{2}-2x+1=4(2x-5)=8x-20\]\[x^{2}-10x+21=0\;\Rightarrow\;(x-3)(x-7)=0\]So \(x=3\) or \(x=7\). Check both in the original equation:
- \(x=3:\ \sqrt{9}-\sqrt{1}=3-1=2\) ✓
- \(x=7:\ \sqrt{25}-\sqrt{9}=5-3=2\) ✓
Both are valid, so \(x=3\) or \(x=7\).
(b) Area enclosed by \(y^{2}=4x\) and \(y+x=0\).
The line is \(x=-y\). Substitute into the parabola:
\[y^{2}=4(-y)\;\Rightarrow\;y^{2}+4y=0\;\Rightarrow\;y(y+4)=0\]So \(y=0\) (giving \((0,0)\)) and \(y=-4\) (giving \((4,-4)\)). Integrating with respect to \(y\), the line \(x=-y\) lies to the right of the parabola \(x=\tfrac{y^{2}}{4}\) on \(-4\le y\le 0\):
\[A=\int_{-4}^{0}\left(-y-\frac{y^{2}}{4}\right)\,dy=\left[-\frac{y^{2}}{2}-\frac{y^{3}}{12}\right]_{-4}^{0}\]\[=0-\left(-\frac{16}{2}-\frac{-64}{12}\right)=0-\left(-8+\frac{16}{3}\right)=\frac{8}{3}\]The finite area is \(\dfrac{8}{3}\approx 2.67\) square units.
Akọwa Nkọwa
(a) Solve \(\sqrt{4x-3}-\sqrt{2x-5}=2\).
Isolate one surd and square:
\[\sqrt{4x-3}=2+\sqrt{2x-5}\]\[4x-3=4+4\sqrt{2x-5}+(2x-5)\]\[4x-3=2x-1+4\sqrt{2x-5}\]\[2x-2=4\sqrt{2x-5}\;\Rightarrow\;x-1=2\sqrt{2x-5}\]Square again:
\[x^{2}-2x+1=4(2x-5)=8x-20\]\[x^{2}-10x+21=0\;\Rightarrow\;(x-3)(x-7)=0\]So \(x=3\) or \(x=7\). Check both in the original equation:
- \(x=3:\ \sqrt{9}-\sqrt{1}=3-1=2\) ✓
- \(x=7:\ \sqrt{25}-\sqrt{9}=5-3=2\) ✓
Both are valid, so \(x=3\) or \(x=7\).
(b) Area enclosed by \(y^{2}=4x\) and \(y+x=0\).
The line is \(x=-y\). Substitute into the parabola:
\[y^{2}=4(-y)\;\Rightarrow\;y^{2}+4y=0\;\Rightarrow\;y(y+4)=0\]So \(y=0\) (giving \((0,0)\)) and \(y=-4\) (giving \((4,-4)\)). Integrating with respect to \(y\), the line \(x=-y\) lies to the right of the parabola \(x=\tfrac{y^{2}}{4}\) on \(-4\le y\le 0\):
\[A=\int_{-4}^{0}\left(-y-\frac{y^{2}}{4}\right)\,dy=\left[-\frac{y^{2}}{2}-\frac{y^{3}}{12}\right]_{-4}^{0}\]\[=0-\left(-\frac{16}{2}-\frac{-64}{12}\right)=0-\left(-8+\frac{16}{3}\right)=\frac{8}{3}\]The finite area is \(\dfrac{8}{3}\approx 2.67\) square units.
Ajụjụ 6 Ripọtì
The table shows the marks obtained by some candidates in Physics (y) and Mathematics (x) tests.
| Mathematics | 43 | 46 | 48 | 39 | 30 | 60 | 8 | 45 | 40 |
| Physics | 54 | 53 | 63 | 30 | 44 | 75 | 20 | 33 | 49 |
(a)(i) Represent this information on a scatter diagram.
(ii) Find \(\bar{x}\) and \(\bar{y}\), the mean of x and y respectively.
(iii) Draw the line of best fit to pass through (x, y).
(b) Find the equation of the line in a(iii).
(c) Use your equation in (b) to find, correct to one decimal place, the mark in Physics for a candidate who scored 28 in Mathematics.
(a)(i) Scatter diagram. Using a scale of 2 cm to 10 marks on each axis, the nine pairs \((43,54),(46,53),(48,63),(39,30),(30,44),(60,75),(8,20),(45,33),(40,49)\) are plotted with Mathematics \((x)\) on the horizontal axis and Physics \((y)\) on the vertical axis. The line of best fit required in part (a)(iii) is drawn on the same diagram.
(a)(ii) The means.
\[\bar{x}=\frac{43+46+48+39+30+60+8+45+40}{9}=\frac{359}{9}=39.9\] \[\bar{y}=\frac{54+53+63+30+44+75+20+33+49}{9}=\frac{421}{9}=46.8\]So the mean point is \((\bar{x},\bar{y})=(39.9,\,46.8)\).
(a)(iii) Line of best fit. A single straight line is drawn through the mean point \((39.9,46.8)\), oriented so that the points are balanced roughly equally above and below it (shown as the straight line on the scatter diagram above).
(b) Equation of the line. Set out the sums needed for the least-squares gradient:
| \(x\) | \(y\) | \(xy\) | \(x^{2}\) |
| 43 | 54 | 2322 | 1849 |
| 46 | 53 | 2438 | 2116 |
| 48 | 63 | 3024 | 2304 |
| 39 | 30 | 1170 | 1521 |
| 30 | 44 | 1320 | 900 |
| 60 | 75 | 4500 | 3600 |
| 8 | 20 | 160 | 64 |
| 45 | 33 | 1485 | 2025 |
| 40 | 49 | 1960 | 1600 |
| \(\sum x=359\) | \(\sum y=421\) | \(\sum xy=18379\) | \(\sum x^{2}=15979\) |
With \(n=9\), the gradient is
\[b=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}=\frac{9(18379)-359(421)}{9(15979)-359^{2}}=\frac{165411-151139}{143811-128881}=\frac{14272}{14930}=0.96.\]The line passes through \((\bar{x},\bar{y})\), so using \(y-\bar{y}=b(x-\bar{x})\):
\[y-46.8=0.96(x-39.9)\] \[y=0.96x-38.3+46.8\] \[y=0.96x+8.6\](c) Physics mark when Mathematics \(=28\).
\[y=0.96(28)+8.6=26.9+8.6=35.5\]The predicted mark in Physics for a candidate who scored 28 in Mathematics is 35.5 (to one decimal place).
Akọwa Nkọwa
(a)(i) Scatter diagram. Using a scale of 2 cm to 10 marks on each axis, the nine pairs \((43,54),(46,53),(48,63),(39,30),(30,44),(60,75),(8,20),(45,33),(40,49)\) are plotted with Mathematics \((x)\) on the horizontal axis and Physics \((y)\) on the vertical axis. The line of best fit required in part (a)(iii) is drawn on the same diagram.
(a)(ii) The means.
\[\bar{x}=\frac{43+46+48+39+30+60+8+45+40}{9}=\frac{359}{9}=39.9\] \[\bar{y}=\frac{54+53+63+30+44+75+20+33+49}{9}=\frac{421}{9}=46.8\]So the mean point is \((\bar{x},\bar{y})=(39.9,\,46.8)\).
(a)(iii) Line of best fit. A single straight line is drawn through the mean point \((39.9,46.8)\), oriented so that the points are balanced roughly equally above and below it (shown as the straight line on the scatter diagram above).
(b) Equation of the line. Set out the sums needed for the least-squares gradient:
| \(x\) | \(y\) | \(xy\) | \(x^{2}\) |
| 43 | 54 | 2322 | 1849 |
| 46 | 53 | 2438 | 2116 |
| 48 | 63 | 3024 | 2304 |
| 39 | 30 | 1170 | 1521 |
| 30 | 44 | 1320 | 900 |
| 60 | 75 | 4500 | 3600 |
| 8 | 20 | 160 | 64 |
| 45 | 33 | 1485 | 2025 |
| 40 | 49 | 1960 | 1600 |
| \(\sum x=359\) | \(\sum y=421\) | \(\sum xy=18379\) | \(\sum x^{2}=15979\) |
With \(n=9\), the gradient is
\[b=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}=\frac{9(18379)-359(421)}{9(15979)-359^{2}}=\frac{165411-151139}{143811-128881}=\frac{14272}{14930}=0.96.\]The line passes through \((\bar{x},\bar{y})\), so using \(y-\bar{y}=b(x-\bar{x})\):
\[y-46.8=0.96(x-39.9)\] \[y=0.96x-38.3+46.8\] \[y=0.96x+8.6\](c) Physics mark when Mathematics \(=28\).
\[y=0.96(28)+8.6=26.9+8.6=35.5\]The predicted mark in Physics for a candidate who scored 28 in Mathematics is 35.5 (to one decimal place).
Ajụjụ 7 Ripọtì
(a) If \(^{18}C_{r} = ^{18}C_{r + 2}\), find \(^{r}C_{5}\).
(b) In a community, 10% of the people tested positive to the HIV virus. If 6 persons from the community are selected at random, one after the other with replacement, calculate, correct to four decimal places, the probability that : (i) exactly 5 (ii) none (iii) at most 2; tested positive to the virus.
(a) If \({}^{18}C_r={}^{18}C_{r+2}\), find \({}^{r}C_5\).
Since \({}^{n}C_a={}^{n}C_b\) with \(a\ne b\) requires \(a+b=n\):
\[r+(r+2)=18\;\Rightarrow\;2r+2=18\;\Rightarrow\;r=8\]\[{}^{r}C_5={}^{8}C_5=\frac{8!}{5!\,3!}=56\](b) Binomial with \(p=0.1,\ q=0.9,\ n=6\).
(i) Exactly 5:
\[{}^{6}C_5(0.1)^{5}(0.9)^{1}=6(0.00001)(0.9)=0.000054\approx 0.0001\](ii) None:
\[(0.9)^{6}=0.531441\approx 0.5314\](iii) At most 2 (0, 1 or 2):
\[P(0)=(0.9)^{6}=0.531441\]\[P(1)={}^{6}C_1(0.1)(0.9)^{5}=6(0.1)(0.59049)=0.354294\]\[P(2)={}^{6}C_2(0.1)^{2}(0.9)^{4}=15(0.01)(0.6561)=0.098415\]\[P(\le 2)=0.531441+0.354294+0.098415=0.98415\approx 0.9842\]Akọwa Nkọwa
(a) If \({}^{18}C_r={}^{18}C_{r+2}\), find \({}^{r}C_5\).
Since \({}^{n}C_a={}^{n}C_b\) with \(a\ne b\) requires \(a+b=n\):
\[r+(r+2)=18\;\Rightarrow\;2r+2=18\;\Rightarrow\;r=8\]\[{}^{r}C_5={}^{8}C_5=\frac{8!}{5!\,3!}=56\](b) Binomial with \(p=0.1,\ q=0.9,\ n=6\).
(i) Exactly 5:
\[{}^{6}C_5(0.1)^{5}(0.9)^{1}=6(0.00001)(0.9)=0.000054\approx 0.0001\](ii) None:
\[(0.9)^{6}=0.531441\approx 0.5314\](iii) At most 2 (0, 1 or 2):
\[P(0)=(0.9)^{6}=0.531441\]\[P(1)={}^{6}C_1(0.1)(0.9)^{5}=6(0.1)(0.59049)=0.354294\]\[P(2)={}^{6}C_2(0.1)^{2}(0.9)^{4}=15(0.01)(0.6561)=0.098415\]\[P(\le 2)=0.531441+0.354294+0.098415=0.98415\approx 0.9842\]Ajụjụ 8 Ripọtì
An object is projected vertically upwards. Its height, h m, at time t seconds is given by \(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\). Find
(a) the time at which it is momentarily at rest (b) correct to two decimal places, the maximum height reached by the object.
Height: \(h = 20t - \tfrac{3}{2}t^2 - \tfrac{2}{3}t^3\). Velocity is \(v = \dfrac{dh}{dt}\):
\[v = 20 - 3t - 2t^2\](a) Momentarily at rest when \(v = 0\):
\[2t^2 + 3t - 20 = 0 \ \Rightarrow\ t = \frac{-3 \pm \sqrt{9 + 160}}{4} = \frac{-3 \pm 13}{4}\]Taking the positive root, \(t = \dfrac{10}{4} = 2.5\,\text{s}\).
(b) Maximum height at \(t = 2.5\,\text{s}\):
\[h = 20(2.5) - \tfrac{3}{2}(2.5)^2 - \tfrac{2}{3}(2.5)^3\] \[= 50 - \tfrac{3}{2}(6.25) - \tfrac{2}{3}(15.625) = 50 - 9.375 - 10.4167\] \[= 30.21\,\text{m}\ (\text{to 2 decimal places})\]Akọwa Nkọwa
Height: \(h = 20t - \tfrac{3}{2}t^2 - \tfrac{2}{3}t^3\). Velocity is \(v = \dfrac{dh}{dt}\):
\[v = 20 - 3t - 2t^2\](a) Momentarily at rest when \(v = 0\):
\[2t^2 + 3t - 20 = 0 \ \Rightarrow\ t = \frac{-3 \pm \sqrt{9 + 160}}{4} = \frac{-3 \pm 13}{4}\]Taking the positive root, \(t = \dfrac{10}{4} = 2.5\,\text{s}\).
(b) Maximum height at \(t = 2.5\,\text{s}\):
\[h = 20(2.5) - \tfrac{3}{2}(2.5)^2 - \tfrac{2}{3}(2.5)^3\] \[= 50 - \tfrac{3}{2}(6.25) - \tfrac{2}{3}(15.625) = 50 - 9.375 - 10.4167\] \[= 30.21\,\text{m}\ (\text{to 2 decimal places})\]Ajụjụ 9 Ripọtì
Four boys participated in a competition in which their respective chances of winning prizes are \(\frac{1}{5}, \frac{1}{4}, \frac{1}{3}\) and \(\frac{1}{2}\). What is the probability that at most two of them win prizes?
Let the winning probabilities be \(p_1 = \tfrac{1}{5},\ p_2 = \tfrac{1}{4},\ p_3 = \tfrac{1}{3},\ p_4 = \tfrac{1}{2}\), with losing probabilities \(q_1 = \tfrac{4}{5},\ q_2 = \tfrac{3}{4},\ q_3 = \tfrac{2}{3},\ q_4 = \tfrac{1}{2}\).
"At most two win" is easier via the complement: \(1 - P(3\text{ win}) - P(4\text{ win})\).
All four win:
\[P(4) = \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} = \frac{1}{120}\]Exactly three win (one loses at a time, summed):
\[\tfrac{4}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{3}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{2}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2}\] \[= \frac{4 + 3 + 2 + 1}{120} = \frac{10}{120}\]So \(P(3\text{ or }4) = \dfrac{10}{120} + \dfrac{1}{120} = \dfrac{11}{120}\), and:
\[P(\text{at most 2}) = 1 - \frac{11}{120} = \frac{109}{120} \approx 0.908\]Akọwa Nkọwa
Let the winning probabilities be \(p_1 = \tfrac{1}{5},\ p_2 = \tfrac{1}{4},\ p_3 = \tfrac{1}{3},\ p_4 = \tfrac{1}{2}\), with losing probabilities \(q_1 = \tfrac{4}{5},\ q_2 = \tfrac{3}{4},\ q_3 = \tfrac{2}{3},\ q_4 = \tfrac{1}{2}\).
"At most two win" is easier via the complement: \(1 - P(3\text{ win}) - P(4\text{ win})\).
All four win:
\[P(4) = \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} = \frac{1}{120}\]Exactly three win (one loses at a time, summed):
\[\tfrac{4}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{3}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{2}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2}\] \[= \frac{4 + 3 + 2 + 1}{120} = \frac{10}{120}\]So \(P(3\text{ or }4) = \dfrac{10}{120} + \dfrac{1}{120} = \dfrac{11}{120}\), and:
\[P(\text{at most 2}) = 1 - \frac{11}{120} = \frac{109}{120} \approx 0.908\]Ajụjụ 10 Ripọtì
(a) Using the trapezium rule with seven ordinates, evaluate \(\int_{0}^{3} \frac{\mathrm d x}{x^{2} + 1}\), correct to two decimal places.
(b) Using matrix method, solve \(-2x + y = 3; - x + 4y = 1\).
(a) Trapezium rule, seven ordinates, \(\displaystyle\int_{0}^{3}\frac{dx}{x^{2}+1}\).
Seven ordinates means six strips, so \(h=\dfrac{3-0}{6}=0.5\). With \(y=\dfrac{1}{x^{2}+1}\):
| x | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|
| y | 1.0000 | 0.8000 | 0.5000 | 0.3077 | 0.2000 | 0.1379 | 0.1000 |
(b) Matrix method: \(-2x+y=3,\ -x+4y=1\).
\[\begin{pmatrix}-2&1\\-1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}\]Determinant \(=(-2)(4)-(1)(-1)=-8+1=-7\). Inverse:
\[\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}4&-1\\1&-2\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}12-1\\3-2\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}11\\1\end{pmatrix}\]\[x=-\frac{11}{7},\qquad y=-\frac{1}{7}\]Akọwa Nkọwa
(a) Trapezium rule, seven ordinates, \(\displaystyle\int_{0}^{3}\frac{dx}{x^{2}+1}\).
Seven ordinates means six strips, so \(h=\dfrac{3-0}{6}=0.5\). With \(y=\dfrac{1}{x^{2}+1}\):
| x | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|
| y | 1.0000 | 0.8000 | 0.5000 | 0.3077 | 0.2000 | 0.1379 | 0.1000 |
(b) Matrix method: \(-2x+y=3,\ -x+4y=1\).
\[\begin{pmatrix}-2&1\\-1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}\]Determinant \(=(-2)(4)-(1)(-1)=-8+1=-7\). Inverse:
\[\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}4&-1\\1&-2\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}12-1\\3-2\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}11\\1\end{pmatrix}\]\[x=-\frac{11}{7},\qquad y=-\frac{1}{7}\]Ajụjụ 11 Ripọtì
(a) The roots of the equation \(x^{2} + mx + 11 = 0\) are \(\alpha\) and \(\beta\), where m is a constant. If \(\alpha^{2} + \beta^{2} = 27\), find the values of m.
(b) The line \(2x + 3y = 1\) intersects the circle \(2x^{2} + 2y^{2} + 4x + 9y - 9 = 0\) at points P and Q where Q lies in the fourth quadrant. Find the coordinates of P and Q.
(a) For \(x^2 + mx + 11 = 0\) with roots \(\alpha,\beta\): \(\alpha + \beta = -m\) and \(\alpha\beta = 11\).
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = m^2 - 22\]Setting this equal to 27:
\[m^2 - 22 = 27 \ \Rightarrow\ m^2 = 49 \ \Rightarrow\ m = \pm 7\](b) From the line \(2x + 3y = 1\), \(x = \dfrac{1 - 3y}{2}\). Substitute into the circle \(2x^2 + 2y^2 + 4x + 9y - 9 = 0\) (dividing by 2 first: \(x^2 + y^2 + 2x + 4.5y - 4.5 = 0\)):
\[\frac{(1 - 3y)^2}{4} + y^2 + (1 - 3y) + 4.5y - 4.5 = 0\]Multiply through by 4 and simplify:
\[13y^2 - 13 = 0 \ \Rightarrow\ y^2 = 1 \ \Rightarrow\ y = \pm 1\]For \(y = 1\): \(x = \dfrac{1 - 3}{2} = -1\), giving \((-1, 1)\).
For \(y = -1\): \(x = \dfrac{1 + 3}{2} = 2\), giving \((2, -1)\), which lies in the fourth quadrant.
Therefore \(P = (-1, 1)\) and \(Q = (2, -1)\).
Akọwa Nkọwa
(a) For \(x^2 + mx + 11 = 0\) with roots \(\alpha,\beta\): \(\alpha + \beta = -m\) and \(\alpha\beta = 11\).
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = m^2 - 22\]Setting this equal to 27:
\[m^2 - 22 = 27 \ \Rightarrow\ m^2 = 49 \ \Rightarrow\ m = \pm 7\](b) From the line \(2x + 3y = 1\), \(x = \dfrac{1 - 3y}{2}\). Substitute into the circle \(2x^2 + 2y^2 + 4x + 9y - 9 = 0\) (dividing by 2 first: \(x^2 + y^2 + 2x + 4.5y - 4.5 = 0\)):
\[\frac{(1 - 3y)^2}{4} + y^2 + (1 - 3y) + 4.5y - 4.5 = 0\]Multiply through by 4 and simplify:
\[13y^2 - 13 = 0 \ \Rightarrow\ y^2 = 1 \ \Rightarrow\ y = \pm 1\]For \(y = 1\): \(x = \dfrac{1 - 3}{2} = -1\), giving \((-1, 1)\).
For \(y = -1\): \(x = \dfrac{1 + 3}{2} = 2\), giving \((2, -1)\), which lies in the fourth quadrant.
Therefore \(P = (-1, 1)\) and \(Q = (2, -1)\).
Ajụjụ 12 Ripọtì
(a) Find, correct to one decimal place, the angle between \(p = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\) and \(q = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\).
(b) ABCD is a square with vertices at A(0, 0), B(2, 0), C(2, 2) and D(0, 2). Forces of magnitude 10 N, 15 N, 20 N and 5 N act along \(\overrightarrow{BA}, \overrightarrow{BC}, \overrightarrow{DC}\) and \(\overrightarrow{AD}\) respectively. Find the (i) magnitude (ii) direction; of the resultant.
(a) Angle between \(p=\binom{3}{-1}\) and \(q=\binom{3}{4}\).
\[\cos\theta=\frac{p\cdot q}{|p|\,|q|}=\frac{(3)(3)+(-1)(4)}{\sqrt{3^{2}+(-1)^{2}}\,\sqrt{3^{2}+4^{2}}}=\frac{9-4}{\sqrt{10}\cdot 5}=\frac{5}{5\sqrt{10}}=\frac{1}{\sqrt{10}}\]\[\theta=\cos^{-1}(0.3162)=71.6^{\circ}\ (\text{1 d.p.})\](b) Forces on square A(0,0), B(2,0), C(2,2), D(0,2).
Write each force as its magnitude times the unit vector of its line of action:
- \(10\,\text{N}\) along \(\overrightarrow{BA}=(-1,0):\ (-10,0)\)
- \(15\,\text{N}\) along \(\overrightarrow{BC}=(0,1):\ (0,15)\)
- \(20\,\text{N}\) along \(\overrightarrow{DC}=(1,0):\ (20,0)\)
- \(5\,\text{N}\) along \(\overrightarrow{AD}=(0,1):\ (0,5)\)
Resultant components:
\[R_x=-10+20=10,\qquad R_y=15+5=20\](i) Magnitude:
\[|R|=\sqrt{10^{2}+20^{2}}=\sqrt{500}=10\sqrt{5}\approx 22.4\,\text{N}\](ii) Direction (measured from the positive x-axis, i.e. along AB):
\[\tan\alpha=\frac{20}{10}=2\;\Rightarrow\;\alpha=63.4^{\circ}\]The resultant is about \(22.4\,\text{N}\) at \(63.4^{\circ}\) to AB.
Akọwa Nkọwa
(a) Angle between \(p=\binom{3}{-1}\) and \(q=\binom{3}{4}\).
\[\cos\theta=\frac{p\cdot q}{|p|\,|q|}=\frac{(3)(3)+(-1)(4)}{\sqrt{3^{2}+(-1)^{2}}\,\sqrt{3^{2}+4^{2}}}=\frac{9-4}{\sqrt{10}\cdot 5}=\frac{5}{5\sqrt{10}}=\frac{1}{\sqrt{10}}\]\[\theta=\cos^{-1}(0.3162)=71.6^{\circ}\ (\text{1 d.p.})\](b) Forces on square A(0,0), B(2,0), C(2,2), D(0,2).
Write each force as its magnitude times the unit vector of its line of action:
- \(10\,\text{N}\) along \(\overrightarrow{BA}=(-1,0):\ (-10,0)\)
- \(15\,\text{N}\) along \(\overrightarrow{BC}=(0,1):\ (0,15)\)
- \(20\,\text{N}\) along \(\overrightarrow{DC}=(1,0):\ (20,0)\)
- \(5\,\text{N}\) along \(\overrightarrow{AD}=(0,1):\ (0,5)\)
Resultant components:
\[R_x=-10+20=10,\qquad R_y=15+5=20\](i) Magnitude:
\[|R|=\sqrt{10^{2}+20^{2}}=\sqrt{500}=10\sqrt{5}\approx 22.4\,\text{N}\](ii) Direction (measured from the positive x-axis, i.e. along AB):
\[\tan\alpha=\frac{20}{10}=2\;\Rightarrow\;\alpha=63.4^{\circ}\]The resultant is about \(22.4\,\text{N}\) at \(63.4^{\circ}\) to AB.
Ajụjụ 13 Ripọtì
Solve \(x^{\frac{2}{3}} - 5x^{\frac{1}{3}} + 6 = 0\).
Let \(u = x^{1/3}\), so \(x^{2/3} = u^2\). The equation becomes a quadratic:
\[u^2 - 5u + 6 = 0 \ \Rightarrow\ (u - 2)(u - 3) = 0\] \[u = 2 \quad\text{or}\quad u = 3\]Return to \(x\):
\[x^{1/3} = 2 \ \Rightarrow\ x = 2^3 = 8\] \[x^{1/3} = 3 \ \Rightarrow\ x = 3^3 = 27\]So \(x = 8\) or \(x = 27\).
Akọwa Nkọwa
Let \(u = x^{1/3}\), so \(x^{2/3} = u^2\). The equation becomes a quadratic:
\[u^2 - 5u + 6 = 0 \ \Rightarrow\ (u - 2)(u - 3) = 0\] \[u = 2 \quad\text{or}\quad u = 3\]Return to \(x\):
\[x^{1/3} = 2 \ \Rightarrow\ x = 2^3 = 8\] \[x^{1/3} = 3 \ \Rightarrow\ x = 3^3 = 27\]So \(x = 8\) or \(x = 27\).
Ajụjụ 14 Ripọtì
The normal to the curve \(y = 2x^{2} + x - 3\) at the point (2, 7) meets the x- axis at the point P. Find the coordinates of P.
Differentiate the curve \(y = 2x^2 + x - 3\):
\[\frac{dy}{dx} = 4x + 1\]At \((2, 7)\) the gradient of the tangent is \(4(2) + 1 = 9\). The normal is perpendicular, so its gradient is:
\[m = -\frac{1}{9}\]Equation of the normal through \((2, 7)\):
\[y - 7 = -\tfrac{1}{9}(x - 2)\]The normal meets the \(x\)-axis where \(y = 0\):
\[-7 = -\tfrac{1}{9}(x - 2) \ \Rightarrow\ 63 = x - 2 \ \Rightarrow\ x = 65\]Therefore \(P = (65,\ 0)\).
Akọwa Nkọwa
Differentiate the curve \(y = 2x^2 + x - 3\):
\[\frac{dy}{dx} = 4x + 1\]At \((2, 7)\) the gradient of the tangent is \(4(2) + 1 = 9\). The normal is perpendicular, so its gradient is:
\[m = -\frac{1}{9}\]Equation of the normal through \((2, 7)\):
\[y - 7 = -\tfrac{1}{9}(x - 2)\]The normal meets the \(x\)-axis where \(y = 0\):
\[-7 = -\tfrac{1}{9}(x - 2) \ \Rightarrow\ 63 = x - 2 \ \Rightarrow\ x = 65\]Therefore \(P = (65,\ 0)\).
Ajụjụ 15 Ripọtì
The table below shows the distribution of ages of workers in a company.
| Age/ yr | 17 - 21 | 22 - 26 | 27 - 31 | 32 - 36 | 37 - 41 | 42 - 46 | 47 - 51 | 52 - 56 |
| Workers | 12 | 24 | 30 | 37 | 45 | 25 | 10 | 7 |
(a) Using an assumed mean of 39, calculate the (i) mean (ii) standard deviation; of the distribution.
(b) If a worker is selected at random from the company for an award, what is the probability that he is at most 36 years old?
Class width \(=5\); mid-values \(19,24,\dots,54\). Code with \(u=\dfrac{x-39}{5}\), \(A=39\).
| Age | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 17-21 | 19 | 12 | -4 | -48 | 192 |
| 22-26 | 24 | 24 | -3 | -72 | 216 |
| 27-31 | 29 | 30 | -2 | -60 | 120 |
| 32-36 | 34 | 37 | -1 | -37 | 37 |
| 37-41 | 39 | 45 | 0 | 0 | 0 |
| 42-46 | 44 | 25 | 1 | 25 | 25 |
| 47-51 | 49 | 10 | 2 | 20 | 40 |
| 52-56 | 54 | 7 | 3 | 21 | 63 |
| Total | 190 | -151 | 693 |
(a)(i) Mean.\[\bar{x}=39+\frac{\sum fu}{\sum f}\times 5=39+\frac{-151}{190}\times 5=39-3.97=35.0\text{ years}\]
(a)(ii) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=5\sqrt{\frac{693}{190}-\left(\frac{-151}{190}\right)^2}\]\[=5\sqrt{3.647-0.632}=5\sqrt{3.015}=5(1.737)=8.68\text{ years}\]
(b) Probability at most 36 years old. Ages \(\le 36\) cover the classes \(17-21,\,22-26,\,27-31,\,32-36\), giving \(12+24+30+37=103\) workers out of 190.\[P(\text{at most }36)=\frac{103}{190}=0.542\]
Mean \(\approx\) 35.0 years, standard deviation \(\approx\) 8.68 years, and \(P=0.542\).
Akọwa Nkọwa
Class width \(=5\); mid-values \(19,24,\dots,54\). Code with \(u=\dfrac{x-39}{5}\), \(A=39\).
| Age | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 17-21 | 19 | 12 | -4 | -48 | 192 |
| 22-26 | 24 | 24 | -3 | -72 | 216 |
| 27-31 | 29 | 30 | -2 | -60 | 120 |
| 32-36 | 34 | 37 | -1 | -37 | 37 |
| 37-41 | 39 | 45 | 0 | 0 | 0 |
| 42-46 | 44 | 25 | 1 | 25 | 25 |
| 47-51 | 49 | 10 | 2 | 20 | 40 |
| 52-56 | 54 | 7 | 3 | 21 | 63 |
| Total | 190 | -151 | 693 |
(a)(i) Mean.\[\bar{x}=39+\frac{\sum fu}{\sum f}\times 5=39+\frac{-151}{190}\times 5=39-3.97=35.0\text{ years}\]
(a)(ii) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=5\sqrt{\frac{693}{190}-\left(\frac{-151}{190}\right)^2}\]\[=5\sqrt{3.647-0.632}=5\sqrt{3.015}=5(1.737)=8.68\text{ years}\]
(b) Probability at most 36 years old. Ages \(\le 36\) cover the classes \(17-21,\,22-26,\,27-31,\,32-36\), giving \(12+24+30+37=103\) workers out of 190.\[P(\text{at most }36)=\frac{103}{190}=0.542\]
Mean \(\approx\) 35.0 years, standard deviation \(\approx\) 8.68 years, and \(P=0.542\).
Ajụjụ 16 Ripọtì
Five digit numbers are formed from digits 4, 5, 6, 7 and 8.
(a)How many such numbers can be formed if repitition of digits is (i) allowed (ii) not allowed?
(b) How many of the numbers are odd, if repetition of digits is not allowed?
Five different digits are available: \(4, 5, 6, 7, 8\).
(a)(i) Repetition allowed: each of the 5 positions can be any of the 5 digits:
\[5^5 = 3125\](ii) Repetition not allowed: arrange all 5 distinct digits:
\[5! = 120\](b) A number is odd if its last digit is odd. Among the digits, the odd ones are \(5\) and \(7\), giving 2 choices for the units position. With no repetition, the remaining 4 positions are filled by the remaining 4 digits in \(4!\) ways:
\[2 \times 4! = 2 \times 24 = 48\]So 48 of the numbers are odd.
Akọwa Nkọwa
Five different digits are available: \(4, 5, 6, 7, 8\).
(a)(i) Repetition allowed: each of the 5 positions can be any of the 5 digits:
\[5^5 = 3125\](ii) Repetition not allowed: arrange all 5 distinct digits:
\[5! = 120\](b) A number is odd if its last digit is odd. Among the digits, the odd ones are \(5\) and \(7\), giving 2 choices for the units position. With no repetition, the remaining 4 positions are filled by the remaining 4 digits in \(4!\) ways:
\[2 \times 4! = 2 \times 24 = 48\]So 48 of the numbers are odd.
