WAEC SSCE - Further Mathematics - 2011 (Essay)
Question 1 Report
If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 7x + 4 = 0\), find the equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\).
For \(2x^2 - 7x + 4 = 0\), the sum and product of the roots \(\alpha,\beta\) are
\[\alpha + \beta = \frac{7}{2},\qquad \alpha\beta = \frac{4}{2} = 2.\]
Sum of the new roots \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}\). Using \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\):
\[\alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2(2) = \frac{49}{4} - 4 = \frac{33}{4}.\]
\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{33/4}{2} = \frac{33}{8}.\]
Product of the new roots \(\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha} = 1\).
The required equation is \(x^2 - (\text{sum})x + (\text{product}) = 0\):
\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]
Answer Details
For \(2x^2 - 7x + 4 = 0\), the sum and product of the roots \(\alpha,\beta\) are
\[\alpha + \beta = \frac{7}{2},\qquad \alpha\beta = \frac{4}{2} = 2.\]
Sum of the new roots \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}\). Using \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\):
\[\alpha^2 + \beta^2 = \left(\frac{7}{2}\right)^2 - 2(2) = \frac{49}{4} - 4 = \frac{33}{4}.\]
\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{33/4}{2} = \frac{33}{8}.\]
Product of the new roots \(\dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha} = 1\).
The required equation is \(x^2 - (\text{sum})x + (\text{product}) = 0\):
\[x^2 - \frac{33}{8}x + 1 = 0 \;\Rightarrow\; 8x^2 - 33x + 8 = 0.\]
Question 2 Report
A tyre manufacturing company researched into the life span of one type of their motorcycle tyres. The results were as follows :
| Distance (100km) |
10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 |
| Number of tyres | 30 | 69 | 93 | 57 | 36 | 15 |
(a) Draw a histogram for the distribution.
(b) Use the histogram to estimate the mode.
(a) Histogram
Since all the class intervals have the same width, namely 10 (in units of 100 km), the bar heights are proportional to the frequencies. Use continuous class boundaries as follows:
| Distance (100 km) | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 30 |
| 20–29 | 19.5–29.5 | 69 |
| 30–39 | 29.5–39.5 | 93 |
| 40–49 | 39.5–49.5 | 57 |
| 50–59 | 49.5–59.5 | 36 |
| 60–69 | 59.5–69.5 | 15 |
(b) Estimate of the mode
The modal class is \(30\text{–}39\), with boundaries \(29.5\text{–}39.5\). From the intersection of the diagonals drawn in the modal rectangle on the histogram, the modal value is approximately \(33.5\).
Equivalently,
\[\text{Mode}=29.5+\frac{93-69}{(93-69)+(93-57)}\times 10\]
\[=29.5+\frac{24}{60}\times10=\boxed{33.5}\]
Thus, the estimated modal tyre life span is \(\boxed{33.5\times100=3350\text{ km}}\).
Answer Details
(a) Histogram
Since all the class intervals have the same width, namely 10 (in units of 100 km), the bar heights are proportional to the frequencies. Use continuous class boundaries as follows:
| Distance (100 km) | Class boundaries | Frequency |
|---|---|---|
| 10–19 | 9.5–19.5 | 30 |
| 20–29 | 19.5–29.5 | 69 |
| 30–39 | 29.5–39.5 | 93 |
| 40–49 | 39.5–49.5 | 57 |
| 50–59 | 49.5–59.5 | 36 |
| 60–69 | 59.5–69.5 | 15 |
(b) Estimate of the mode
The modal class is \(30\text{–}39\), with boundaries \(29.5\text{–}39.5\). From the intersection of the diagonals drawn in the modal rectangle on the histogram, the modal value is approximately \(33.5\).
Equivalently,
\[\text{Mode}=29.5+\frac{93-69}{(93-69)+(93-57)}\times 10\]
\[=29.5+\frac{24}{60}\times10=\boxed{33.5}\]
Thus, the estimated modal tyre life span is \(\boxed{33.5\times100=3350\text{ km}}\).
Question 3 Report
(a) Find the equation of the tangent to curve \(\frac{x^{2}}{4} + y^{2} = 1\) at point \(1, \frac{\sqrt{3}}{2}\).
(b) Express \(\frac{3x + 2}{x^{2} + x - 2}\) in partial fractions.
(a) Differentiate \(\dfrac{x^2}{4}+y^2=1\) implicitly:
\[\frac{2x}{4}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{4y}\]
At \(\left(1,\tfrac{\sqrt3}{2}\right)\): \(\dfrac{dy}{dx}=-\dfrac{1}{4\cdot\frac{\sqrt3}{2}}=-\dfrac{1}{2\sqrt3}=-\dfrac{\sqrt3}{6}\).
Tangent: \(y-\dfrac{\sqrt3}{2}=-\dfrac{\sqrt3}{6}(x-1)\). Multiplying through by 6 and simplifying:
\[\sqrt3\,x+6y=4\sqrt3\quad\text{or}\quad x+2\sqrt3\,y=4\]
(b) Factorise the denominator: \(x^2+x-2=(x+2)(x-1)\). Write
\[\frac{3x+2}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\Rightarrow 3x+2=A(x-1)+B(x+2)\]
Put \(x=1\): \(5=3B\Rightarrow B=\tfrac53\). Put \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).
\[\frac{3x+2}{x^2+x-2}=\frac{4}{3(x+2)}+\frac{5}{3(x-1)}\]
Answer Details
(a) Differentiate \(\dfrac{x^2}{4}+y^2=1\) implicitly:
\[\frac{2x}{4}+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{4y}\]
At \(\left(1,\tfrac{\sqrt3}{2}\right)\): \(\dfrac{dy}{dx}=-\dfrac{1}{4\cdot\frac{\sqrt3}{2}}=-\dfrac{1}{2\sqrt3}=-\dfrac{\sqrt3}{6}\).
Tangent: \(y-\dfrac{\sqrt3}{2}=-\dfrac{\sqrt3}{6}(x-1)\). Multiplying through by 6 and simplifying:
\[\sqrt3\,x+6y=4\sqrt3\quad\text{or}\quad x+2\sqrt3\,y=4\]
(b) Factorise the denominator: \(x^2+x-2=(x+2)(x-1)\). Write
\[\frac{3x+2}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\Rightarrow 3x+2=A(x-1)+B(x+2)\]
Put \(x=1\): \(5=3B\Rightarrow B=\tfrac53\). Put \(x=-2\): \(-4=-3A\Rightarrow A=\tfrac43\).
\[\frac{3x+2}{x^2+x-2}=\frac{4}{3(x+2)}+\frac{5}{3(x-1)}\]
Question 4 Report
(a) Write down the binomial expansion of \((2 - x)^{5}\) in ascending powers of x.
(b) Use your expansion in (a) to evaluate \((1.98)^{5}\) correct to four decimal places.
(a) Using \((a+b)^5=\sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\) with \(a=2\) and \(b=-x\):
\[(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5\]
The coefficients come from \(\binom{5}{k}2^{5-k}(-1)^k\): \(32,\,-80,\,80,\,-40,\,10,\,-1\).
(b) Choose \(x\) so that \(2-x=1.98\), i.e. \(x=0.02\). Substitute:
- \(32\)
- \(-80(0.02)=-1.6\)
- \(+80(0.02)^2=+0.032\)
- \(-40(0.02)^3=-0.00032\)
- \(+10(0.02)^4=+0.0000016\)
- \(-(0.02)^5\approx 0\)
Adding: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).
\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]
Answer Details
(a) Using \((a+b)^5=\sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\) with \(a=2\) and \(b=-x\):
\[(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5\]
The coefficients come from \(\binom{5}{k}2^{5-k}(-1)^k\): \(32,\,-80,\,80,\,-40,\,10,\,-1\).
(b) Choose \(x\) so that \(2-x=1.98\), i.e. \(x=0.02\). Substitute:
- \(32\)
- \(-80(0.02)=-1.6\)
- \(+80(0.02)^2=+0.032\)
- \(-40(0.02)^3=-0.00032\)
- \(+10(0.02)^4=+0.0000016\)
- \(-(0.02)^5\approx 0\)
Adding: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).
\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]
Question 5 Report
The table shows the frequency distribution of marks scored by some candidates in an examination.
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Freq | 2 | 5 | 8 | 18 | 20 | 15 | 5 | 4 | 2 | 1 |
(a) Draw the cumulative frequency curve of the distribution.
(b) Use your graph to estimate the :
(i) semi-interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.
(a) Cumulative frequency table
| Marks | Frequency | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 0–9 | 2 | 9.5 | 2 |
| 10–19 | 5 | 19.5 | 7 |
| 20–29 | 8 | 29.5 | 15 |
| 30–39 | 18 | 39.5 | 33 |
| 40–49 | 20 | 49.5 | 53 |
| 50–59 | 15 | 59.5 | 68 |
| 60–69 | 5 | 69.5 | 73 |
| 70–79 | 4 | 79.5 | 77 |
| 80–89 | 2 | 89.5 | 79 |
| 90–99 | 1 | 99.5 | 80 |
The total number of candidates is \(N=80\). Plot cumulative frequency against the upper class boundaries and draw a smooth increasing ogive.
(b)(i) Semi-interquartile range
\[Q_1\text{ is the }\frac{N}{4}=\frac{80}{4}=20\text{th value.}\]
From the ogive, \(Q_1\approx 32.5\).
\[Q_3\text{ is the }\frac{3N}{4}=\frac{3(80)}{4}=60\text{th value.}\]
From the ogive, \(Q_3\approx 53.0\).
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{53.0-32.5}{2}=\boxed{10.25\text{ marks}}\]
(b)(ii) Percentage passing with distinction
From the ogive, the cumulative frequency at a mark of \(72\) is approximately \(74\).
\[\text{Number scoring at least }72=80-74=6\]
\[\text{Percentage with distinction}=\frac{6}{80}\times100\%=\boxed{7.5\%}\]
Answer Details
(a) Cumulative frequency table
| Marks | Frequency | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 0–9 | 2 | 9.5 | 2 |
| 10–19 | 5 | 19.5 | 7 |
| 20–29 | 8 | 29.5 | 15 |
| 30–39 | 18 | 39.5 | 33 |
| 40–49 | 20 | 49.5 | 53 |
| 50–59 | 15 | 59.5 | 68 |
| 60–69 | 5 | 69.5 | 73 |
| 70–79 | 4 | 79.5 | 77 |
| 80–89 | 2 | 89.5 | 79 |
| 90–99 | 1 | 99.5 | 80 |
The total number of candidates is \(N=80\). Plot cumulative frequency against the upper class boundaries and draw a smooth increasing ogive.
(b)(i) Semi-interquartile range
\[Q_1\text{ is the }\frac{N}{4}=\frac{80}{4}=20\text{th value.}\]
From the ogive, \(Q_1\approx 32.5\).
\[Q_3\text{ is the }\frac{3N}{4}=\frac{3(80)}{4}=60\text{th value.}\]
From the ogive, \(Q_3\approx 53.0\).
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{53.0-32.5}{2}=\boxed{10.25\text{ marks}}\]
(b)(ii) Percentage passing with distinction
From the ogive, the cumulative frequency at a mark of \(72\) is approximately \(74\).
\[\text{Number scoring at least }72=80-74=6\]
\[\text{Percentage with distinction}=\frac{6}{80}\times100\%=\boxed{7.5\%}\]
Question 6 Report
(a) Evaluate \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\) correct to two decimal places.
(b) A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done if one particular tutor must be on the committee and two particular pupils must not be on the committee?
(a) Evaluate each term.
\({}^{9}P_{3}=9\times8\times7=504\), and \({}^{15}C_{3}=\dfrac{15\times14\times13}{3!}=455\), so \(\dfrac{{}^{9}P_{3}}{{}^{15}C_{3}}=\dfrac{504}{455}=1.1077\).
\({}^{5}C_{3}=10\), and \({}^{3}P_{2}=3\times2=6\), so \(\dfrac{{}^{5}C_{3}}{{}^{3}P_{2}}=\dfrac{10}{6}=1.6667\).
Sum \(=1.1077+1.6667=2.7744\approx 2.77\) (2 d.p.).
(b) We form a committee of 2 tutors and 5 pupils from 6 tutors and 10 pupils.
Tutors: one particular tutor is already chosen, so pick 1 more tutor from the remaining 5: \({}^{5}C_{1}=5\).
Pupils: two particular pupils are excluded, so choose 5 pupils from the remaining 8: \({}^{8}C_{5}=56\).
By the multiplication principle:
\[5\times56=280\ \text{ways}\]
Answer Details
(a) Evaluate each term.
\({}^{9}P_{3}=9\times8\times7=504\), and \({}^{15}C_{3}=\dfrac{15\times14\times13}{3!}=455\), so \(\dfrac{{}^{9}P_{3}}{{}^{15}C_{3}}=\dfrac{504}{455}=1.1077\).
\({}^{5}C_{3}=10\), and \({}^{3}P_{2}=3\times2=6\), so \(\dfrac{{}^{5}C_{3}}{{}^{3}P_{2}}=\dfrac{10}{6}=1.6667\).
Sum \(=1.1077+1.6667=2.7744\approx 2.77\) (2 d.p.).
(b) We form a committee of 2 tutors and 5 pupils from 6 tutors and 10 pupils.
Tutors: one particular tutor is already chosen, so pick 1 more tutor from the remaining 5: \({}^{5}C_{1}=5\).
Pupils: two particular pupils are excluded, so choose 5 pupils from the remaining 8: \({}^{8}C_{5}=56\).
By the multiplication principle:
\[5\times56=280\ \text{ways}\]
Question 7 Report
Solve \(2^{(2y + 2)} - 9(2^{y}) = -2\).
Write everything in terms of \(u=2^{y}\). Note \(2^{2y+2}=2^{2}\cdot 2^{2y}=4(2^{y})^{2}=4u^{2}\).
The equation \(2^{2y+2}-9(2^{y})=-2\) becomes:
\[4u^{2}-9u+2=0\]
Factorising (or using the formula \(u=\frac{9\pm\sqrt{81-32}}{8}=\frac{9\pm 7}{8}\)):
\[(4u-1)(u-2)=0\Rightarrow u=\tfrac14\ \text{or}\ u=2\]
Since \(u=2^{y}\):
- \(2^{y}=2\Rightarrow y=1\)
- \(2^{y}=\tfrac14=2^{-2}\Rightarrow y=-2\)
\[\boxed{y=1\ \text{or}\ y=-2}\]
Answer Details
Write everything in terms of \(u=2^{y}\). Note \(2^{2y+2}=2^{2}\cdot 2^{2y}=4(2^{y})^{2}=4u^{2}\).
The equation \(2^{2y+2}-9(2^{y})=-2\) becomes:
\[4u^{2}-9u+2=0\]
Factorising (or using the formula \(u=\frac{9\pm\sqrt{81-32}}{8}=\frac{9\pm 7}{8}\)):
\[(4u-1)(u-2)=0\Rightarrow u=\tfrac14\ \text{or}\ u=2\]
Since \(u=2^{y}\):
- \(2^{y}=2\Rightarrow y=1\)
- \(2^{y}=\tfrac14=2^{-2}\Rightarrow y=-2\)
\[\boxed{y=1\ \text{or}\ y=-2}\]
Question 8 Report
(a) An object P of mass 6.5kg is suspended by two light inextensible strings, AP and BP. The strings make angles 50° and 60° respectively with the downward vertical.
(i) Express the forces acting on P in component form; (ii) If P is at rest, write down the vector equation connecting all the forces; (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5 kg moves with initial velocity \(\frac{1}{2} m/s\) and final velocity \(\frac{3}{4} m/s \). Find the magnitude of its change in momentum.
(a) Weight \(W=6.5\times 10=65\ \text{N}\). Let string \(AP\) (\(50^\circ\) to the downward vertical) carry tension \(T_1\) up to the left and \(BP\) (\(60^\circ\)) carry \(T_2\) up to the right.
(i) Component form:
\[\vec{T_1}=\begin{pmatrix}-T_1\sin 50^\circ\\ T_1\cos 50^\circ\end{pmatrix},\quad \vec{T_2}=\begin{pmatrix}T_2\sin 60^\circ\\ T_2\cos 60^\circ\end{pmatrix},\quad \vec{W}=\begin{pmatrix}0\\-65\end{pmatrix}.\](ii) Equilibrium of P: \(\vec{T_1}+\vec{T_2}+\vec{W}=\vec{0}\).
(iii) Horizontal: \(T_1\sin 50^\circ=T_2\sin 60^\circ\). Using Lami's theorem (angles \(110^\circ,120^\circ,130^\circ\)):
\[T_1=\frac{65\sin 120^\circ}{\sin 110^\circ}=\frac{65(0.8660)}{0.9397}=59.9\ \text{N},\quad T_2=\frac{65\sin 130^\circ}{\sin 110^\circ}=\frac{65(0.7660)}{0.9397}=53.0\ \text{N}.\]Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).
(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).
Answer Details
(a) Weight \(W=6.5\times 10=65\ \text{N}\). Let string \(AP\) (\(50^\circ\) to the downward vertical) carry tension \(T_1\) up to the left and \(BP\) (\(60^\circ\)) carry \(T_2\) up to the right.
(i) Component form:
\[\vec{T_1}=\begin{pmatrix}-T_1\sin 50^\circ\\ T_1\cos 50^\circ\end{pmatrix},\quad \vec{T_2}=\begin{pmatrix}T_2\sin 60^\circ\\ T_2\cos 60^\circ\end{pmatrix},\quad \vec{W}=\begin{pmatrix}0\\-65\end{pmatrix}.\](ii) Equilibrium of P: \(\vec{T_1}+\vec{T_2}+\vec{W}=\vec{0}\).
(iii) Horizontal: \(T_1\sin 50^\circ=T_2\sin 60^\circ\). Using Lami's theorem (angles \(110^\circ,120^\circ,130^\circ\)):
\[T_1=\frac{65\sin 120^\circ}{\sin 110^\circ}=\frac{65(0.8660)}{0.9397}=59.9\ \text{N},\quad T_2=\frac{65\sin 130^\circ}{\sin 110^\circ}=\frac{65(0.7660)}{0.9397}=53.0\ \text{N}.\]Tension in \(AP\approx\textbf{59.9 N}\), tension in \(BP\approx\textbf{53.0 N}\).
(b) Change in momentum \(=m(v-u)=5\left(\tfrac34-\tfrac12\right)=5\times\tfrac14=\textbf{1.25 kg m/s}\).
Question 9 Report
(a) \(m \begin{pmatrix} 2 \\ 1 \end{pmatrix} + n \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}\) where m and n are scalars. Find the value of (m + n).
(b) A(-1, 3), B(2, -1) and C(5, 3) are the vertices of \(\Delta\) ABC.
(i) Express in column notation, the unit vectors parallel to AB and AC.
(ii) Use a dot product to calculate \(\stackrel \frown{BAC}\), correct to the nearest degree.
(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):
\[2m-n=5,\qquad m+2n=-4.\]From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).
\[m+n=\frac{6}{5}-\frac{13}{5}=-\frac{7}{5}=-1.4.\](b) \(A(-1,3),B(2,-1),C(5,3)\). \(\vec{AB}=\begin{pmatrix}3\\-4\end{pmatrix},\ |\vec{AB}|=5\); \(\vec{AC}=\begin{pmatrix}6\\0\end{pmatrix},\ |\vec{AC}|=6\).
(i) Unit vectors: \(\hat{AB}=\begin{pmatrix}3/5\\-4/5\end{pmatrix},\quad \hat{AC}=\begin{pmatrix}1\\0\end{pmatrix}.\)
(ii) \(\vec{AB}\cdot\vec{AC}=3(6)+(-4)(0)=18\), so
\[\cos(\widehat{BAC})=\frac{18}{5\times 6}=\frac{18}{30}=0.6\Rightarrow \widehat{BAC}=53^\circ.\]Answer Details
(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):
\[2m-n=5,\qquad m+2n=-4.\]From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).
\[m+n=\frac{6}{5}-\frac{13}{5}=-\frac{7}{5}=-1.4.\](b) \(A(-1,3),B(2,-1),C(5,3)\). \(\vec{AB}=\begin{pmatrix}3\\-4\end{pmatrix},\ |\vec{AB}|=5\); \(\vec{AC}=\begin{pmatrix}6\\0\end{pmatrix},\ |\vec{AC}|=6\).
(i) Unit vectors: \(\hat{AB}=\begin{pmatrix}3/5\\-4/5\end{pmatrix},\quad \hat{AC}=\begin{pmatrix}1\\0\end{pmatrix}.\)
(ii) \(\vec{AB}\cdot\vec{AC}=3(6)+(-4)(0)=18\), so
\[\cos(\widehat{BAC})=\frac{18}{5\times 6}=\frac{18}{30}=0.6\Rightarrow \widehat{BAC}=53^\circ.\]Question 10 Report
The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the :
(a) matrix A ; (b) inverse of A ; (c) point whose image is (-1, 1).
Let \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The two mappings give:
\(A\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}\Rightarrow 2a-3b=3,\ 2c-3d=4\)
\(A\begin{pmatrix}4\\5\end{pmatrix}=\begin{pmatrix}5\\6\end{pmatrix}\Rightarrow 4a+5b=5,\ 4c+5d=6\)
(a) Solve for \(a,b\): from \(2a-3b=3\) and \(4a+5b=5\), eliminating \(a\) gives \(11b=-1\Rightarrow b=-\tfrac{1}{11}\), then \(a=\tfrac{15}{11}\). Solve for \(c,d\): \(11d=-2\Rightarrow d=-\tfrac{2}{11}\), then \(c=\tfrac{19}{11}\).
\[A=\frac{1}{11}\begin{pmatrix}15&-1\\19&-2\end{pmatrix}\]
(b) \(\det A=\dfrac{(15)(-2)-(-1)(19)}{11^2}=\dfrac{-11}{121}=-\dfrac{1}{11}\).
\[A^{-1}=\frac{1}{\det A}\cdot\frac{1}{11}\begin{pmatrix}-2&1\\-19&15\end{pmatrix}=\begin{pmatrix}2&-1\\19&-15\end{pmatrix}\]
(c) The point mapping to \((-1,1)\) is \(A^{-1}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}2(-1)-1(1)\\19(-1)-15(1)\end{pmatrix}=\begin{pmatrix}-3\\-34\end{pmatrix}\).
So the point is \((-3,-34)\).
Answer Details
Let \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The two mappings give:
\(A\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}\Rightarrow 2a-3b=3,\ 2c-3d=4\)
\(A\begin{pmatrix}4\\5\end{pmatrix}=\begin{pmatrix}5\\6\end{pmatrix}\Rightarrow 4a+5b=5,\ 4c+5d=6\)
(a) Solve for \(a,b\): from \(2a-3b=3\) and \(4a+5b=5\), eliminating \(a\) gives \(11b=-1\Rightarrow b=-\tfrac{1}{11}\), then \(a=\tfrac{15}{11}\). Solve for \(c,d\): \(11d=-2\Rightarrow d=-\tfrac{2}{11}\), then \(c=\tfrac{19}{11}\).
\[A=\frac{1}{11}\begin{pmatrix}15&-1\\19&-2\end{pmatrix}\]
(b) \(\det A=\dfrac{(15)(-2)-(-1)(19)}{11^2}=\dfrac{-11}{121}=-\dfrac{1}{11}\).
\[A^{-1}=\frac{1}{\det A}\cdot\frac{1}{11}\begin{pmatrix}-2&1\\-19&15\end{pmatrix}=\begin{pmatrix}2&-1\\19&-15\end{pmatrix}\]
(c) The point mapping to \((-1,1)\) is \(A^{-1}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}2(-1)-1(1)\\19(-1)-15(1)\end{pmatrix}=\begin{pmatrix}-3\\-34\end{pmatrix}\).
So the point is \((-3,-34)\).
Question 11 Report
A bag contains 4 red, 6 blue and 8 green identical marbles.
(a) If three marbles are drawn at random, without replacement, calculate the probability that :
(i) all will be green ; (ii) all will have the same colour.
(b) If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.
Total marbles \(=4R+6B+8G=18\).
(a) Without replacement.
(i) all green:
\[\frac{8}{18}\cdot\frac{7}{17}\cdot\frac{6}{16}=\frac{336}{4896}=\frac{7}{102}\approx 0.069.\](ii) all the same colour = all red + all blue + all green:
\[\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}+\frac{6}{18}\cdot\frac{5}{17}\cdot\frac{4}{16}+\frac{8}{18}\cdot\frac{7}{17}\cdot\frac{6}{16}=\frac{24+120+336}{4896}=\frac{480}{4896}=\frac{5}{51}\approx 0.098.\](b) With replacement (each draw from all 18):
\[\left(\frac{4}{18}\right)^3+\left(\frac{6}{18}\right)^3+\left(\frac{8}{18}\right)^3=\left(\frac{2}{9}\right)^3+\left(\frac13\right)^3+\left(\frac49\right)^3=\frac{8+27+64}{729}=\frac{99}{729}=\frac{11}{81}\approx 0.136.\]Answer Details
Total marbles \(=4R+6B+8G=18\).
(a) Without replacement.
(i) all green:
\[\frac{8}{18}\cdot\frac{7}{17}\cdot\frac{6}{16}=\frac{336}{4896}=\frac{7}{102}\approx 0.069.\](ii) all the same colour = all red + all blue + all green:
\[\frac{4}{18}\cdot\frac{3}{17}\cdot\frac{2}{16}+\frac{6}{18}\cdot\frac{5}{17}\cdot\frac{4}{16}+\frac{8}{18}\cdot\frac{7}{17}\cdot\frac{6}{16}=\frac{24+120+336}{4896}=\frac{480}{4896}=\frac{5}{51}\approx 0.098.\](b) With replacement (each draw from all 18):
\[\left(\frac{4}{18}\right)^3+\left(\frac{6}{18}\right)^3+\left(\frac{8}{18}\right)^3=\left(\frac{2}{9}\right)^3+\left(\frac13\right)^3+\left(\frac49\right)^3=\frac{8+27+64}{729}=\frac{99}{729}=\frac{11}{81}\approx 0.136.\]Question 12 Report
A particle of mass 400g is moving under the action of two forces \(F_{1} = (35N, 210°), F_{2} = (35\sqrt{3} N, 300°)\) and a resistance of 40N. Find the magnitude of the
(a) resultant of \(F_{1}\) and \(F_{2}\).
(b) resultant force acting on the particle.
Resolve each force into components \((F\cos\theta,\ F\sin\theta)\).
\(F_{1}=(35,210^{\circ})\): \(\cos210^{\circ}=-\frac{\sqrt3}{2},\ \sin210^{\circ}=-\frac12\), so \(F_{1}=\left(-\tfrac{35\sqrt3}{2},\,-\tfrac{35}{2}\right)\approx(-30.31,\,-17.5)\).
\(F_{2}=(35\sqrt3,300^{\circ})\): \(\cos300^{\circ}=\frac12,\ \sin300^{\circ}=-\frac{\sqrt3}{2}\), so \(F_{2}=\left(\tfrac{35\sqrt3}{2},\,-\tfrac{35\cdot3}{2}\right)\approx(30.31,\,-52.5)\).
(a) Resultant of \(F_{1}\) and \(F_{2}\):
\[R=\big(-30.31+30.31,\ -17.5-52.5\big)=(0,\,-70)\]
Magnitude \(=\sqrt{0^{2}+(-70)^{2}}=70\text{N}\), directed vertically downward (along \(270^{\circ}\)).
(b) The 40N resistance opposes the motion, i.e. it acts opposite to this 70N resultant. The net (resultant) force acting on the particle is therefore
\[70-40=30\text{N}\]
directed the same way as the \(70\)N resultant (\(270^{\circ}\)). Its magnitude is \(\boxed{30\text{N}}\).
Answer Details
Resolve each force into components \((F\cos\theta,\ F\sin\theta)\).
\(F_{1}=(35,210^{\circ})\): \(\cos210^{\circ}=-\frac{\sqrt3}{2},\ \sin210^{\circ}=-\frac12\), so \(F_{1}=\left(-\tfrac{35\sqrt3}{2},\,-\tfrac{35}{2}\right)\approx(-30.31,\,-17.5)\).
\(F_{2}=(35\sqrt3,300^{\circ})\): \(\cos300^{\circ}=\frac12,\ \sin300^{\circ}=-\frac{\sqrt3}{2}\), so \(F_{2}=\left(\tfrac{35\sqrt3}{2},\,-\tfrac{35\cdot3}{2}\right)\approx(30.31,\,-52.5)\).
(a) Resultant of \(F_{1}\) and \(F_{2}\):
\[R=\big(-30.31+30.31,\ -17.5-52.5\big)=(0,\,-70)\]
Magnitude \(=\sqrt{0^{2}+(-70)^{2}}=70\text{N}\), directed vertically downward (along \(270^{\circ}\)).
(b) The 40N resistance opposes the motion, i.e. it acts opposite to this 70N resultant. The net (resultant) force acting on the particle is therefore
\[70-40=30\text{N}\]
directed the same way as the \(70\)N resultant (\(270^{\circ}\)). Its magnitude is \(\boxed{30\text{N}}\).
Question 13 Report
(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the :
(i) acceleration of the particle ; (ii) velocity of the particle after 3 seconds.
(b) \(F_{1} = (2i + 3j)N, F_{2} = (-5j)N \) and \(F_{3} = (6i - 4j)N\) act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.
(a) Resolve the polar forces into components.
\[F_1=(3,210^\circ):\ (3\cos210^\circ,3\sin210^\circ)=(-2.598,-1.5).\] \[F_2=(4,120^\circ):\ (4\cos120^\circ,4\sin120^\circ)=(-2,3.464).\]Resultant \(R=(-4.598,\ 1.964)\), so \(|R|=\sqrt{(-4.598)^2+(1.964)^2}=\sqrt{25.0}=5\ \text{N}\), at direction \(\theta=180^\circ-\tan^{-1}\!\frac{1.964}{4.598}=156.9^\circ\).
(i) Acceleration: \(a=\dfrac{|R|}{m}=\dfrac{5}{7}=0.71\ \text{m/s}^2\) in the direction \(156.9^\circ\).
(ii) Velocity after 3 s (from rest): \(v=at=0.71\times 3=2.14\ \text{m/s}\), along \(156.9^\circ\).
(b) Sum of the three forces: \(F_1+F_2+F_3=(2+0+6,\ 3-5-4)=(8,-6)\). The fourth force for equilibrium is \(F_4=-(8,-6)=(-8,6)\).
\[|F_4|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10\ \text{N},\quad \theta=180^\circ-\tan^{-1}\!\tfrac{6}{8}=143.1^\circ.\]The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.
Answer Details
(a) Resolve the polar forces into components.
\[F_1=(3,210^\circ):\ (3\cos210^\circ,3\sin210^\circ)=(-2.598,-1.5).\] \[F_2=(4,120^\circ):\ (4\cos120^\circ,4\sin120^\circ)=(-2,3.464).\]Resultant \(R=(-4.598,\ 1.964)\), so \(|R|=\sqrt{(-4.598)^2+(1.964)^2}=\sqrt{25.0}=5\ \text{N}\), at direction \(\theta=180^\circ-\tan^{-1}\!\frac{1.964}{4.598}=156.9^\circ\).
(i) Acceleration: \(a=\dfrac{|R|}{m}=\dfrac{5}{7}=0.71\ \text{m/s}^2\) in the direction \(156.9^\circ\).
(ii) Velocity after 3 s (from rest): \(v=at=0.71\times 3=2.14\ \text{m/s}\), along \(156.9^\circ\).
(b) Sum of the three forces: \(F_1+F_2+F_3=(2+0+6,\ 3-5-4)=(8,-6)\). The fourth force for equilibrium is \(F_4=-(8,-6)=(-8,6)\).
\[|F_4|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10\ \text{N},\quad \theta=180^\circ-\tan^{-1}\!\tfrac{6}{8}=143.1^\circ.\]The fourth force is 10 N at \(143.1^\circ\) to the positive x-axis.
Question 14 Report
A survey conducted revealed that four out of every twenty taxi drivers do not have a valid driving license. If 6 drivers are selected at random, calculate, correct to three decimal places, the probability that
(a) exactly 2 ;
(b) more than 3 ;
(c) at least 5; have valid driving license.
Four out of twenty lack a valid licence, so \(P(\text{no licence})=\tfrac{4}{20}=0.2\) and \(P(\text{valid})=0.8\). With \(n=6\), \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\) (\(X\) = number with valid licences).
(a) exactly 2 valid:
\[\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.015.\](b) more than 3 valid \(=P(4)+P(5)+P(6)\):
\[P(4)=\binom{6}{4}(0.8)^4(0.2)^2=15(0.4096)(0.04)=0.24576,\] \[P(5)=\binom{6}{5}(0.8)^5(0.2)=6(0.32768)(0.2)=0.393216,\quad P(6)=(0.8)^6=0.262144.\] \[\text{Sum}=0.24576+0.393216+0.262144=0.901.\](c) at least 5 valid \(=P(5)+P(6)=0.393216+0.262144=0.655\).
Answer Details
Four out of twenty lack a valid licence, so \(P(\text{no licence})=\tfrac{4}{20}=0.2\) and \(P(\text{valid})=0.8\). With \(n=6\), \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\) (\(X\) = number with valid licences).
(a) exactly 2 valid:
\[\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.015.\](b) more than 3 valid \(=P(4)+P(5)+P(6)\):
\[P(4)=\binom{6}{4}(0.8)^4(0.2)^2=15(0.4096)(0.04)=0.24576,\] \[P(5)=\binom{6}{5}(0.8)^5(0.2)=6(0.32768)(0.2)=0.393216,\quad P(6)=(0.8)^6=0.262144.\] \[\text{Sum}=0.24576+0.393216+0.262144=0.901.\](c) at least 5 valid \(=P(5)+P(6)=0.393216+0.262144=0.655\).
Question 15 Report
(a) If \(A = \begin{pmatrix} -2 & 5 \\ 4 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} 3 & 1 \\ 2 & 3 \end{pmatrix}\), find the values of x and y such that \(BA = 2\begin{pmatrix} 3 & 7 \\ -2 & x \end{pmatrix} + \begin{pmatrix} y & 4 \\ 12 & -3 \end{pmatrix}\).
(b) Two functions, f and g are defined by \(f : x \to \frac{1}{2}x + 1\) and \(g : x \to \frac{5x - 1}{3}\). Find :
(i) \(g^{-1}\) ; (ii) \(g^{-1} \circ f\).
(a) Compute \(BA\):
\[BA=\begin{pmatrix}3&1\\2&3\end{pmatrix}\begin{pmatrix}-2&5\\4&3\end{pmatrix}=\begin{pmatrix}-6+4&15+3\\-4+12&10+9\end{pmatrix}=\begin{pmatrix}-2&18\\8&19\end{pmatrix}\]
Now the right-hand side:
\[2\begin{pmatrix}3&7\\-2&x\end{pmatrix}+\begin{pmatrix}y&4\\12&-3\end{pmatrix}=\begin{pmatrix}6+y&18\\8&2x-3\end{pmatrix}\]
Equating entries: top-left \(-2=6+y\Rightarrow y=-8\); bottom-right \(19=2x-3\Rightarrow x=11\). (The other entries \(18=18,\ 8=8\) check out.)
\[\boxed{x=11,\ y=-8}\]
(b)(i) Let \(y=\dfrac{5x-1}{3}\). Then \(3y=5x-1\Rightarrow x=\dfrac{3y+1}{5}\), so
\[g^{-1}(x)=\frac{3x+1}{5}\]
(ii) \(g^{-1}\circ f(x)=g^{-1}\!\left(\tfrac12 x+1\right)=\dfrac{3\left(\tfrac12 x+1\right)+1}{5}=\dfrac{\tfrac32 x+4}{5}=\dfrac{3x+8}{10}\)
Answer Details
(a) Compute \(BA\):
\[BA=\begin{pmatrix}3&1\\2&3\end{pmatrix}\begin{pmatrix}-2&5\\4&3\end{pmatrix}=\begin{pmatrix}-6+4&15+3\\-4+12&10+9\end{pmatrix}=\begin{pmatrix}-2&18\\8&19\end{pmatrix}\]
Now the right-hand side:
\[2\begin{pmatrix}3&7\\-2&x\end{pmatrix}+\begin{pmatrix}y&4\\12&-3\end{pmatrix}=\begin{pmatrix}6+y&18\\8&2x-3\end{pmatrix}\]
Equating entries: top-left \(-2=6+y\Rightarrow y=-8\); bottom-right \(19=2x-3\Rightarrow x=11\). (The other entries \(18=18,\ 8=8\) check out.)
\[\boxed{x=11,\ y=-8}\]
(b)(i) Let \(y=\dfrac{5x-1}{3}\). Then \(3y=5x-1\Rightarrow x=\dfrac{3y+1}{5}\), so
\[g^{-1}(x)=\frac{3x+1}{5}\]
(ii) \(g^{-1}\circ f(x)=g^{-1}\!\left(\tfrac12 x+1\right)=\dfrac{3\left(\tfrac12 x+1\right)+1}{5}=\dfrac{\tfrac32 x+4}{5}=\dfrac{3x+8}{10}\)
Question 16 Report
(a) Find, from first principles, the derivative of \(f(x) = (2x + 3)^{2}\).
(b) Evaluate : \(\int_{1} ^{2} \frac{(x + 1)(x^{2} - 2x + 2)}{x^{2}} \mathrm {d} x\)
(a) First principles: \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\) with \(f(x)=(2x+3)^2\).
\[f(x+h)-f(x)=\big[(2x+3)+2h\big]^2-(2x+3)^2\]
\[=2(2x+3)(2h)+(2h)^2=4h(2x+3)+4h^2\]
Divide by \(h\): \(4(2x+3)+4h\). Taking the limit as \(h\to0\):
\[f'(x)=4(2x+3)=8x+12\]
(b) First expand the numerator: \((x+1)(x^2-2x+2)=x^3-x^2+2\). Then
\[\frac{x^3-x^2+2}{x^2}=x-1+\frac{2}{x^2}=x-1+2x^{-2}\]
\[\int_{1}^{2}\left(x-1+2x^{-2}\right)dx=\left[\frac{x^2}{2}-x-\frac{2}{x}\right]_{1}^{2}\]
At \(x=2\): \(2-2-1=-1\). At \(x=1\): \(\tfrac12-1-2=-2.5\).
\[=-1-(-2.5)=\tfrac32\]
Answer Details
(a) First principles: \(f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\) with \(f(x)=(2x+3)^2\).
\[f(x+h)-f(x)=\big[(2x+3)+2h\big]^2-(2x+3)^2\]
\[=2(2x+3)(2h)+(2h)^2=4h(2x+3)+4h^2\]
Divide by \(h\): \(4(2x+3)+4h\). Taking the limit as \(h\to0\):
\[f'(x)=4(2x+3)=8x+12\]
(b) First expand the numerator: \((x+1)(x^2-2x+2)=x^3-x^2+2\). Then
\[\frac{x^3-x^2+2}{x^2}=x-1+\frac{2}{x^2}=x-1+2x^{-2}\]
\[\int_{1}^{2}\left(x-1+2x^{-2}\right)dx=\left[\frac{x^2}{2}-x-\frac{2}{x}\right]_{1}^{2}\]
At \(x=2\): \(2-2-1=-1\). At \(x=1\): \(\tfrac12-1-2=-2.5\).
\[=-1-(-2.5)=\tfrac32\]
Question 17 Report
(a) Write the following as column vectors: \(r = (10N, 090°) ; q = (8N, 135°)\).
(b) Use your answer in (a) to find \((r + q)\).
(a) A force \((F,\theta)\) resolves into the column vector \(\begin{pmatrix}F\cos\theta\\ F\sin\theta\end{pmatrix}\), where \(\theta\) is measured anticlockwise from the positive x-axis.
For \(r=(10\text{N},090^{\circ})\): \(\cos90^{\circ}=0,\ \sin90^{\circ}=1\).
\[r=\begin{pmatrix}0\\ 10\end{pmatrix}\]
For \(q=(8\text{N},135^{\circ})\): \(\cos135^{\circ}=-\frac{\sqrt2}{2},\ \sin135^{\circ}=\frac{\sqrt2}{2}\).
\[q=\begin{pmatrix}8(-\frac{\sqrt2}{2})\\ 8(\frac{\sqrt2}{2})\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 4\sqrt2\end{pmatrix}\]
(b) Add corresponding components:
\[r+q=\begin{pmatrix}0-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}\approx\begin{pmatrix}-5.66\\ 15.66\end{pmatrix}\text{N}\]
Answer Details
(a) A force \((F,\theta)\) resolves into the column vector \(\begin{pmatrix}F\cos\theta\\ F\sin\theta\end{pmatrix}\), where \(\theta\) is measured anticlockwise from the positive x-axis.
For \(r=(10\text{N},090^{\circ})\): \(\cos90^{\circ}=0,\ \sin90^{\circ}=1\).
\[r=\begin{pmatrix}0\\ 10\end{pmatrix}\]
For \(q=(8\text{N},135^{\circ})\): \(\cos135^{\circ}=-\frac{\sqrt2}{2},\ \sin135^{\circ}=\frac{\sqrt2}{2}\).
\[q=\begin{pmatrix}8(-\frac{\sqrt2}{2})\\ 8(\frac{\sqrt2}{2})\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 4\sqrt2\end{pmatrix}\]
(b) Add corresponding components:
\[r+q=\begin{pmatrix}0-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}=\begin{pmatrix}-4\sqrt2\\ 10+4\sqrt2\end{pmatrix}\approx\begin{pmatrix}-5.66\\ 15.66\end{pmatrix}\text{N}\]
Question 18 Report
If \(f(x) = 6x^{3} + 13x^{2} + 2x - 5\) and \(f(-1) = 0\), find the factors of f(x).
Since \(f(-1) = 0\), \((x + 1)\) is a factor of \(f(x) = 6x^3 + 13x^2 + 2x - 5\).
Divide \(f(x)\) by \((x + 1)\):
\[6x^3 + 13x^2 + 2x - 5 = (x + 1)(6x^2 + 7x - 5).\]
Now factor the quadratic \(6x^2 + 7x - 5\). We need two numbers multiplying to \(6\times(-5) = -30\) and adding to \(7\): these are \(10\) and \(-3\):
\[6x^2 + 10x - 3x - 5 = 2x(3x + 5) - 1(3x + 5) = (3x + 5)(2x - 1).\]
Therefore
\[f(x) = (x + 1)(2x - 1)(3x + 5).\]
Answer Details
Since \(f(-1) = 0\), \((x + 1)\) is a factor of \(f(x) = 6x^3 + 13x^2 + 2x - 5\).
Divide \(f(x)\) by \((x + 1)\):
\[6x^3 + 13x^2 + 2x - 5 = (x + 1)(6x^2 + 7x - 5).\]
Now factor the quadratic \(6x^2 + 7x - 5\). We need two numbers multiplying to \(6\times(-5) = -30\) and adding to \(7\): these are \(10\) and \(-3\):
\[6x^2 + 10x - 3x - 5 = 2x(3x + 5) - 1(3x + 5) = (3x + 5)(2x - 1).\]
Therefore
\[f(x) = (x + 1)(2x - 1)(3x + 5).\]
