WAEC SSCE - Further Mathematics - 2007 (Essay)
Question 1 Rapport
(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):
\[P(0)=(0.9)^{6}=0.531441\]\[P(1)={}^{6}C_1(0.1)(0.9)^{5}=6(0.1)(0.59049)=0.354294\]\[P(2)={}^{6}C_2(0.1)^{2}(0.9)^{4}=15(0.01)(0.6561)=0.098415\]\[P(\le 2)=0.531441+0.354294+0.098415=0.98415\approx 0.9842\]Détails de la réponse
(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):
\[P(0)=(0.9)^{6}=0.531441\]\[P(1)={}^{6}C_1(0.1)(0.9)^{5}=6(0.1)(0.59049)=0.354294\]\[P(2)={}^{6}C_2(0.1)^{2}(0.9)^{4}=15(0.01)(0.6561)=0.098415\]\[P(\le 2)=0.531441+0.354294+0.098415=0.98415\approx 0.9842\]Question 2 Rapport
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 \(t = 2.5\,\text{s}\):
\[h = 20(2.5) - \tfrac{3}{2}(2.5)^2 - \tfrac{2}{3}(2.5)^3\] \[= 50 - \tfrac{3}{2}(6.25) - \tfrac{2}{3}(15.625) = 50 - 9.375 - 10.4167\] \[= 30.21\,\text{m}\ (\text{to 2 decimal places})\]Détails de la réponse
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 \(t = 2.5\,\text{s}\):
\[h = 20(2.5) - \tfrac{3}{2}(2.5)^2 - \tfrac{2}{3}(2.5)^3\] \[= 50 - \tfrac{3}{2}(6.25) - \tfrac{2}{3}(15.625) = 50 - 9.375 - 10.4167\] \[= 30.21\,\text{m}\ (\text{to 2 decimal places})\]Question 3 Rapport
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 \(P = (65,\ 0)\).
Détails de la réponse
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 \(P = (65,\ 0)\).
Question 4 Rapport
(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.
Détails de la réponse
(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.
Question 5 Rapport
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.
Détails de la réponse
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.
Question 6 Rapport
(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. \(x = 0.02\):
\[(1.99)^5 \approx 32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3\] \[= 32 - 0.8 + 0.008 - 0.00004 = 31.20796 \approx 31.21\]Détails de la réponse
(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. \(x = 0.02\):
\[(1.99)^5 \approx 32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3\] \[= 32 - 0.8 + 0.008 - 0.00004 = 31.20796 \approx 31.21\]Question 7 Rapport
(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:
\[\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}4&-1\\1&-2\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}12-1\\3-2\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}11\\1\end{pmatrix}\]\[x=-\frac{11}{7},\qquad y=-\frac{1}{7}\]Détails de la réponse
(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:
\[\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}4&-1\\1&-2\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}12-1\\3-2\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}11\\1\end{pmatrix}\]\[x=-\frac{11}{7},\qquad y=-\frac{1}{7}\]Question 8 Rapport
(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 \((-1,-3)\):
\[y+3=-1(x+1)\;\Rightarrow\;y=-x-4\quad\text{i.e. }x+y+4=0\]Détails de la réponse
(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 \((-1,-3)\):
\[y+3=-1(x+1)\;\Rightarrow\;y=-x-4\quad\text{i.e. }x+y+4=0\]Question 9 Rapport
(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 \(Q = (2, -1)\).
Détails de la réponse
(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 \(Q = (2, -1)\).
Question 10 Rapport
(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.
Détails de la réponse
(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.
Question 11 Rapport
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 \(x = 27\).
Détails de la réponse
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 \(x = 27\).
Question 12 Rapport
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:
\[P(\text{at most 2}) = 1 - \frac{11}{120} = \frac{109}{120} \approx 0.908\]Détails de la réponse
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:
\[P(\text{at most 2}) = 1 - \frac{11}{120} = \frac{109}{120} \approx 0.908\]Question 13 Rapport
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).
Détails de la réponse
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).
Question 14 Rapport
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 \(r = \dfrac{2}{5}\).
Détails de la réponse
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 \(r = \dfrac{2}{5}\).
Question 15 Rapport
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 \(P=0.542\).
Détails de la réponse
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 \(P=0.542\).
Question 16 Rapport
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).
Détails de la réponse
(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).
