WAEC SSCE - Further Mathematics - 2023 (Essay)
Pergunta 1 Relatório
(a) A see-saw pivoted at the middle is kept in balance by weights of Richard, John and Philip such that only Richard whose mass is 60 kg sits on one side. If they sit at distances 2 m , 3 m , and 4 m respectively from the pivot and Philip is 15 kg, find the mass of John.
(bi) A body of mass 12 kg rests on a rough plane inclined at an angle of 30º to the horizontal. The coefficient of friction between the body and the plane is \(\frac{2}{3}\). A force of magnitude P Newton acts on the body along the inclined plane. Find the value of P, if the body is at the point of moving:
down the plane;
[Take \(g = 10 ms ^{-2}\)]
(bii) A body of mass 12 kg rests on a rough plane inclined at an angle of 30º to the horizontal. The coefficient of friction between the body and the plane is \(\frac{2}{3}\). A force of magnitude P Newton acts on the body along the inclined plane. Find the value of P, if the body is at the point of moving:
up the plane;
[Take \(g = 10 ms ^{-2}\)]
(a) 
∑ clockwise moments = ∑ anti-clockwise moments
= 3 x mJ + 4 x 15 = 60 x 2
= 3mJ + 60 = 120
= 3mJ = 120 - 60
= 3mJ = 60
\(∴mJ=\frac{60}{3}=20kg\)
(bi)
\(∑f_y=0==>N-mgcosθ=0\)
\(=N-12(10)cos30^o=0\)
\(=N=120cos 30^o\)
\(=N=60√3N\)
At the point of moving down,
\(∑f_x=0==>μN-P-mgsinθ=0\)
\(=\frac{2}{3}(60√3)-P-12(10)sin30^o=0\)
\(=40√3-P-60=0\)
\(=P=40√3-60\)
\(∴P=9.28N\)
(bii)
\(∑f_y=0==>N-mgcosθ=0\)
\(=N-12(10)cos 30^o=0\)
\(=N=120 cos 30^o\)
\(=N=60√3N\)
At the point of moving up,
\(∑f_x=0==>P-μN-mgsinθ=0\)
\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)
\(=P-40√3-60=0\)
\(=P=40√3+60\)
\(∴P=129.28N\)
Detalhes da Resposta
(a)
∑ clockwise moments = ∑ anti-clockwise moments
= 3 x mJ + 4 x 15 = 60 x 2
= 3mJ + 60 = 120
= 3mJ = 120 - 60
= 3mJ = 60
\(∴mJ=\frac{60}{3}=20kg\)
(bi)
\(∑f_y=0==>N-mgcosθ=0\)
\(=N-12(10)cos30^o=0\)
\(=N=120cos 30^o\)
\(=N=60√3N\)
At the point of moving down,
\(∑f_x=0==>μN-P-mgsinθ=0\)
\(=\frac{2}{3}(60√3)-P-12(10)sin30^o=0\)
\(=40√3-P-60=0\)
\(=P=40√3-60\)
\(∴P=9.28N\)
(bii)
\(∑f_y=0==>N-mgcosθ=0\)
\(=N-12(10)cos 30^o=0\)
\(=N=120 cos 30^o\)
\(=N=60√3N\)
At the point of moving up,
\(∑f_x=0==>P-μN-mgsinθ=0\)
\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)
\(=P-40√3-60=0\)
\(=P=40√3+60\)
\(∴P=129.28N\)
Pergunta 2 Relatório
(a) Find the derivative of \(4x-\frac{7}{x^2}\)with respect to \(x\), from first principle.
(b) Given that tan \(P =\frac{3}{x - 1}\) and tan \(Q\) =\frac{2}{x + 1}\), find tan \(( P - Q )\)
(a) \(y=4x-\frac{7}{x^2}\)
\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}\)
\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}-(4x-\frac{7}{x^2})\)
\(dy=4dx-\frac{7}{(x + dx)^2}+\frac{7}{x^2}\)
\(dy=4dx-(\frac{7}{(x + dx)^2}-\frac{7}{x^2})\)
\(dy=4dx-\frac{7x^2 - 7(x + dx)^2}{x^2(x + dx)^2)}\)
\(dy=4dx-\frac{(7(x^2 - (x + dx)^2)}{x^2(x + dx)^2)}\)
\(a^a-b^2=(a+b)(a-b)\)
\(dy=4dx-(\frac{7((x+x+dx)(x-x-dx))}{x^2(x+dx)^2})\)
\(dy=4dx-(\frac{7((2x+dx)(-dx))}{x^2(x+dx)^2})\)
\(dy=dx(4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2}))\)
\(dy=4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2})\)
As \(dx \to 0 \frac{dy}{dx}=4-(\frac{7((2x)(-1))}{x^2(x)^2})\)
\(\frac{dy}{dx}=4-(\frac{-14x}{x^4}\)
\(\therefore \frac{dy}{dx}=4+\frac{14}{x^3}\)
(b) tan \(P= \frac{3}{x - 1},tan Q={2}{x + 1}\)
\(tan (P-Q)=\frac{tan P - tan Q}{1 + tan P tan Q}\)
tan P-tan Q=\(\frac{3}{x - 1}-\frac{2}{x + 1}=\frac{3(x + 1) - 2(x - 1)}{(x + 1)(x - 1)}\)
\(=\frac{3x + 3 - 2x + 2}{(x + 1)(x - 1)}=\frac{x + 5}{(x + 1)(x - 1)}\)
1+tan P tan Q =\(1+(\frac{3}{x - 1})(\frac{2}{x + 1})=1+\frac{6}{(x + 1)(x - 1)}\)
=\(\frac{(x + 1)(x - 1) + 6}{(x + 1)(x - 1)}\)
\(\frac{tan P - tan Q}{1 + tan Ptan Q}=\frac{x + 5}{(x + 1)(x - 1)}÷\frac{(x + 1)(x - 1)+6}{(x + 1)(x - 1)}\)
\(=\frac{x + 5}{(x + 1)(x - 1)}\times\frac{(x + 1)(x - 1)}{(x + 1)(x - 1) + 6}\)
\(=\frac{x + 5}{(x + 1)(x - 1) + 6}=\frac{x + 5}{x^2 - 1 + 6}\)
∴tan(P-Q)\(=\frac{x + 5}{x^2 + 5}\)
Detalhes da Resposta
(a) \(y=4x-\frac{7}{x^2}\)
\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}\)
\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}-(4x-\frac{7}{x^2})\)
\(dy=4dx-\frac{7}{(x + dx)^2}+\frac{7}{x^2}\)
\(dy=4dx-(\frac{7}{(x + dx)^2}-\frac{7}{x^2})\)
\(dy=4dx-\frac{7x^2 - 7(x + dx)^2}{x^2(x + dx)^2)}\)
\(dy=4dx-\frac{(7(x^2 - (x + dx)^2)}{x^2(x + dx)^2)}\)
\(a^a-b^2=(a+b)(a-b)\)
\(dy=4dx-(\frac{7((x+x+dx)(x-x-dx))}{x^2(x+dx)^2})\)
\(dy=4dx-(\frac{7((2x+dx)(-dx))}{x^2(x+dx)^2})\)
\(dy=dx(4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2}))\)
\(dy=4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2})\)
As \(dx \to 0 \frac{dy}{dx}=4-(\frac{7((2x)(-1))}{x^2(x)^2})\)
\(\frac{dy}{dx}=4-(\frac{-14x}{x^4}\)
\(\therefore \frac{dy}{dx}=4+\frac{14}{x^3}\)
(b) tan \(P= \frac{3}{x - 1},tan Q={2}{x + 1}\)
\(tan (P-Q)=\frac{tan P - tan Q}{1 + tan P tan Q}\)
tan P-tan Q=\(\frac{3}{x - 1}-\frac{2}{x + 1}=\frac{3(x + 1) - 2(x - 1)}{(x + 1)(x - 1)}\)
\(=\frac{3x + 3 - 2x + 2}{(x + 1)(x - 1)}=\frac{x + 5}{(x + 1)(x - 1)}\)
1+tan P tan Q =\(1+(\frac{3}{x - 1})(\frac{2}{x + 1})=1+\frac{6}{(x + 1)(x - 1)}\)
=\(\frac{(x + 1)(x - 1) + 6}{(x + 1)(x - 1)}\)
\(\frac{tan P - tan Q}{1 + tan Ptan Q}=\frac{x + 5}{(x + 1)(x - 1)}÷\frac{(x + 1)(x - 1)+6}{(x + 1)(x - 1)}\)
\(=\frac{x + 5}{(x + 1)(x - 1)}\times\frac{(x + 1)(x - 1)}{(x + 1)(x - 1) + 6}\)
\(=\frac{x + 5}{(x + 1)(x - 1) + 6}=\frac{x + 5}{x^2 - 1 + 6}\)
∴tan(P-Q)\(=\frac{x + 5}{x^2 + 5}\)
Pergunta 3 Relatório
(ai) A quadratic polynomial, g (x) has (2x + 1) as a factor. If g (x) is divided by (x - 1) and (x - 2), the remainder are -6 and -5 respectively. Find;
g (x);
(aii) A quadratic polynomial, g (x) has (2x + 1) as a factor. If g (x) is divided by (x - 1) and (x - 2), the remainder are -6 and -5 respectively. Find;
the zeros of g (x).
(b) Find the third term when (\(\frac{x}{2}-1\))\(^8\)is expanded in descending powers of \(x\).
(ai) Let the quadratic equation be \(ax^2+bx+c\)
\(g(x)=ax^2+bx+c\)
Since \((2x+1)\) is a factor,\(x=-\frac{1}{2}\)is a root
\(∴g(-\frac{1}{2})=a(-\frac{1}{2})^2+b(-\frac{1}{2})+c=0\)
\(=a(\frac{1}{4})-b(\frac{1}{2})+c=0\)
Multiply through by 4
=a-2b+4c=0-----(i)
For(x-1)and(x-2),x=1 and x = 2 respectively
So,
\(g(1)=a(1)^2+b(1)+c=-6\)
=a+b+c=-6-----(ii)
and
\(g(2)=a(2)^2+b(2)+c=-5\)
= 4a+2b+c=-5
Adding equations (i) and (iii) gives;
=5a+5b=-5
Divide through by 5
=a+b=-1......(iv)
Adding equation (i) to 2 times equation (iv) gives:
3a+6c=-12.....(v)
Equation (v) minus three times equation (iv) gives:
=3c=-12-(-3)
=3c=-9
=c\(\frac{-9}{3}=-3\)
Substitute (-3) for c in equation (v)
=3a+6(-3)=-12
=3a-18=-12
=3a=-12+18
=3a=6
=a=\(\frac{6}{3}=2\)
Substitute 2 for a and (-3) for c in equation (ii)
=2+6-3=-6
=b-1=-6
=b=-6+1
=b=-5
\(\therefore g(x)=2x^2-5x-3\)
(aii) \(2\times2 - 5x - 3\)
\(= 2\times2 - 6x + x - 3\)
\(= 2x(x - 3) + 1(x - 3)\)
\(= (x - 3)(2x + 1)\)
zeros of \(g(x)\) are \(2x + 1 = 0\) and \(x - 3 = 0\)
∴ zeros of \(g(x)\) are -\frac{1}{2}\) and \(3\)
(b) \((\frac{x}{2}-1)^8\)
rth term is given as \(^nC_r-1 a^n-(r-1)b^r-1\)
\(a=\frac{x}{2},b=-1,n=8,r=3,r-1=2\)
3rd term = \(^8C_2(\frac{x}{2})^8-2(-1)^2\)
= \(28×(\frac{x}{2})^6\times1\)
∴3rd term = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)
Detalhes da Resposta
(ai) Let the quadratic equation be \(ax^2+bx+c\)
\(g(x)=ax^2+bx+c\)
Since \((2x+1)\) is a factor,\(x=-\frac{1}{2}\)is a root
\(∴g(-\frac{1}{2})=a(-\frac{1}{2})^2+b(-\frac{1}{2})+c=0\)
\(=a(\frac{1}{4})-b(\frac{1}{2})+c=0\)
Multiply through by 4
=a-2b+4c=0-----(i)
For(x-1)and(x-2),x=1 and x = 2 respectively
So,
\(g(1)=a(1)^2+b(1)+c=-6\)
=a+b+c=-6-----(ii)
and
\(g(2)=a(2)^2+b(2)+c=-5\)
= 4a+2b+c=-5
Adding equations (i) and (iii) gives;
=5a+5b=-5
Divide through by 5
=a+b=-1......(iv)
Adding equation (i) to 2 times equation (iv) gives:
3a+6c=-12.....(v)
Equation (v) minus three times equation (iv) gives:
=3c=-12-(-3)
=3c=-9
=c\(\frac{-9}{3}=-3\)
Substitute (-3) for c in equation (v)
=3a+6(-3)=-12
=3a-18=-12
=3a=-12+18
=3a=6
=a=\(\frac{6}{3}=2\)
Substitute 2 for a and (-3) for c in equation (ii)
=2+6-3=-6
=b-1=-6
=b=-6+1
=b=-5
\(\therefore g(x)=2x^2-5x-3\)
(aii) \(2\times2 - 5x - 3\)
\(= 2\times2 - 6x + x - 3\)
\(= 2x(x - 3) + 1(x - 3)\)
\(= (x - 3)(2x + 1)\)
zeros of \(g(x)\) are \(2x + 1 = 0\) and \(x - 3 = 0\)
∴ zeros of \(g(x)\) are -\frac{1}{2}\) and \(3\)
(b) \((\frac{x}{2}-1)^8\)
rth term is given as \(^nC_r-1 a^n-(r-1)b^r-1\)
\(a=\frac{x}{2},b=-1,n=8,r=3,r-1=2\)
3rd term = \(^8C_2(\frac{x}{2})^8-2(-1)^2\)
= \(28×(\frac{x}{2})^6\times1\)
∴3rd term = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)
Pergunta 4 Relatório
(a) The first term of an Arithmetic Progression is -8, the last term is 52 and the sum of terms is 286. Find the:
number of terms in the series;
(b) The first term of an Arithmetic Progression is -8, the last term is 52 and the sum of terms is 286. Find the:
common difference.
(a) a=8;l=52;n=?
\(S_n=\frac{n}{2}(a+l)=286\)
\(=\frac{n}{2}(-8+52)=286\)
\(=\frac{n}{2}(44)=286\)
=22n=286
\(∴n=\frac{286}{22}=13\)
(b) l = a + (n - 1)d = 52
= -8 + (13 - 1)d = 52
= -8 + 12d = 52
= 12d = 52 + 8
= 12d = 60
\(∴d=\frac{60}{12}=5\)
Detalhes da Resposta
(a) a=8;l=52;n=?
\(S_n=\frac{n}{2}(a+l)=286\)
\(=\frac{n}{2}(-8+52)=286\)
\(=\frac{n}{2}(44)=286\)
=22n=286
\(∴n=\frac{286}{22}=13\)
(b) l = a + (n - 1)d = 52
= -8 + (13 - 1)d = 52
= -8 + 12d = 52
= 12d = 52 + 8
= 12d = 60
\(∴d=\frac{60}{12}=5\)
Pergunta 5 Relatório
The volume of a cube is increasing at the rate of \(3\frac{1}{2} cm ^3 s^{ -1}\). Find the rate of change of the side of the base when its length is 6 cm .
\(\frac{dV}{dt}=3\frac{1}{2}cm^3s^{-1}=3.5cm^3s^{-1}\)
L = 6cm
\(\frac{dL}{dt}=?\)
\(V = L^3\)
\(\frac{dV}{dL}=3L^2\)
\(\frac{dL}{dt}=(\frac{dV}{dL})^{-1}\times \frac{dV}{dt}\)
\(\frac{dL}{dt}=(3L^2)^{-1}\times 3.5\)
At L= 6cm
\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)
\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)
\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)
\(\therefore \frac{dL}{dt}0.032cms^{-1}\)
Detalhes da Resposta
\(\frac{dV}{dt}=3\frac{1}{2}cm^3s^{-1}=3.5cm^3s^{-1}\)
L = 6cm
\(\frac{dL}{dt}=?\)
\(V = L^3\)
\(\frac{dV}{dL}=3L^2\)
\(\frac{dL}{dt}=(\frac{dV}{dL})^{-1}\times \frac{dV}{dt}\)
\(\frac{dL}{dt}=(3L^2)^{-1}\times 3.5\)
At L= 6cm
\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)
\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)
\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)
\(\therefore \frac{dL}{dt}0.032cms^{-1}\)
Pergunta 6 Relatório
There are 6 boys and 8 girls in a class. If five students are selected from the class, find the probability that more girls than boys are selected
Total number of students =6 boys + 8 girls = 14 students.
Total ways to choose 5 students out of 14:
Total ways = \(^{14}C_5=\frac{14!}{5!*(14 - 5)!}=2002.\)
The number of ways to have more girls than boys:
(Selecting 3 girls and 2 boys) or (Selecting 4 girls and 1 boy)
Selecting 3 girls and 2 boys:\(^8C_3\times ^6C_2\)
=(\(\frac{8!}{3!*(8 - 3)!})\times(\frac{6!}{2! * (6 - 2)!})=56\times 15=840.\)
Selecting 4 girls and 1 boy:\(^8C_4\times^6C_1\)
\(=(\frac{8!}{4!*(8 - 4)!})\times(\frac{6!}{1! * (6 - 1)!})=70\times 6=420.\)
Total favorable cases =840+420=1260.
Finally, the probability is given by:
Probability (more girls than boys)
= \(\frac{Total Favorable Cases}{Total Ways}=\frac{1260}{2002}=\frac{90}{143}≈0.629.\)
The probability that more girls than boys are selected is approximately 0.629, or about 62.9%.
Detalhes da Resposta
Total number of students =6 boys + 8 girls = 14 students.
Total ways to choose 5 students out of 14:
Total ways = \(^{14}C_5=\frac{14!}{5!*(14 - 5)!}=2002.\)
The number of ways to have more girls than boys:
(Selecting 3 girls and 2 boys) or (Selecting 4 girls and 1 boy)
Selecting 3 girls and 2 boys:\(^8C_3\times ^6C_2\)
=(\(\frac{8!}{3!*(8 - 3)!})\times(\frac{6!}{2! * (6 - 2)!})=56\times 15=840.\)
Selecting 4 girls and 1 boy:\(^8C_4\times^6C_1\)
\(=(\frac{8!}{4!*(8 - 4)!})\times(\frac{6!}{1! * (6 - 1)!})=70\times 6=420.\)
Total favorable cases =840+420=1260.
Finally, the probability is given by:
Probability (more girls than boys)
= \(\frac{Total Favorable Cases}{Total Ways}=\frac{1260}{2002}=\frac{90}{143}≈0.629.\)
The probability that more girls than boys are selected is approximately 0.629, or about 62.9%.
Pergunta 7 Relatório
(a) A particle of mass 2 kg moves under the action of a constant force, F N , with an initial velocity \((3 i + 2 j ) ms^{ -1}\) and a velocity of \((15 i - 4 j ) ms^{ -1}\) after 4 seconds . Find the:
acceleration of the particle;
(b) A particle of mass 2 kg moves under the action of a constant force, F N , with an initial velocity \((3 i + 2 j ) ms^{ -1}\) and a velocity of \((15 i - 4 j ) ms^{ -1}\) after 4 seconds . Find the:
magnitude of the force F ;
(c) A particle of mass 2 kg moves under the action of a constant force, F N , with an initial velocity \((3 i + 2 j ) ms^{ -1}\) and a velocity of \((15 i - 4 j ) ms^{ -1}\) after 4 seconds . Find the:
magnitude of the velocity of the particle after 8 seconds , correct to three decimal places.
(a) \(m=2kg;u=(3i+2j)ms^{-1};v=(15i-4j)ms^{-1};t=4s;a=?\)
\(a=\frac{v - u}{t}=\frac{(15i - 4j) - (3i + 2j)}{4}\)
\(a=\frac{15i - 4j - 3i - 2j}{4}=\frac{12i - 6j}{4}\)
\(∴a=3i-\frac{3}{2}j ms^{-2}\)
(b) \(m=2kg;a=3i-\frac{3}{2} j\)
\(F=ma=2(3i-\frac{3}{2}j)\)
F = 6i - 3j
\(|F| = √(6^2 + (-3)^2)\)
\(|F| = √(36 + 9) = √45\)
\(∴ |F| = 3√5 N = 6.71 N\)
(c) \(u=3i+2jms^{-1};a=3i-\frac{3}{2}j ms^{-2};t=8s\)
v = u + at
\(v=(3i+2j)+8(3i-\frac{3}{2} j)\)
\(v = 3i + 2j + 24i - 12j\)
\(v = 27i - 10j\)
\(|v| = √(27^2 + (-10)^2)\)
\(|v| = √(729 + 100) = √829\)
\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)
Detalhes da Resposta
(a) \(m=2kg;u=(3i+2j)ms^{-1};v=(15i-4j)ms^{-1};t=4s;a=?\)
\(a=\frac{v - u}{t}=\frac{(15i - 4j) - (3i + 2j)}{4}\)
\(a=\frac{15i - 4j - 3i - 2j}{4}=\frac{12i - 6j}{4}\)
\(∴a=3i-\frac{3}{2}j ms^{-2}\)
(b) \(m=2kg;a=3i-\frac{3}{2} j\)
\(F=ma=2(3i-\frac{3}{2}j)\)
F = 6i - 3j
\(|F| = √(6^2 + (-3)^2)\)
\(|F| = √(36 + 9) = √45\)
\(∴ |F| = 3√5 N = 6.71 N\)
(c) \(u=3i+2jms^{-1};a=3i-\frac{3}{2}j ms^{-2};t=8s\)
v = u + at
\(v=(3i+2j)+8(3i-\frac{3}{2} j)\)
\(v = 3i + 2j + 24i - 12j\)
\(v = 27i - 10j\)
\(|v| = √(27^2 + (-10)^2)\)
\(|v| = √(729 + 100) = √829\)
\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)
Pergunta 8 Relatório
(a) The inverse of a function \(f\) is given by \(f^{-1}(x)=\frac{5x - 6}{4 - x},x ≠ 4\).Find the:
function, \(f (x)\)
(b) The inverse of a function \(f\) is given by \(f^{-1}(x)=\frac{5x - 6}{4 - x},x ≠ 4\).Find the:
value of x for which \(f (x) = 5\)
(a) \((f^{-1})^{-1}(x)=f(x)\)
\(f^{-1}(x)=\frac{5x - 6}{4 - x},x ≠ 4\)
Let \(y=\frac{5x - 6}{4 - x}\)
\(=y(4-x)=5x-6\)
\(=4y-xy=5x-6\)
\(=-xy-5x=-6-4y\)
\(=x(-y-5)=-6-4y\)
\(=x=\frac{-6 - 4y}{-y - 5}=\frac{-(6 + 4y)}{-(y + 5)}\)
\(=x=\frac{6 + 4y}{y + 5}\)
\(∴f(x)=\frac{6 + 4x}{x + 5},x≠-5\)
(b) \(f(x)=5\)
\(=\frac{6 + 4x}{x + 5}=5\)
\(=6+4x=5(x+5)\)
\(=6+4x=5x+25\)
\(=4x-5x=25-6\)
\(=-x=19\)
\(∴x=-19\)
Detalhes da Resposta
(a) \((f^{-1})^{-1}(x)=f(x)\)
\(f^{-1}(x)=\frac{5x - 6}{4 - x},x ≠ 4\)
Let \(y=\frac{5x - 6}{4 - x}\)
\(=y(4-x)=5x-6\)
\(=4y-xy=5x-6\)
\(=-xy-5x=-6-4y\)
\(=x(-y-5)=-6-4y\)
\(=x=\frac{-6 - 4y}{-y - 5}=\frac{-(6 + 4y)}{-(y + 5)}\)
\(=x=\frac{6 + 4y}{y + 5}\)
\(∴f(x)=\frac{6 + 4x}{x + 5},x≠-5\)
(b) \(f(x)=5\)
\(=\frac{6 + 4x}{x + 5}=5\)
\(=6+4x=5(x+5)\)
\(=6+4x=5x+25\)
\(=4x-5x=25-6\)
\(=-x=19\)
\(∴x=-19\)
Pergunta 9 Relatório
(a)The table shows the distribution of heights ( cm ) of 60 seedlings in a vegetable garden.
| Heights(cm) | 0.1 - 0.3 | 0.4 - 0.6 | 0.7 - 0.9 | 1.0 - 1.4 | 1.5 - 1.9 | 2.0 - 22 | 2.3 - 2.5 |
| Frequency | 6 | 9 | 12 | 15 | 3 | 6 | 9 |
Draw a histogram for the distribution.
(b) The table shows the distribution of heights ( cm ) of 60 seedlings in a vegetable garden.
| Heights(cm) | 0.1 - 0.3 | 0.4 - 0.6 | 0.7 - 0.9 | 1.0 - 1.4 | 1.5 - 1.9 | 2.0 - 2.2 | 2.3 - 2.5 |
| Frequency | 6 | 9 | 12 | 15 | 3 | 6 | 9 |
Use the histogram to estimate the modal height of the seedlings.
(a)
| Class intervals | Class boundaries | Frequency |
| 0.1 - 0.3 | 0.05 - 0.35 | 6 |
| 0.4 - 0.6 | 0.35 - 0.65 | 9 |
| 0.7 - 0.9 | 0.65 - 0.95 | 12 |
| 1.0 - 1.4 | 0.95 - 1.45 | 15 |
| 1.5 - 1.9 | 1.45 - 1.95 | 3 |
| 2.0 - 2.2 | 1.95 - 2.25 | 6 |
| 2.3 - 2.5 | 2.25 - 2.55 | 9 |
(b) The estimate of the modal height of the seedlings is 1.08
Detalhes da Resposta
(a)
| Class intervals | Class boundaries | Frequency |
| 0.1 - 0.3 | 0.05 - 0.35 | 6 |
| 0.4 - 0.6 | 0.35 - 0.65 | 9 |
| 0.7 - 0.9 | 0.65 - 0.95 | 12 |
| 1.0 - 1.4 | 0.95 - 1.45 | 15 |
| 1.5 - 1.9 | 1.45 - 1.95 | 3 |
| 2.0 - 2.2 | 1.95 - 2.25 | 6 |
| 2.3 - 2.5 | 2.25 - 2.55 | 9 |
(b) The estimate of the modal height of the seedlings is 1.08
Pergunta 10 Relatório
(ai) A bag contains 16 identical balls of which 4 are green. A boy picks a ball at random from the bag and replaces it. If this is repeated 5 times, what is the probability that he:
did not pick a green ball;
(aii) A bag contains 16 identical balls of which 4 are green. A boy picks a ball at random from the bag and replaces it. If this is repeated 5 times, what is the probability that he:
picked a green ball at least three times?
(b) The deviations from a mean of values from a set of data are \(-2, ( m - 1), ( m ^2 + 1), -1, 2, (2 m - 1)\) and \(-2\). Find the possible values of \(m\) .
(ai) \(p=\frac{4}{16}=\frac{1}{4}\)
\(∴q=1-\frac{1}{4}=\frac{3}{4}\)
P(The probability that he did not pick a green ball) = \(^nC_r p^rq^{n - r}\)
Where n = 5 and r = 0
\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^{5 - 0}\)
\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^5\)
\(=1\times1\times\frac{243}{1024}\)
\(=\frac{243}{1024}\)
(aii) Pr(at least three) = Pr(for 3 green balls) + Pr (for 4 green balls) + Pr (for 5 green balls)
\(^5C_3(\frac{1}{4})^3(\frac{3}{4})^{5 - 3}+^5C_4(\frac{1}{4})^4(\frac{3}{4})^{5 - 4}+^5C_5(\frac{1}{4})^5(\frac{3}{4})^{5 - 5}\)
\(=^5C_3(\frac{1}{4})^3(\frac{3}{4})^2+^5C_4(\frac{1}{4})^4(\frac{3}{4})^1+^5C_5(\frac{1}{4})^5(\frac{3}{4})^0\)
\(=10\times\frac{1}{6}4\times\frac{9}{16}+5\times\frac{1}{256}\times\frac{3}{4}+1\times\frac{1}{1024}\times1\)
\(=\frac{45}{512}+\frac{15}{1024}+\frac{1}{1024}\)
\(=\frac{53}{512}\)
(b) The sum of deviations from the mean is always equal to 0. This is a fundamental property of deviations and the definition of the mean.
\(= -2 + (m - 1) + (m^2 + 1) + (-1) + 2 + (2m - 1) + (-2) = 0\)
\(= -2 + m - 1 + m^2 + 1 - 1 + 2 + 2m - 1 - 2 = 0\)
\(= m^2 + 3m - 4 = 0\)
\(= m^2 + 4m - m - 4 = 0\)
= m(m + 4) - 1(m + 4) = 0
= (m + 4)(m - 1) = 0
This gives us two possible values for m: -4 and 1
So, the possible values of m are -4 and 1.
Detalhes da Resposta
(ai) \(p=\frac{4}{16}=\frac{1}{4}\)
\(∴q=1-\frac{1}{4}=\frac{3}{4}\)
P(The probability that he did not pick a green ball) = \(^nC_r p^rq^{n - r}\)
Where n = 5 and r = 0
\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^{5 - 0}\)
\(=^5C_0(\frac{1}{4})^o(\frac{3}{4})^5\)
\(=1\times1\times\frac{243}{1024}\)
\(=\frac{243}{1024}\)
(aii) Pr(at least three) = Pr(for 3 green balls) + Pr (for 4 green balls) + Pr (for 5 green balls)
\(^5C_3(\frac{1}{4})^3(\frac{3}{4})^{5 - 3}+^5C_4(\frac{1}{4})^4(\frac{3}{4})^{5 - 4}+^5C_5(\frac{1}{4})^5(\frac{3}{4})^{5 - 5}\)
\(=^5C_3(\frac{1}{4})^3(\frac{3}{4})^2+^5C_4(\frac{1}{4})^4(\frac{3}{4})^1+^5C_5(\frac{1}{4})^5(\frac{3}{4})^0\)
\(=10\times\frac{1}{6}4\times\frac{9}{16}+5\times\frac{1}{256}\times\frac{3}{4}+1\times\frac{1}{1024}\times1\)
\(=\frac{45}{512}+\frac{15}{1024}+\frac{1}{1024}\)
\(=\frac{53}{512}\)
(b) The sum of deviations from the mean is always equal to 0. This is a fundamental property of deviations and the definition of the mean.
\(= -2 + (m - 1) + (m^2 + 1) + (-1) + 2 + (2m - 1) + (-2) = 0\)
\(= -2 + m - 1 + m^2 + 1 - 1 + 2 + 2m - 1 - 2 = 0\)
\(= m^2 + 3m - 4 = 0\)
\(= m^2 + 4m - m - 4 = 0\)
= m(m + 4) - 1(m + 4) = 0
= (m + 4)(m - 1) = 0
This gives us two possible values for m: -4 and 1
So, the possible values of m are -4 and 1.
Pergunta 11 Relatório
(a) A bus travels with a velocity of \(6 ms ^{-1}\). It then accelerates uniformly and travels a distance of 70 m. If the final velocity is \(20 ms ^{-1}\), find, correct to one decimal place, the:
acceleration;
(b) A bus travels with a velocity of \(6 ms ^{-1}\). It then accelerates uniformly and travels a distance of 70 m. If the final velocity is \(20 ms ^{-1}\), find, correct to one decimal place, the:
time to travel this distance.
(a) \(u=6ms^{-1};s=70m;v=20ms^{-1}a=?\)
\(v^2=u^2+2as\)
⇒\(a=\frac{v^2 - u^2}{2s}=\frac{20^2 - 6^2}{2(70)}\)
\(a=\frac{400 - 36}{140}=\frac{364}{140}\)
\(∴a=2.6ms^{-2}\)(to 1 d.p)
(b) \(u=6ms^{-1};s=70m;v=20ms^{-1}t=?\)
v=u+at
⇒\(t=\frac{v - u}{a}=\frac{20 - 6}{2.6}\)
\(∴t=\frac{14}{2.6}=5.4s\)
Detalhes da Resposta
(a) \(u=6ms^{-1};s=70m;v=20ms^{-1}a=?\)
\(v^2=u^2+2as\)
⇒\(a=\frac{v^2 - u^2}{2s}=\frac{20^2 - 6^2}{2(70)}\)
\(a=\frac{400 - 36}{140}=\frac{364}{140}\)
\(∴a=2.6ms^{-2}\)(to 1 d.p)
(b) \(u=6ms^{-1};s=70m;v=20ms^{-1}t=?\)
v=u+at
⇒\(t=\frac{v - u}{a}=\frac{20 - 6}{2.6}\)
\(∴t=\frac{14}{2.6}=5.4s\)
Pergunta 12 Relatório
If \(^9C_x = 4[^7C_{x - 1}]\), find the values of \(x\)
\(^9C_x = 4[^7C_{x - 1}]\)
\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)
\(\frac{9!}{x!(9 - x)!} = 4*\frac{7!}{(x - 1)!(8 - x)!}\)
\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)
Cancel out the common terms
\(=\frac{9 \times 8}{x(9 - x)}=\frac{4}{1}\)
\(=4x(9-x) = 9*8\)
\(=36x- 4x^2=72\)
divide thru by 4
\(=x^2 - 9x+18 = 0\)
\(=x^2-6x-3x+18=0\)
\(=x(x-6)-3(x-6)=0\)
\(=(x-6)(x-3)=0\)
\(\therefore x\) = 6 or 3
Detalhes da Resposta
\(^9C_x = 4[^7C_{x - 1}]\)
\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)
\(\frac{9!}{x!(9 - x)!} = 4*\frac{7!}{(x - 1)!(8 - x)!}\)
\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)
Cancel out the common terms
\(=\frac{9 \times 8}{x(9 - x)}=\frac{4}{1}\)
\(=4x(9-x) = 9*8\)
\(=36x- 4x^2=72\)
divide thru by 4
\(=x^2 - 9x+18 = 0\)
\(=x^2-6x-3x+18=0\)
\(=x(x-6)-3(x-6)=0\)
\(=(x-6)(x-3)=0\)
\(\therefore x\) = 6 or 3
