WAEC SSCE - Further Mathematics - 2020 (Essay)
Tambaya 1 Rahoto
The diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal level. A body of mass 800g is hung on the string at a point O, 2.4m from Y. If the system is kept in equilibrium by a horizontal force P acting on the body and the tensions are equal, calculate:
(a) < XOY;
(b) the magnitude of the force P;
(c) the tension T in the string.
The geometry of the string gives \(XO=4.2-2.4=1.8\,\text{m}\), \(OY=2.4\,\text{m}\), and \(XY=3.0\,\text{m}\).
(a) Angle \(XOY\)
\[ XO^2+OY^2=1.8^2+2.4^2=3.24+5.76=9=3^2=XY^2. \]
Therefore, by the converse of Pythagoras’ theorem,
\[ \angle XOY=90^\circ. \]
(b) Horizontal force \(P\), and (c) tension \(T\)
The two tensions have equal magnitude \(T\). From the right-angled triangle:
- For \(OX\), the horizontal and vertical components of tension are \(0.6T\) to the left and \(0.8T\) upwards.
- For \(OY\), the horizontal and vertical components of tension are \(0.8T\) to the right and \(0.6T\) upwards.
The mass is \(0.800\,\text{kg}\), so its weight is \(W=0.8g\).
Resolving vertically for equilibrium:
\[ 0.8T+0.6T=0.8g \]
\[ 1.4T=0.8g \]
\[ T=\frac{0.8g}{1.4}=\frac{4g}{7}\,\text{N}. \]
Taking \(g=10\,\text{m s}^{-2}\),
\[ T=\frac{40}{7}=5.71\,\text{N}. \]
Horizontally, the rightward component is \(0.8T\) and the leftward component is \(0.6T\). Their resultant is \(0.2T\) to the right, so \(P\) must act to the left:
\[ P=0.8T-0.6T=0.2T. \]
\[ P=0.2\left(\frac{40}{7}\right)=\frac{8}{7}=1.14\,\text{N}. \]
Thus:
- \(\angle XOY=90^\circ\)
- \(P=1.14\,\text{N}\), acting horizontally towards \(X\)
- \(T=5.71\,\text{N}\)
The supplied school reference answer is not consistent with equilibrium: it treats \(P\) as though it were found directly from a string length, and then combines \(P\) and the weight to obtain one tension. There are two equal tension forces, so their horizontal and vertical components must both be included when resolving forces.
Examination reminder: Draw the force directions at the hanging body, resolve both tensions into horizontal and vertical components, and apply equilibrium separately in each direction.
Bayanin Amsa
The geometry of the string gives \(XO=4.2-2.4=1.8\,\text{m}\), \(OY=2.4\,\text{m}\), and \(XY=3.0\,\text{m}\).
(a) Angle \(XOY\)
\[ XO^2+OY^2=1.8^2+2.4^2=3.24+5.76=9=3^2=XY^2. \]
Therefore, by the converse of Pythagoras’ theorem,
\[ \angle XOY=90^\circ. \]
(b) Horizontal force \(P\), and (c) tension \(T\)
The two tensions have equal magnitude \(T\). From the right-angled triangle:
- For \(OX\), the horizontal and vertical components of tension are \(0.6T\) to the left and \(0.8T\) upwards.
- For \(OY\), the horizontal and vertical components of tension are \(0.8T\) to the right and \(0.6T\) upwards.
The mass is \(0.800\,\text{kg}\), so its weight is \(W=0.8g\).
Resolving vertically for equilibrium:
\[ 0.8T+0.6T=0.8g \]
\[ 1.4T=0.8g \]
\[ T=\frac{0.8g}{1.4}=\frac{4g}{7}\,\text{N}. \]
Taking \(g=10\,\text{m s}^{-2}\),
\[ T=\frac{40}{7}=5.71\,\text{N}. \]
Horizontally, the rightward component is \(0.8T\) and the leftward component is \(0.6T\). Their resultant is \(0.2T\) to the right, so \(P\) must act to the left:
\[ P=0.8T-0.6T=0.2T. \]
\[ P=0.2\left(\frac{40}{7}\right)=\frac{8}{7}=1.14\,\text{N}. \]
Thus:
- \(\angle XOY=90^\circ\)
- \(P=1.14\,\text{N}\), acting horizontally towards \(X\)
- \(T=5.71\,\text{N}\)
The supplied school reference answer is not consistent with equilibrium: it treats \(P\) as though it were found directly from a string length, and then combines \(P\) and the weight to obtain one tension. There are two equal tension forces, so their horizontal and vertical components must both be included when resolving forces.
Examination reminder: Draw the force directions at the hanging body, resolve both tensions into horizontal and vertical components, and apply equilibrium separately in each direction.
Tambaya 2 Rahoto
Evaluate; \(\int^3_1(3x - 2)^5 dx\)
Integrate \( (3x - 2)^5 \) by reversing the chain rule (raise the power, divide by the new power and by the derivative of the bracket, \(3\)):
\[ \int (3x - 2)^5\,dx = \frac{(3x - 2)^6}{6 \times 3} = \frac{(3x - 2)^6}{18}. \]
Evaluate between \(1\) and \(3\):
\[ \left[\frac{(3x - 2)^6}{18}\right]_1^3 = \frac{(7)^6}{18} - \frac{(1)^6}{18} = \frac{117649 - 1}{18} = \frac{117648}{18} = 6536. \]
The value of the integral is \( 6536 \).
Bayanin Amsa
Integrate \( (3x - 2)^5 \) by reversing the chain rule (raise the power, divide by the new power and by the derivative of the bracket, \(3\)):
\[ \int (3x - 2)^5\,dx = \frac{(3x - 2)^6}{6 \times 3} = \frac{(3x - 2)^6}{18}. \]
Evaluate between \(1\) and \(3\):
\[ \left[\frac{(3x - 2)^6}{18}\right]_1^3 = \frac{(7)^6}{18} - \frac{(1)^6}{18} = \frac{117649 - 1}{18} = \frac{117648}{18} = 6536. \]
The value of the integral is \( 6536 \).
Tambaya 3 Rahoto
The essays of 10 candidates were ranked by three examiners as shown in the table.
| candidates | A | B | C | D | E | F | G | H | I | J |
| Examiner I | 1st | 3rd | 6th | 2nd | 10th | 9th | 7th | 4th | 8th | 5th |
| Examiner II | 2nd | 1st | 3rd | 9th | 7th | 4th | 8th | 10th | 5th | 6th |
| Examiner III | 3rd | 2nd | 1st | 6th | 9th | 8th | 7th | 5th | 4th | 10th |
a) Calculate the Spearman's rank correlation coefficient of the ranks assigned by:
(i) Examiners I and lI;
(ii) Examiners I and III
(iii) Examiners II and II.
(b) Using the results in (a), state which two examiners agree most.
Convert each ordinal rank to a number and apply \(r_s=1-\dfrac{6\sum d^2}{n(n^2-1)}\) with \(n=10\), so \(n(n^2-1)=990\).
| Cand. | I | II | III |
|---|---|---|---|
| A | 1 | 2 | 3 |
| B | 3 | 1 | 2 |
| C | 6 | 3 | 1 |
| D | 2 | 9 | 6 |
| E | 10 | 7 | 9 |
| F | 9 | 4 | 8 |
| G | 7 | 8 | 7 |
| H | 4 | 10 | 5 |
| I | 8 | 5 | 4 |
| J | 5 | 6 | 10 |
(a)(i) Examiners I and II. \(d=I-II:\ -1,2,3,-7,3,5,-1,-6,3,-1\); \(\sum d^2=1+4+9+49+9+25+1+36+9+1=144\).\[r_s=1-\frac{6(144)}{990}=1-\frac{864}{990}=0.127\]
(a)(ii) Examiners I and III. \(d=I-III:\ -2,1,5,-4,1,1,0,-1,4,-5\); \(\sum d^2=4+1+25+16+1+1+0+1+16+25=90\).\[r_s=1-\frac{6(90)}{990}=1-\frac{540}{990}=0.455\]
(a)(iii) Examiners II and III. \(d=II-III:\ -1,-1,2,3,-2,-4,1,5,1,-4\); \(\sum d^2=1+1+4+9+4+16+1+25+1+16=78\).\[r_s=1-\frac{6(78)}{990}=1-\frac{468}{990}=0.527\]
(b) Agreement. The largest coefficient is \(0.527\), so Examiners II and III agree most.
Bayanin Amsa
Convert each ordinal rank to a number and apply \(r_s=1-\dfrac{6\sum d^2}{n(n^2-1)}\) with \(n=10\), so \(n(n^2-1)=990\).
| Cand. | I | II | III |
|---|---|---|---|
| A | 1 | 2 | 3 |
| B | 3 | 1 | 2 |
| C | 6 | 3 | 1 |
| D | 2 | 9 | 6 |
| E | 10 | 7 | 9 |
| F | 9 | 4 | 8 |
| G | 7 | 8 | 7 |
| H | 4 | 10 | 5 |
| I | 8 | 5 | 4 |
| J | 5 | 6 | 10 |
(a)(i) Examiners I and II. \(d=I-II:\ -1,2,3,-7,3,5,-1,-6,3,-1\); \(\sum d^2=1+4+9+49+9+25+1+36+9+1=144\).\[r_s=1-\frac{6(144)}{990}=1-\frac{864}{990}=0.127\]
(a)(ii) Examiners I and III. \(d=I-III:\ -2,1,5,-4,1,1,0,-1,4,-5\); \(\sum d^2=4+1+25+16+1+1+0+1+16+25=90\).\[r_s=1-\frac{6(90)}{990}=1-\frac{540}{990}=0.455\]
(a)(iii) Examiners II and III. \(d=II-III:\ -1,-1,2,3,-2,-4,1,5,1,-4\); \(\sum d^2=1+1+4+9+4+16+1+25+1+16=78\).\[r_s=1-\frac{6(78)}{990}=1-\frac{468}{990}=0.527\]
(b) Agreement. The largest coefficient is \(0.527\), so Examiners II and III agree most.
Tambaya 4 Rahoto
Forces F\(_1\)(10N, 090°) and F\(_2\)(20N, 210\(^o\)) and (4N,330°) act on a particle, Find, correct to one decimal place, the magnitude of the resultant force.
Resolve each force into components using bearings measured clockwise from north, with \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).
- \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
- \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
- \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)
Add 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).
Bayanin Amsa
Resolve each force into components using bearings measured clockwise from north, with \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).
- \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
- \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
- \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)
Add 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).
Tambaya 5 Rahoto
(a) A bag contains 10 red and 8 green identical balls. Two balls are drawn at random from the bag, one after the other, without replacement. Find the probability that one is red and the other is green.
(b) There are 20% defective bulbs in a large box. If 12 bulbs are selected randomly from the box, calculate the probability that between two and five are defective.
(a) Bag: 10 red, 8 green (18 balls), two drawn without replacement. One red and one green can occur as (red then green) or (green then red):
\[ P = \frac{10}{18}\cdot\frac{8}{17} + \frac{8}{18}\cdot\frac{10}{17} = 2\cdot\frac{80}{306} = \frac{160}{306} = \frac{80}{153} \approx 0.523. \]
(b) Defective rate \( p = 0.2,\ q = 0.8,\ n = 12 \). Required \( P(2 \le X \le 5) = \sum_{k=2}^{5}\binom{12}{k}(0.2)^k(0.8)^{12-k} \):
- \( P(2) = \binom{12}{2}(0.2)^2(0.8)^{10} \approx 0.2835 \)
- \( P(3) = \binom{12}{3}(0.2)^3(0.8)^{9} \approx 0.2362 \)
- \( P(4) = \binom{12}{4}(0.2)^4(0.8)^{8} \approx 0.1329 \)
- \( P(5) = \binom{12}{5}(0.2)^5(0.8)^{7} \approx 0.0531 \)
\[ P(2 \le X \le 5) \approx 0.2835 + 0.2362 + 0.1329 + 0.0531 = 0.706. \]
Bayanin Amsa
(a) Bag: 10 red, 8 green (18 balls), two drawn without replacement. One red and one green can occur as (red then green) or (green then red):
\[ P = \frac{10}{18}\cdot\frac{8}{17} + \frac{8}{18}\cdot\frac{10}{17} = 2\cdot\frac{80}{306} = \frac{160}{306} = \frac{80}{153} \approx 0.523. \]
(b) Defective rate \( p = 0.2,\ q = 0.8,\ n = 12 \). Required \( P(2 \le X \le 5) = \sum_{k=2}^{5}\binom{12}{k}(0.2)^k(0.8)^{12-k} \):
- \( P(2) = \binom{12}{2}(0.2)^2(0.8)^{10} \approx 0.2835 \)
- \( P(3) = \binom{12}{3}(0.2)^3(0.8)^{9} \approx 0.2362 \)
- \( P(4) = \binom{12}{4}(0.2)^4(0.8)^{8} \approx 0.1329 \)
- \( P(5) = \binom{12}{5}(0.2)^5(0.8)^{7} \approx 0.0531 \)
\[ P(2 \le X \le 5) \approx 0.2835 + 0.2362 + 0.1329 + 0.0531 = 0.706. \]
Tambaya 6 Rahoto
(a) Find the derivative of y = x\(^2\) (1 + x)\(^{\frac{3}{2}}\) with respect to x.
(b) The centre of a circle lies on the line 2y - x = 3. If the circle passes through P(2,3) and Q(6,7), find its equation.
(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:
\[\frac{dy}{dx} = 2x(1 + x)^{3/2} + x^2 \cdot \tfrac{3}{2}(1 + x)^{1/2}\]Factor out \((1 + x)^{1/2}\):
\[\frac{dy}{dx} = (1 + x)^{1/2}\left[2x(1 + x) + \tfrac{3}{2}x^2\right] = (1 + x)^{1/2}\left[2x + \tfrac{7}{2}x^2\right]\] \[\frac{dy}{dx} = \frac{x(4 + 7x)}{2}\sqrt{1 + x}\](b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):
\[(h - 2)^2 + (k - 3)^2 = (h - 6)^2 + (k - 7)^2\]Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):
\[2k - 3 + k = 9 \ \Rightarrow\ k = 4,\quad h = 5\]Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).
\[(x - 5)^2 + (y - 4)^2 = 10 \quad\text{or}\quad x^2 + y^2 - 10x - 8y + 31 = 0\]Bayanin Amsa
(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:
\[\frac{dy}{dx} = 2x(1 + x)^{3/2} + x^2 \cdot \tfrac{3}{2}(1 + x)^{1/2}\]Factor out \((1 + x)^{1/2}\):
\[\frac{dy}{dx} = (1 + x)^{1/2}\left[2x(1 + x) + \tfrac{3}{2}x^2\right] = (1 + x)^{1/2}\left[2x + \tfrac{7}{2}x^2\right]\] \[\frac{dy}{dx} = \frac{x(4 + 7x)}{2}\sqrt{1 + x}\](b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):
\[(h - 2)^2 + (k - 3)^2 = (h - 6)^2 + (k - 7)^2\]Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):
\[2k - 3 + k = 9 \ \Rightarrow\ k = 4,\quad h = 5\]Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).
\[(x - 5)^2 + (y - 4)^2 = 10 \quad\text{or}\quad x^2 + y^2 - 10x - 8y + 31 = 0\]Tambaya 7 Rahoto
(a) An association is made up of 6 farmers and 8 traders. If an executive body of 4 members is to be formed, find the probability that it will consist of at least two farmers. (b) The probability of an accident occurring in a given month in factories X, Y, and Z are \(\frac{1}{5}, \frac{1}{12} \) and \(\frac{1}{6}\) respectively.
Find the probability that the accident will occur in:
i) none of the factories;
(ii) all the factories;
(iii) at least one factory.
(a) Total members \(= 6 + 8 = 14\). Ways to choose 4 \(= \binom{14}{4} = 1001\).
Use the complement of "fewer than 2 farmers":
\[\text{0 farmers} = \binom{6}{0}\binom{8}{4} = 70,\qquad \text{1 farmer} = \binom{6}{1}\binom{8}{3} = 6 \times 56 = 336\] \[P(\text{at least 2 farmers}) = \frac{1001 - 70 - 336}{1001} = \frac{595}{1001} = \frac{85}{143} \approx 0.594\](b) With \(P(X) = \tfrac{1}{5},\ P(Y) = \tfrac{1}{12},\ P(Z) = \tfrac{1}{6}\):
(i) None:
\[\left(\tfrac{4}{5}\right)\left(\tfrac{11}{12}\right)\left(\tfrac{5}{6}\right) = \frac{220}{360} = \frac{11}{18}\](ii) All:
\[\left(\tfrac{1}{5}\right)\left(\tfrac{1}{12}\right)\left(\tfrac{1}{6}\right) = \frac{1}{360}\](iii) At least one:
\[1 - \frac{11}{18} = \frac{7}{18}\]Bayanin Amsa
(a) Total members \(= 6 + 8 = 14\). Ways to choose 4 \(= \binom{14}{4} = 1001\).
Use the complement of "fewer than 2 farmers":
\[\text{0 farmers} = \binom{6}{0}\binom{8}{4} = 70,\qquad \text{1 farmer} = \binom{6}{1}\binom{8}{3} = 6 \times 56 = 336\] \[P(\text{at least 2 farmers}) = \frac{1001 - 70 - 336}{1001} = \frac{595}{1001} = \frac{85}{143} \approx 0.594\](b) With \(P(X) = \tfrac{1}{5},\ P(Y) = \tfrac{1}{12},\ P(Z) = \tfrac{1}{6}\):
(i) None:
\[\left(\tfrac{4}{5}\right)\left(\tfrac{11}{12}\right)\left(\tfrac{5}{6}\right) = \frac{220}{360} = \frac{11}{18}\](ii) All:
\[\left(\tfrac{1}{5}\right)\left(\tfrac{1}{12}\right)\left(\tfrac{1}{6}\right) = \frac{1}{360}\](iii) At least one:
\[1 - \frac{11}{18} = \frac{7}{18}\]Tambaya 8 Rahoto
(a) If (x + 2) is a factor of g(x) = 2x\(^3\) +11x\(^2\) - x - 30, find the zeros of g(x).
(b) Solve 3(2\(^x\)) +3\(^{y - 2}\) = 25 and 2x - 3\(^{y + 1}\) = -19 simultaneously.
(a) Since \((x + 2)\) is a factor, divide \(g(x) = 2x^3 + 11x^2 - x - 30\) by \((x + 2)\).
Using synthetic division with root \(x = -2\):
\[2x^3 + 11x^2 - x - 30 = (x + 2)(2x^2 + 7x - 15)\]Factorise the quadratic:
\[2x^2 + 7x - 15 = (2x - 3)(x + 5)\]So \(g(x) = (x + 2)(2x - 3)(x + 5)\). The zeros are:
\[x = -2,\qquad x = \tfrac{3}{2},\qquad x = -5\](b) Let \(a = 2^x\) and \(b = 3^y\). The equations become:
\[3a + \tfrac{b}{9} = 25 \ \Rightarrow\ 27a + b = 225\] \[a - 3b = -19\]From the second equation \(a = 3b - 19\). Substitute:
\[27(3b - 19) + b = 225 \ \Rightarrow\ 82b - 513 = 225 \ \Rightarrow\ b = 9\]So \(3^y = 9 \Rightarrow y = 2\), and \(a = 3(9) - 19 = 8\), so \(2^x = 8 \Rightarrow x = 3\).
Check: \(3(8) + 3^{0} = 25\) and \(8 - 3^{3} = -19\). Hence \(x = 3,\ y = 2\).
Bayanin Amsa
(a) Since \((x + 2)\) is a factor, divide \(g(x) = 2x^3 + 11x^2 - x - 30\) by \((x + 2)\).
Using synthetic division with root \(x = -2\):
\[2x^3 + 11x^2 - x - 30 = (x + 2)(2x^2 + 7x - 15)\]Factorise the quadratic:
\[2x^2 + 7x - 15 = (2x - 3)(x + 5)\]So \(g(x) = (x + 2)(2x - 3)(x + 5)\). The zeros are:
\[x = -2,\qquad x = \tfrac{3}{2},\qquad x = -5\](b) Let \(a = 2^x\) and \(b = 3^y\). The equations become:
\[3a + \tfrac{b}{9} = 25 \ \Rightarrow\ 27a + b = 225\] \[a - 3b = -19\]From the second equation \(a = 3b - 19\). Substitute:
\[27(3b - 19) + b = 225 \ \Rightarrow\ 82b - 513 = 225 \ \Rightarrow\ b = 9\]So \(3^y = 9 \Rightarrow y = 2\), and \(a = 3(9) - 19 = 8\), so \(2^x = 8 \Rightarrow x = 3\).
Check: \(3(8) + 3^{0} = 25\) and \(8 - 3^{3} = -19\). Hence \(x = 3,\ y = 2\).
Tambaya 9 Rahoto
(a) Two functions p and q are defined on the set of real numbers, R, by p : y \(\to\) 2y +3 and q : y -> y - 2. Find QOP
(b) How many four digits odd numbers greater than 4000 can be formed from 1,7,3,8,2 if repetition is allowed?
(a) \( p: y \to 2y + 3 \) and \( q: y \to y - 2 \). The composite \( q \circ p \) (q of p) means apply p first, then q:
\[ (q \circ p)(y) = q(p(y)) = q(2y + 3) = (2y + 3) - 2 = 2y + 1. \]
So \( q \circ p : y \to 2y + 1 \).
(b) Form four-digit odd numbers greater than 4000 from \( \{1, 7, 3, 8, 2\} \) with repetition allowed.
First (thousands) digit: must be \(\ge 4\) for the number to exceed 4000. Only \(7\) and \(8\) qualify \( \Rightarrow 2 \) choices.
Last (units) digit: must be odd, from \( \{1, 3, 7\} \Rightarrow 3 \) choices.
Middle two digits: any of the 5 digits, repetition allowed \( \Rightarrow 5 \times 5 = 25 \) choices.
\[ \text{Total} = 2 \times 5 \times 5 \times 3 = 150. \]
Bayanin Amsa
(a) \( p: y \to 2y + 3 \) and \( q: y \to y - 2 \). The composite \( q \circ p \) (q of p) means apply p first, then q:
\[ (q \circ p)(y) = q(p(y)) = q(2y + 3) = (2y + 3) - 2 = 2y + 1. \]
So \( q \circ p : y \to 2y + 1 \).
(b) Form four-digit odd numbers greater than 4000 from \( \{1, 7, 3, 8, 2\} \) with repetition allowed.
First (thousands) digit: must be \(\ge 4\) for the number to exceed 4000. Only \(7\) and \(8\) qualify \( \Rightarrow 2 \) choices.
Last (units) digit: must be odd, from \( \{1, 3, 7\} \Rightarrow 3 \) choices.
Middle two digits: any of the 5 digits, repetition allowed \( \Rightarrow 5 \times 5 = 25 \) choices.
\[ \text{Total} = 2 \times 5 \times 5 \times 3 = 150. \]
Tambaya 10 Rahoto
Given that w = 8i + 3j, x = 6i - 5j, y = 2i + 3j and |z| = 41. find z in the direction of w + x - 2y.
First find the direction vector \(w + x - 2y\):
\[w + x - 2y = (8i + 3j) + (6i - 5j) - 2(2i + 3j)\] \[= (8 + 6 - 4)i + (3 - 5 - 6)j = 10i - 10j\]Its magnitude is:
\[|10i - 10j| = \sqrt{10^2 + (-10)^2} = \sqrt{200} = 10\sqrt{2}\]The unit vector in this direction is:
\[\hat{u} = \frac{10i - 10j}{10\sqrt{2}} = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}}j\]Since \(|z| = 41\) and \(z\) points in this direction:
\[z = 41\hat{u} = \frac{41}{\sqrt{2}}i - \frac{41}{\sqrt{2}}j = \frac{41\sqrt{2}}{2}(i - j)\]Therefore \(z \approx 29.0i - 29.0j\).
Bayanin Amsa
First find the direction vector \(w + x - 2y\):
\[w + x - 2y = (8i + 3j) + (6i - 5j) - 2(2i + 3j)\] \[= (8 + 6 - 4)i + (3 - 5 - 6)j = 10i - 10j\]Its magnitude is:
\[|10i - 10j| = \sqrt{10^2 + (-10)^2} = \sqrt{200} = 10\sqrt{2}\]The unit vector in this direction is:
\[\hat{u} = \frac{10i - 10j}{10\sqrt{2}} = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}}j\]Since \(|z| = 41\) and \(z\) points in this direction:
\[z = 41\hat{u} = \frac{41}{\sqrt{2}}i - \frac{41}{\sqrt{2}}j = \frac{41\sqrt{2}}{2}(i - j)\]Therefore \(z \approx 29.0i - 29.0j\).
Tambaya 11 Rahoto
(a) Simplify; \(\frac{log_2 ^8 + log_2 ^{16} - 4 log_2 ^2}{log_4^{16}}\)
(b) The first, third, and seventh terms of an Arithmetic Progression (A.P) from three consecutive terms of a Geometric Progression (G.P). If the sum of the first two terms of the A.P is 6, find its:
(I) first term; (ii) common difference.
(a) Evaluate each logarithm to base 2:
\[\log_2 8 = 3,\quad \log_2 16 = 4,\quad 4\log_2 2 = 4,\quad \log_4 16 = 2\] \[\frac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16} = \frac{3 + 4 - 4}{2} = \frac{3}{2}\](b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:
\[(a + 2d)^2 = a(a + 6d)\] \[a^2 + 4ad + 4d^2 = a^2 + 6ad \ \Rightarrow\ 4d^2 - 2ad = 0 \ \Rightarrow\ 2d(2d - a) = 0\]Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:
\[a + (a + d) = 2a + d = 6\]Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).
(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).
Bayanin Amsa
(a) Evaluate each logarithm to base 2:
\[\log_2 8 = 3,\quad \log_2 16 = 4,\quad 4\log_2 2 = 4,\quad \log_4 16 = 2\] \[\frac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16} = \frac{3 + 4 - 4}{2} = \frac{3}{2}\](b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:
\[(a + 2d)^2 = a(a + 6d)\] \[a^2 + 4ad + 4d^2 = a^2 + 6ad \ \Rightarrow\ 4d^2 - 2ad = 0 \ \Rightarrow\ 2d(2d - a) = 0\]Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:
\[a + (a + d) = 2a + d = 6\]Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).
(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).
Tambaya 12 Rahoto
If \(\frac{3x^2 + 3x - 2}{(x - 1)(x + 1)}\) = P + \(\frac{Q}{x - 1} + \frac{R}{x - 1}\)
Find the value of Q and R
The fraction \( \dfrac{3x^2 + 3x - 2}{(x-1)(x+1)} \) is improper (numerator and denominator both degree 2), so it resolves as \[ \frac{3x^2 + 3x - 2}{(x-1)(x+1)} = P + \frac{Q}{x-1} + \frac{R}{x+1}. \]
Find P. The denominator expands to \( x^2 - 1 \). Divide: \[ 3x^2 + 3x - 2 = 3(x^2 - 1) + (3x + 1), \] so \( P = 3 \) and the proper remainder is \( \dfrac{3x + 1}{(x-1)(x+1)} \).
Find Q and R from \( 3x + 1 = Q(x+1) + R(x-1) \).
At \( x = 1 \): \( 3(1) + 1 = Q(2) \Rightarrow 4 = 2Q \Rightarrow Q = 2 \).
At \( x = -1 \): \( 3(-1) + 1 = R(-2) \Rightarrow -2 = -2R \Rightarrow R = 1 \).
Hence \( Q = 2 \) and \( R = 1 \) (with \( P = 3 \)).
Bayanin Amsa
The fraction \( \dfrac{3x^2 + 3x - 2}{(x-1)(x+1)} \) is improper (numerator and denominator both degree 2), so it resolves as \[ \frac{3x^2 + 3x - 2}{(x-1)(x+1)} = P + \frac{Q}{x-1} + \frac{R}{x+1}. \]
Find P. The denominator expands to \( x^2 - 1 \). Divide: \[ 3x^2 + 3x - 2 = 3(x^2 - 1) + (3x + 1), \] so \( P = 3 \) and the proper remainder is \( \dfrac{3x + 1}{(x-1)(x+1)} \).
Find Q and R from \( 3x + 1 = Q(x+1) + R(x-1) \).
At \( x = 1 \): \( 3(1) + 1 = Q(2) \Rightarrow 4 = 2Q \Rightarrow Q = 2 \).
At \( x = -1 \): \( 3(-1) + 1 = R(-2) \Rightarrow -2 = -2R \Rightarrow R = 1 \).
Hence \( Q = 2 \) and \( R = 1 \) (with \( P = 3 \)).
Tambaya 13 Rahoto
A binary operation * is defined on the set of real numbers R, by p*q = p + q - \(\frac{pq}{2}\), where p, q \(\in\) R. Find the:
(a) inverse of -1 under * given that the identity clement is zero.
(b) truth set of m* 7 = m* 5,
The operation is \( p * q = p + q - \dfrac{pq}{2} \), with identity element \(0\).
(a) Inverse of \(-1\). Let \(x\) be the inverse of \(-1\), so \( (-1) * x = 0 \):
\[ -1 + x - \frac{(-1)x}{2} = 0 \Rightarrow -1 + x + \frac{x}{2} = 0 \Rightarrow \frac{3x}{2} = 1 \Rightarrow x = \frac{2}{3}. \]
The inverse of \(-1\) is \( \dfrac{2}{3} \).
(b) Truth set of \( m * 7 = m * 5 \).
\( m * 7 = m + 7 - \dfrac{7m}{2} \) and \( m * 5 = m + 5 - \dfrac{5m}{2} \). Setting them equal:
\[ 7 - \frac{7m}{2} = 5 - \frac{5m}{2} \Rightarrow 7 - 5 = \frac{7m}{2} - \frac{5m}{2} \Rightarrow 2 = m. \]
The truth set is \( \{2\} \).
Bayanin Amsa
The operation is \( p * q = p + q - \dfrac{pq}{2} \), with identity element \(0\).
(a) Inverse of \(-1\). Let \(x\) be the inverse of \(-1\), so \( (-1) * x = 0 \):
\[ -1 + x - \frac{(-1)x}{2} = 0 \Rightarrow -1 + x + \frac{x}{2} = 0 \Rightarrow \frac{3x}{2} = 1 \Rightarrow x = \frac{2}{3}. \]
The inverse of \(-1\) is \( \dfrac{2}{3} \).
(b) Truth set of \( m * 7 = m * 5 \).
\( m * 7 = m + 7 - \dfrac{7m}{2} \) and \( m * 5 = m + 5 - \dfrac{5m}{2} \). Setting them equal:
\[ 7 - \frac{7m}{2} = 5 - \frac{5m}{2} \Rightarrow 7 - 5 = \frac{7m}{2} - \frac{5m}{2} \Rightarrow 2 = m. \]
The truth set is \( \{2\} \).
Tambaya 14 Rahoto
| Marks | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 2 | 2 | 2 | 8 | 13 | 11 | 12 | 10 | 4 |
The table shows the distribution of marks scored by 64 students in a test
(a) Draw a histogram for the distribution.
(b) Use the histogram to estimate the modal score.
The class widths are all equal to 10, so the heights of the histogram bars are the frequencies.
| Marks | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 2 |
| 20–29 | 19.5–29.5 | 2 |
| 30–39 | 29.5–39.5 | 2 |
| 40–49 | 39.5–49.5 | 8 |
| 50–59 | 49.5–59.5 | 13 |
| 60–69 | 59.5–69.5 | 11 |
| 70–79 | 69.5–79.5 | 12 |
| 80–89 | 79.5–89.5 | 10 |
| 90–99 | 89.5–99.5 | 4 |
(a) Histogram:
(b) The modal class is 50–59, since it has the greatest frequency, 13.
Using the intersecting diagonals from the top corners of the modal rectangle to the top corners of the adjacent rectangles gives
\[\text{Mode}=L+\frac{f_m-f_1}{(f_m-f_1)+(f_m-f_2)}\times c\]
where \(L=49.5\), \(f_m=13\), \(f_1=8\), \(f_2=11\), and \(c=10\).
\[\text{Mode}=49.5+\frac{13-8}{(13-8)+(13-11)}\times10\]
\[=49.5+\frac{5}{7}\times10=56.64\]
Hence, the estimated modal score is \(\boxed{57\text{ marks}}\).
Bayanin Amsa
The class widths are all equal to 10, so the heights of the histogram bars are the frequencies.
| Marks | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 2 |
| 20–29 | 19.5–29.5 | 2 |
| 30–39 | 29.5–39.5 | 2 |
| 40–49 | 39.5–49.5 | 8 |
| 50–59 | 49.5–59.5 | 13 |
| 60–69 | 59.5–69.5 | 11 |
| 70–79 | 69.5–79.5 | 12 |
| 80–89 | 79.5–89.5 | 10 |
| 90–99 | 89.5–99.5 | 4 |
(a) Histogram:
(b) The modal class is 50–59, since it has the greatest frequency, 13.
Using the intersecting diagonals from the top corners of the modal rectangle to the top corners of the adjacent rectangles gives
\[\text{Mode}=L+\frac{f_m-f_1}{(f_m-f_1)+(f_m-f_2)}\times c\]
where \(L=49.5\), \(f_m=13\), \(f_1=8\), \(f_2=11\), and \(c=10\).
\[\text{Mode}=49.5+\frac{13-8}{(13-8)+(13-11)}\times10\]
\[=49.5+\frac{5}{7}\times10=56.64\]
Hence, the estimated modal score is \(\boxed{57\text{ marks}}\).
Tambaya 15 Rahoto
(a) A car is moving with a velocity of 10ms\(^{-1}\) It then accelerates at 0.2ms\(^{-2}\) for 100m. Find, correct to two decimal places the time taken by the car to cover the distance.
(b) A particle moves along a straight line such that its distance S metres from a fixed point O is given by S = t\(^2\) - 5t + 6, where t is the time in seconds. Find its:
(i) initial velocity;
(ii) distance when it is momentarily at rest
(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):
\[100 = 10t + 0.1t^2 \ \Rightarrow\ t^2 + 100t - 1000 = 0\] \[t = \frac{-100 + \sqrt{100^2 + 4000}}{2} = \frac{-100 + \sqrt{14000}}{2} = \frac{-100 + 118.32}{2}\] \[t \approx 9.16\,\text{s}\](b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).
(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).
(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).
\[S = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25\,\text{m}\]The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).
Bayanin Amsa
(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):
\[100 = 10t + 0.1t^2 \ \Rightarrow\ t^2 + 100t - 1000 = 0\] \[t = \frac{-100 + \sqrt{100^2 + 4000}}{2} = \frac{-100 + \sqrt{14000}}{2} = \frac{-100 + 118.32}{2}\] \[t \approx 9.16\,\text{s}\](b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).
(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).
(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).
\[S = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25\,\text{m}\]The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).
