WAEC SSCE - Further Mathematics - 2006 (Essay)
Ajụjụ 1 Ripọtì
(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings \(\overrightarrow{XP}\) and \(\overrightarrow{YP}\). The strings are inclined at 60° and 40° respectively to the downward vertical. Find, correct to two decimal places, the tensionsin the strings. [Take g = \(10 ms^{-2}\)].
(b) The height h metres, of a ball thrown into the air is \(2 + 20t + kt^{2}\), after t seconds. If it takes 2 seconds for the ball to reach its highest point, find
(i) the value of k (ii) its highest point from the point of throw.
(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:
\[\frac{T_{XP}}{\sin 140^\circ}=\frac{T_{YP}}{\sin 120^\circ}=\frac{150}{\sin 100^\circ}.\] \[T_{XP}=\frac{150\sin 140^\circ}{\sin 100^\circ}=\frac{150(0.6428)}{0.9848}=97.91\ \text{N}.\] \[T_{YP}=\frac{150\sin 120^\circ}{\sin 100^\circ}=\frac{150(0.8660)}{0.9848}=131.91\ \text{N}.\]Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).
(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):
\[20+2k(2)=0\Rightarrow 20+4k=0\Rightarrow k=-5.\](ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.
Akọwa Nkọwa
(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:
\[\frac{T_{XP}}{\sin 140^\circ}=\frac{T_{YP}}{\sin 120^\circ}=\frac{150}{\sin 100^\circ}.\] \[T_{XP}=\frac{150\sin 140^\circ}{\sin 100^\circ}=\frac{150(0.6428)}{0.9848}=97.91\ \text{N}.\] \[T_{YP}=\frac{150\sin 120^\circ}{\sin 100^\circ}=\frac{150(0.8660)}{0.9848}=131.91\ \text{N}.\]Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).
(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):
\[20+2k(2)=0\Rightarrow 20+4k=0\Rightarrow k=-5.\](ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.
Ajụjụ 2 Ripọtì
(a) Using the substitution \(u = 5 - x^{2}\), evaluate \(\int_{1}^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\).
(b) If \(y = px^{2} + qx; \frac{\mathrm d y}{\mathrm d x} = 6x + 7\) and \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6\), find the values of p and q.
(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).
\[\int_{1}^{2}\frac{x}{\sqrt{5-x^2}}\,dx=\int_{4}^{1}\frac{-\tfrac12\,du}{\sqrt{u}}=-\frac12\int_{4}^{1}u^{-1/2}\,du=-\frac12\Big[2\sqrt u\Big]_{4}^{1}=-\Big[\sqrt u\Big]_{4}^{1}.\] \[=-\big(\sqrt1-\sqrt4\big)=-(1-2)=1.\]The value of the integral is 1.
(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).
Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.
So \(p=3,\ q=7\).
Akọwa Nkọwa
(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).
\[\int_{1}^{2}\frac{x}{\sqrt{5-x^2}}\,dx=\int_{4}^{1}\frac{-\tfrac12\,du}{\sqrt{u}}=-\frac12\int_{4}^{1}u^{-1/2}\,du=-\frac12\Big[2\sqrt u\Big]_{4}^{1}=-\Big[\sqrt u\Big]_{4}^{1}.\] \[=-\big(\sqrt1-\sqrt4\big)=-(1-2)=1.\]The value of the integral is 1.
(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).
Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.
So \(p=3,\ q=7\).
Ajụjụ 3 Ripọtì
The sum of the 2nd and 5th terms of an arithmetic progression (AP) is 42. If the difference between the 6th and 3rd term is 12, find the
(i) Common difference
(ii) first term
(iii) 20th term.
Let the first term be \(a\) and common difference \(d.\) The \(n\)th term is \(T_n=a+(n-1)d.\)
Set up the equations. Sum of 2nd and 5th terms is 42:
\[(a+d)+(a+4d)=42\Rightarrow 2a+5d=42.\]
Difference between 6th and 3rd terms is 12:
\[(a+5d)-(a+2d)=12\Rightarrow 3d=12.\]
(i) Common difference. \(3d=12\Rightarrow d=4.\)
(ii) First term. Substitute into \(2a+5d=42:\)
\[2a+5(4)=42\Rightarrow 2a=22\Rightarrow a=11.\]
(iii) 20th term.
\[T_{20}=a+19d=11+19(4)=11+76=87.\]
Akọwa Nkọwa
Let the first term be \(a\) and common difference \(d.\) The \(n\)th term is \(T_n=a+(n-1)d.\)
Set up the equations. Sum of 2nd and 5th terms is 42:
\[(a+d)+(a+4d)=42\Rightarrow 2a+5d=42.\]
Difference between 6th and 3rd terms is 12:
\[(a+5d)-(a+2d)=12\Rightarrow 3d=12.\]
(i) Common difference. \(3d=12\Rightarrow d=4.\)
(ii) First term. Substitute into \(2a+5d=42:\)
\[2a+5(4)=42\Rightarrow 2a=22\Rightarrow a=11.\]
(iii) 20th term.
\[T_{20}=a+19d=11+19(4)=11+76=87.\]
Ajụjụ 4 Ripọtì
(a) There are 6 points in a plane. How many triangles can be formed with the points?
(b) A family of 6 is to be seated in a row . In how many ways can this be done if the father and mother are not to be seated together?
(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:
\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]
So 20 triangles can be formed.
(b) Seating 6 people in a row with father and mother NOT together.
Total arrangements of 6 people in a row: \(6!=720.\)
Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:
\[5!\times2!=120\times2=240.\]
Therefore arrangements with them NOT together:
\[720-240=480.\]
Akọwa Nkọwa
(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:
\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]
So 20 triangles can be formed.
(b) Seating 6 people in a row with father and mother NOT together.
Total arrangements of 6 people in a row: \(6!=720.\)
Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:
\[5!\times2!=120\times2=240.\]
Therefore arrangements with them NOT together:
\[720-240=480.\]
Ajụjụ 5 Ripọtì
A body of mass 5 kg resting on a smooth horizontal plane, is acted upon by force 6i + 2j, 5i + 4j and 4i - j. Calculate the:
(a) velocity of the body
(b) Magnitude of its velocity after 4s.
A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.
Resultant force.
\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]
Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)
\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]
(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)
\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]
(b) Magnitude of the velocity after 4 s.
\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]
(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))
Akọwa Nkọwa
A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.
Resultant force.
\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]
Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)
\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]
(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)
\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]
(b) Magnitude of the velocity after 4 s.
\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]
(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))
Ajụjụ 6 Ripọtì
If \(2^{2x - 3y} = 32\) and \(\log_{y} x = 2\), find the values of x and y.
Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)
First equation. Since \(32=2^5,\) equate indices:
\[2x-3y=5.\quad(1)\]
Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)
Substitute (2) into (1):
\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]
Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)
A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence
\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]
Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.
Akọwa Nkọwa
Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)
First equation. Since \(32=2^5,\) equate indices:
\[2x-3y=5.\quad(1)\]
Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)
Substitute (2) into (1):
\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]
Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)
A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence
\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]
Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.
Ajụjụ 7 Ripọtì
(a) Two pupils are chosen at random from a group of 4 boys and 5 girls. Find the probability that the two pupils chosen would be boys.
(b) Twenty percent of the total production of transistors produced by a machine are below standard. If a random sample of 6 transistors produced by the machine is taken, what is the probability of getting (i) exactly 2 standard transistors (ii) exactly 1 standard transistor (iii) at least 2 standard transistors (iv) at most 2 standard transistors?
(a) Choosing 2 from 4 boys and 5 girls (9 pupils):
\[P(\text{both boys})=\frac{\binom{4}{2}}{\binom{9}{2}}=\frac{6}{36}=\frac16.\](b) Let \(p=P(\text{standard})=0.8\) and \(q=P(\text{below standard})=0.2\), with \(n=6\). This is binomial \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\).
(i) exactly 2 standard: \(\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.0154\).
(ii) exactly 1 standard: \(\binom{6}{1}(0.8)(0.2)^5=6(0.8)(0.00032)=0.0015\).
(iii) at least 2 standard: \(1-P(0)-P(1)\), where \(P(0)=(0.2)^6=0.000064\).
\[=1-0.000064-0.001536=0.9984.\](iv) at most 2 standard: \(P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.0170\).
Akọwa Nkọwa
(a) Choosing 2 from 4 boys and 5 girls (9 pupils):
\[P(\text{both boys})=\frac{\binom{4}{2}}{\binom{9}{2}}=\frac{6}{36}=\frac16.\](b) Let \(p=P(\text{standard})=0.8\) and \(q=P(\text{below standard})=0.2\), with \(n=6\). This is binomial \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\).
(i) exactly 2 standard: \(\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.0154\).
(ii) exactly 1 standard: \(\binom{6}{1}(0.8)(0.2)^5=6(0.8)(0.00032)=0.0015\).
(iii) at least 2 standard: \(1-P(0)-P(1)\), where \(P(0)=(0.2)^6=0.000064\).
\[=1-0.000064-0.001536=0.9984.\](iv) at most 2 standard: \(P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.0170\).
Ajụjụ 8 Ripọtì
(a) A man P has 5 red, 3 blue and 2 white buses. Another man Q has 3 red, 2 blue and 4 white buses. A bus owned by P is involved in an accident with a bus belonging to Q. Calculate the probability that the two buses are not of the same color.
(b) A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?
(a) P owns \(5R,3B,2W\) (total 10); Q owns \(3R,2B,4W\) (total 9). One bus of P meets one bus of Q. First find P(same colour):
\[P(\text{both R})=\tfrac{5}{10}\cdot\tfrac{3}{9}=\tfrac{15}{90},\quad P(\text{both B})=\tfrac{3}{10}\cdot\tfrac{2}{9}=\tfrac{6}{90},\quad P(\text{both W})=\tfrac{2}{10}\cdot\tfrac{4}{9}=\tfrac{8}{90}.\] \[P(\text{same})=\frac{15+6+8}{90}=\frac{29}{90}.\] \[P(\text{not same colour})=1-\frac{29}{90}=\frac{61}{90}.\](b) Outward journey: air Nigeria \(\to\) Ghana in 6 ways; ship Ghana \(\to\) Liberia in 4 ways. On the return he must not repeat the airline or shipping line, so: ship Liberia \(\to\) Ghana in \(4-1=3\) ways; air Ghana \(\to\) Nigeria in \(6-1=5\) ways.
\[\text{Total}=6\times 4\times 3\times 5=360\ \text{ways}.\]Akọwa Nkọwa
(a) P owns \(5R,3B,2W\) (total 10); Q owns \(3R,2B,4W\) (total 9). One bus of P meets one bus of Q. First find P(same colour):
\[P(\text{both R})=\tfrac{5}{10}\cdot\tfrac{3}{9}=\tfrac{15}{90},\quad P(\text{both B})=\tfrac{3}{10}\cdot\tfrac{2}{9}=\tfrac{6}{90},\quad P(\text{both W})=\tfrac{2}{10}\cdot\tfrac{4}{9}=\tfrac{8}{90}.\] \[P(\text{same})=\frac{15+6+8}{90}=\frac{29}{90}.\] \[P(\text{not same colour})=1-\frac{29}{90}=\frac{61}{90}.\](b) Outward journey: air Nigeria \(\to\) Ghana in 6 ways; ship Ghana \(\to\) Liberia in 4 ways. On the return he must not repeat the airline or shipping line, so: ship Liberia \(\to\) Ghana in \(4-1=3\) ways; air Ghana \(\to\) Nigeria in \(6-1=5\) ways.
\[\text{Total}=6\times 4\times 3\times 5=360\ \text{ways}.\]Ajụjụ 9 Ripọtì
(a) Simplify \(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\), leaving your answer in the form \(a + b\sqrt{c}\); where a, b and c are rational numbers.
(b) The points (7, 3), (2, 8) and (-3, 3) lie on a circle. Find the (i) equation and (ii) radius of the circle.
(a) Simplify \(\dfrac{\sqrt{75}-3}{\sqrt{3}+1}\).
Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):
\[\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{15-5\sqrt{3}-3\sqrt{3}+3}{3-1}=\frac{18-8\sqrt{3}}{2}=9-4\sqrt{3}.\]So \(a=9,\ b=-4,\ c=3\).
(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).
\((7,3):\ 58+14g+6f+c=0\) (i)
\((-3,3):\ 18-6g+6f+c=0\) (ii)
\((2,8):\ 68+4g+16f+c=0\) (iii)
(i) - (ii): \(40+20g=0\Rightarrow g=-2\).
(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\) (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\) (v).
(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).
(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).
(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).
Akọwa Nkọwa
(a) Simplify \(\dfrac{\sqrt{75}-3}{\sqrt{3}+1}\).
Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):
\[\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{15-5\sqrt{3}-3\sqrt{3}+3}{3-1}=\frac{18-8\sqrt{3}}{2}=9-4\sqrt{3}.\]So \(a=9,\ b=-4,\ c=3\).
(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).
\((7,3):\ 58+14g+6f+c=0\) (i)
\((-3,3):\ 18-6g+6f+c=0\) (ii)
\((2,8):\ 68+4g+16f+c=0\) (iii)
(i) - (ii): \(40+20g=0\Rightarrow g=-2\).
(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\) (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\) (v).
(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).
(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).
(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).
Ajụjụ 10 Ripọtì
If (x + 2) and (x - 1) are factors of \(f(x) = 6x^{4} + mx^{3} - 13x^{2} + nx + 14\), find the
(a) values of m and n.
(b) remainder when f(x) is divided be (x + 1).
Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)
(a) Values of m and n.
\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)
So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)
\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)
Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)
\[m=12,\qquad n=-19.\]
(b) Remainder when f(x) is divided by \((x+1).\)
By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)
\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]
Remainder \(=14.\)
Akọwa Nkọwa
Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)
(a) Values of m and n.
\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)
So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)
\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)
Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)
\[m=12,\qquad n=-19.\]
(b) Remainder when f(x) is divided by \((x+1).\)
By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)
\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]
Remainder \(=14.\)
Ajụjụ 11 Ripọtì
The table shows the distribution of the ages of a group of people in a village.
| Ages (in years) | 15 - 18 | 19 - 22 | 23 - 26 | 27 - 30 | 31 - 34 | 35 - 38 |
| Frequency | 40 | 33 | 25 | 10 | 8 | 4 |
Using an assumed mean of 24.5, calculate the mean of the distribution.
The assumed-mean (working-mean) method for grouped data uses \(\bar{x}=A+\dfrac{\sum fd}{\sum f}\), where \(d=x-A\), \(x\) is the class mid-value and \(A=24.5\) is the assumed mean.
Each class has width 4, so the mid-value is the average of the class limits, e.g. \(\tfrac{15+18}{2}=16.5\).
| Ages | Frequency \(f\) | Mid-value \(x\) | \(d=x-24.5\) | \(fd\) |
|---|---|---|---|---|
| 15 - 18 | 40 | 16.5 | -8 | -320 |
| 19 - 22 | 33 | 20.5 | -4 | -132 |
| 23 - 26 | 25 | 24.5 | 0 | 0 |
| 27 - 30 | 10 | 28.5 | 4 | 40 |
| 31 - 34 | 8 | 32.5 | 8 | 64 |
| 35 - 38 | 4 | 36.5 | 12 | 48 |
| Total | 120 | -300 |
\[\bar{x}=A+\frac{\sum fd}{\sum f}=24.5+\frac{-300}{120}=24.5-2.5=22.0\]
The mean age of the distribution is 22 years.
Akọwa Nkọwa
The assumed-mean (working-mean) method for grouped data uses \(\bar{x}=A+\dfrac{\sum fd}{\sum f}\), where \(d=x-A\), \(x\) is the class mid-value and \(A=24.5\) is the assumed mean.
Each class has width 4, so the mid-value is the average of the class limits, e.g. \(\tfrac{15+18}{2}=16.5\).
| Ages | Frequency \(f\) | Mid-value \(x\) | \(d=x-24.5\) | \(fd\) |
|---|---|---|---|---|
| 15 - 18 | 40 | 16.5 | -8 | -320 |
| 19 - 22 | 33 | 20.5 | -4 | -132 |
| 23 - 26 | 25 | 24.5 | 0 | 0 |
| 27 - 30 | 10 | 28.5 | 4 | 40 |
| 31 - 34 | 8 | 32.5 | 8 | 64 |
| 35 - 38 | 4 | 36.5 | 12 | 48 |
| Total | 120 | -300 |
\[\bar{x}=A+\frac{\sum fd}{\sum f}=24.5+\frac{-300}{120}=24.5-2.5=22.0\]
The mean age of the distribution is 22 years.
Ajụjụ 12 Ripọtì
Find the equation of the tangent to the curve \(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\) at the point (1, 0).
Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)
Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)
\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]
Gradient at the point. At \(x=1:\)
\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]
Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)
\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]
Akọwa Nkọwa
Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)
Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)
\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]
Gradient at the point. At \(x=1:\)
\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]
Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)
\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]
Ajụjụ 13 Ripọtì
(a) A body of mass 5 kg is placed on a smooth plane inclined at an angle 30° to the horizontal. Find the magnitude of the force: (i) acting parallel to the plane (ii) required to keep the body in equilibrium. [Take \(g = 10 ms^{-2}\)].
(b) A uniform plank PQ of length 10m and mass m kg rests on two supports A and B, where \(|PA| = |BQ| = 1m\). A load of mass 8 kg is placed on the plank at point C such that \(|AC| = 3.5 m\). If the reaction at B is 100 N, calculate the (i) value of m (ii) reaction at A. [Take \(g = 10 ms^{-2}\)].
(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).
(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).
(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).
(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).
Take moments about A (clockwise = anticlockwise):
\[R_B(8)=mg(4)+80(3.5)\Rightarrow 100(8)=40m+280.\] \[800-280=40m\Rightarrow m=\frac{520}{40}=\textbf{13 kg}.\](ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).
\[R_A=210-100=\textbf{110 N}.\]Akọwa Nkọwa
(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).
(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).
(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).
(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).
Take moments about A (clockwise = anticlockwise):
\[R_B(8)=mg(4)+80(3.5)\Rightarrow 100(8)=40m+280.\] \[800-280=40m\Rightarrow m=\frac{520}{40}=\textbf{13 kg}.\](ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).
\[R_A=210-100=\textbf{110 N}.\]Ajụjụ 14 Ripọtì
An object is projected vertically upwards with a velocity of 80 m/s. Find the :
(a) Maximum height reached
(b) Time taken to return to the point of projection. [Take g = \(10 ms^{-2}\)].
Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).
(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):
\[H = \frac{u^2}{2g} = \frac{80^2}{2\times 10} = \frac{6400}{20} = 320\ \text{m}.\]The maximum height reached is 320 m.
(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).
\[T = 2t = 2\times 8 = 16\ \text{s}.\]The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)
Akọwa Nkọwa
Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).
(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):
\[H = \frac{u^2}{2g} = \frac{80^2}{2\times 10} = \frac{6400}{20} = 320\ \text{m}.\]The maximum height reached is 320 m.
(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).
\[T = 2t = 2\times 8 = 16\ \text{s}.\]The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)
Ajụjụ 15 Ripọtì
(a) The gradient of the tangent to the curve \(y = 4x^{3}\) at points P and Q is 108. Find the coordinates of P and Q.
(b) Given that \(A = 45°, B = 30°, \sin (A + B) = \sin A \cos B + \sin B \cos A\) and \(\cos (A + B) = \cos A \cos B - \sin A \sin B\)
(i) Show that \(\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}\) and \(\cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4}\)
(ii) hence find \(\tan 15°\).
(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:
\[12x^2=108\Rightarrow x^2=9\Rightarrow x=\pm 3.\]At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).
So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).
(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):
\[\sin 75^\circ=\sin A\cos B+\sin B\cos A=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\] \[\cos 75^\circ=\cos A\cos B-\sin A\sin B=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}.\]Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.
(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:
\[\tan 15^\circ=\frac{(\sqrt6-\sqrt2)^2}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{6-2\sqrt{12}+2}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3.\]So \(\tan 15^\circ=2-\sqrt3\).
Akọwa Nkọwa
(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:
\[12x^2=108\Rightarrow x^2=9\Rightarrow x=\pm 3.\]At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).
So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).
(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):
\[\sin 75^\circ=\sin A\cos B+\sin B\cos A=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\] \[\cos 75^\circ=\cos A\cos B-\sin A\sin B=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}.\]Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.
(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:
\[\tan 15^\circ=\frac{(\sqrt6-\sqrt2)^2}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{6-2\sqrt{12}+2}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3.\]So \(\tan 15^\circ=2-\sqrt3\).
Ajụjụ 16 Ripọtì
(a) A ball P moving with velocity \(2u ms^{-1}\), collides with a similar ball Q, of different mass, which is at rest. After collision, Q moves with \(u ms^{-1}\) and P with velocity \(\frac{1}{2} u ms^{-2}\), in the opposite direction. Find the ratio of the masses of P and Q.
(b) Two forces of magnitudes 3 N and 7 N have a resultant of magnitude 5 N. Calculate, correct to one decimal place, the angle between the two forces.
(c) \(AB = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(CB = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) are two vectors in the XY- plane. If V is the midpoint of AB, find CV.
(a) Let masses be \(m_P\) and \(m_Q\). Taking the original direction of P as positive, after collision P reverses to \(-\tfrac12 u\) and Q moves \(+u\). Conservation of momentum:
\[m_P(2u)+m_Q(0)=m_P\!\left(-\tfrac12 u\right)+m_Q(u).\] \[2m_P u+\tfrac12 m_P u=m_Q u\Rightarrow \tfrac52 m_P=m_Q\Rightarrow \frac{m_P}{m_Q}=\frac{2}{5}.\]Ratio \(m_P:m_Q=\textbf{2:5}\).
(b) For forces 3 N and 7 N with resultant 5 N at angle \(\theta\) between them:
\[R^2=P^2+Q^2+2PQ\cos\theta\Rightarrow 25=9+49+2(3)(7)\cos\theta.\] \[42\cos\theta=25-58=-33\Rightarrow \cos\theta=-0.7857\Rightarrow \theta\approx 141.8^\circ.\](c) \(\vec{AB}=\begin{pmatrix}-4\\6\end{pmatrix},\ \vec{CB}=\begin{pmatrix}2\\-3\end{pmatrix}\). V is the midpoint of AB, so \(\vec{BV}=-\tfrac12\vec{AB}=\begin{pmatrix}2\\-3\end{pmatrix}\). Then
\[\vec{CV}=\vec{CB}+\vec{BV}=\begin{pmatrix}2\\-3\end{pmatrix}+\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}4\\-6\end{pmatrix}.\]Akọwa Nkọwa
(a) Let masses be \(m_P\) and \(m_Q\). Taking the original direction of P as positive, after collision P reverses to \(-\tfrac12 u\) and Q moves \(+u\). Conservation of momentum:
\[m_P(2u)+m_Q(0)=m_P\!\left(-\tfrac12 u\right)+m_Q(u).\] \[2m_P u+\tfrac12 m_P u=m_Q u\Rightarrow \tfrac52 m_P=m_Q\Rightarrow \frac{m_P}{m_Q}=\frac{2}{5}.\]Ratio \(m_P:m_Q=\textbf{2:5}\).
(b) For forces 3 N and 7 N with resultant 5 N at angle \(\theta\) between them:
\[R^2=P^2+Q^2+2PQ\cos\theta\Rightarrow 25=9+49+2(3)(7)\cos\theta.\] \[42\cos\theta=25-58=-33\Rightarrow \cos\theta=-0.7857\Rightarrow \theta\approx 141.8^\circ.\](c) \(\vec{AB}=\begin{pmatrix}-4\\6\end{pmatrix},\ \vec{CB}=\begin{pmatrix}2\\-3\end{pmatrix}\). V is the midpoint of AB, so \(\vec{BV}=-\tfrac12\vec{AB}=\begin{pmatrix}2\\-3\end{pmatrix}\). Then
\[\vec{CV}=\vec{CB}+\vec{BV}=\begin{pmatrix}2\\-3\end{pmatrix}+\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}4\\-6\end{pmatrix}.\]Ajụjụ 17 Ripọtì
The table shows the distribution of hours spent at work by the employees of a factory in a week.
| Time (in hours) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 |
| No of persons | 8 | 11 | 23 | 25 | 8 | 5 |
(a) Draw an ogive for the distribution.
(b) Using your graph, estimate the (i) lower quartile (ii) median (iii) 40th percentile (iv) number of employees that spent at least 50 hours 30 minutes at work.
(a) Cumulative-frequency table
| Time (hours) | Frequency | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 20–29 | 8 | 19.5–29.5 | 8 |
| 30–39 | 11 | 29.5–39.5 | 19 |
| 40–49 | 23 | 39.5–49.5 | 42 |
| 50–59 | 25 | 49.5–59.5 | 67 |
| 60–69 | 8 | 59.5–69.5 | 75 |
| 70–79 | 5 | 69.5–79.5 | 80 |
Plot cumulative frequency against the upper class boundary, beginning at .5, 0. The resulting less-than ogive is:
(b) Total number of employees, N=80.
- The lower quartile corresponds to cumulative frequency \(\frac{N}{4}=20\). From the ogive, \(Q_1\approx 40\) hours.
- The median corresponds to cumulative frequency \(\frac{N}{2}=40\). From the ogive, median \(\approx 49\) hours.
- The 40th percentile corresponds to cumulative frequency \(0.40\times80=32\). From the ogive, \(P_{40}\approx45.5\) hours.
- At 50 hours 30 minutes, that is, 50.5 hours, the cumulative frequency read from the ogive is approximately 45. Hence, the number who spent at least 50.5 hours at work is \[80-45=\boxed{35\text{ employees}}.\]
Akọwa Nkọwa
(a) Cumulative-frequency table
| Time (hours) | Frequency | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 20–29 | 8 | 19.5–29.5 | 8 |
| 30–39 | 11 | 29.5–39.5 | 19 |
| 40–49 | 23 | 39.5–49.5 | 42 |
| 50–59 | 25 | 49.5–59.5 | 67 |
| 60–69 | 8 | 59.5–69.5 | 75 |
| 70–79 | 5 | 69.5–79.5 | 80 |
Plot cumulative frequency against the upper class boundary, beginning at .5, 0. The resulting less-than ogive is:
(b) Total number of employees, N=80.
- The lower quartile corresponds to cumulative frequency \(\frac{N}{4}=20\). From the ogive, \(Q_1\approx 40\) hours.
- The median corresponds to cumulative frequency \(\frac{N}{2}=40\). From the ogive, median \(\approx 49\) hours.
- The 40th percentile corresponds to cumulative frequency \(0.40\times80=32\). From the ogive, \(P_{40}\approx45.5\) hours.
- At 50 hours 30 minutes, that is, 50.5 hours, the cumulative frequency read from the ogive is approximately 45. Hence, the number who spent at least 50.5 hours at work is \[80-45=\boxed{35\text{ employees}}.\]
