WAEC SSCE - Further Mathematics - 2023 (Essay)
Ajụjụ 1 Ripọtì
(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
⇒\(t=\frac{v - u}{a}=\frac{20 - 6}{2.6}\)
\(∴t=\frac{14}{2.6}=5.4s\)
Akọwa Nkọwa
(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
⇒\(t=\frac{v - u}{a}=\frac{20 - 6}{2.6}\)
\(∴t=\frac{14}{2.6}=5.4s\)
Ajụjụ 2 Ripọtì
(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
Akọwa Nkọwa
(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
Ajụjụ 3 Ripọtì
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
Akọwa Nkọwa
\(^9C_x = 4[^7C_{x - 1}]\)
\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)
\(\frac{9!}{x!(9 - x)!} = 4*\frac{7!}{(x - 1)!(8 - x)!}\)
\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)
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
Ajụjụ 4 Ripọtì
(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 = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)
Akọwa Nkọwa
(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 = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)
Ajụjụ 5 Ripọtì
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%.
Akọwa Nkọwa
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%.
Ajụjụ 6 Ripọtì
(a) Express \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}\) in partial fractions.
(b) The coordinates of the centre and circumference of a circle are (-2, 5) and 6π units respectively. Find the equation of the circle.
(a) \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A}{(x - 1)}+\frac{B}{2x + 3}+\frac{C}{(2x + 3)^2}\)
\(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A(2x + 3)2+B(x - 1)(2x + 3)+C(x - 1)}{(x - 1)(2x + 3)^2}\)
\(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)
Put \(x=1\)
\(8(1)^2+8(1)+9=A(2(1)+3)^2+B(1-1)(2(1)+3)+C(1-1)\)
⇒25=25A
=A=\(\frac{25}{25}=1\)
Put \(x=-\frac{3}{2}\)
\(8(-\frac{3}{2})^2+8(-\frac{3}{2})+9=A(2(-\frac{3}{2})+3)^2+B(-\frac{3}{2}-1)(2(-\frac{3}{2})+3)+C(-\frac{3}{2}-1)\)
⇒15=-2.5C
\(=C=-\frac{15}{2.5}=-6\)
Since \(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)
\(⇒8x^2+8x+9=A(4x^2+12x+9)+B(2x^2+x-3)+C(x-1)\)
\(=8x^2+8x+9=4Ax^2+12A+9A+2Bx^2+Bx-3B+Cx-C\)
\(=8x^2+8x+9=4Ax^2+2Bx^2+12Ax+Bx+Cx+9A-3B-C\)
\(=8x^2+8x+9=(4A+2B)x^2+(12A+B+C)x+9A-3B-C\)
By comparing the coefficient of \(x, 8=12A+B+c\)
=8=12(1)+B-6
=8=12+B-6
=8=6+B
=8-6=B
=\(\therefore \frac{8x^2+8x+9}{(x-1)(2x+3)^2}=\frac{1}{x-1}+\frac{2}{2x+3}-\frac{6}{(2x+3)^2}\)
(b) Equation of a circle =\((x - a)^2 + (y - b)^2 = r^2\)
Where "a" and "b" are the coordinate of the center and "r" is the radius
2πr = 6π (given)
∴ r = 3 units
=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)
= \((x + 2)^2 + (y - 5)^2 = 9\)
= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)
= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)
∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)
Akọwa Nkọwa
(a) \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A}{(x - 1)}+\frac{B}{2x + 3}+\frac{C}{(2x + 3)^2}\)
\(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A(2x + 3)2+B(x - 1)(2x + 3)+C(x - 1)}{(x - 1)(2x + 3)^2}\)
\(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)
Put \(x=1\)
\(8(1)^2+8(1)+9=A(2(1)+3)^2+B(1-1)(2(1)+3)+C(1-1)\)
⇒25=25A
=A=\(\frac{25}{25}=1\)
Put \(x=-\frac{3}{2}\)
\(8(-\frac{3}{2})^2+8(-\frac{3}{2})+9=A(2(-\frac{3}{2})+3)^2+B(-\frac{3}{2}-1)(2(-\frac{3}{2})+3)+C(-\frac{3}{2}-1)\)
⇒15=-2.5C
\(=C=-\frac{15}{2.5}=-6\)
Since \(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)
\(⇒8x^2+8x+9=A(4x^2+12x+9)+B(2x^2+x-3)+C(x-1)\)
\(=8x^2+8x+9=4Ax^2+12A+9A+2Bx^2+Bx-3B+Cx-C\)
\(=8x^2+8x+9=4Ax^2+2Bx^2+12Ax+Bx+Cx+9A-3B-C\)
\(=8x^2+8x+9=(4A+2B)x^2+(12A+B+C)x+9A-3B-C\)
By comparing the coefficient of \(x, 8=12A+B+c\)
=8=12(1)+B-6
=8=12+B-6
=8=6+B
=8-6=B
=\(\therefore \frac{8x^2+8x+9}{(x-1)(2x+3)^2}=\frac{1}{x-1}+\frac{2}{2x+3}-\frac{6}{(2x+3)^2}\)
(b) Equation of a circle =\((x - a)^2 + (y - b)^2 = r^2\)
Where "a" and "b" are the coordinate of the center and "r" is the radius
2πr = 6π (given)
∴ r = 3 units
=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)
= \((x + 2)^2 + (y - 5)^2 = 9\)
= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)
= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)
∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)
Ajụjụ 7 Ripọtì
(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) \(f(x)=5\)
\(=\frac{6 + 4x}{x + 5}=5\)
\(=6+4x=5(x+5)\)
\(=6+4x=5x+25\)
\(=4x-5x=25-6\)
\(=-x=19\)
\(∴x=-19\)
Akọwa Nkọwa
(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) \(f(x)=5\)
\(=\frac{6 + 4x}{x + 5}=5\)
\(=6+4x=5(x+5)\)
\(=6+4x=5x+25\)
\(=4x-5x=25-6\)
\(=-x=19\)
\(∴x=-19\)
Ajụjụ 8 Ripọtì
(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)\(=\frac{x + 5}{x^2 + 5}\)
Akọwa Nkọwa
(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)\(=\frac{x + 5}{x^2 + 5}\)
Ajụjụ 9 Ripọtì
(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,
\(∑f_x=0==>P-μN-mgsinθ=0\)
\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)
\(=P-40√3-60=0\)
\(=P=40√3+60\)
\(∴P=129.28N\)
Akọwa Nkọwa
(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,
\(∑f_x=0==>P-μN-mgsinθ=0\)
\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)
\(=P-40√3-60=0\)
\(=P=40√3+60\)
\(∴P=129.28N\)
Ajụjụ 10 Ripọtì
(a) The table shows the distribution of marks scored by some candidates in an examination.
| Marks | 11 - 20 | 21 - 30 | 31 - 40 | 41 - 50 | 51 - 60 | 61 - 70 | 71 - 80 | 81 - 90 | 91 - 100 |
| Num of candidates | 5 | 39 | 14 | 40 | 57 | 25 | 11 | 8 | 1 |
Construct a cumulative frequency table for the distribution.
(b) The table shows the distribution of marks scored by some candidates in an examination.
| Marks | 11 - 20 | 21 - 30 | 31 - 40 | 41 - 50 | 51 - 60 | 61 - 70 | 71 - 80 | 81 - 90 | 91 - 100 |
| Num of candidates | 5 | 39 | 14 | 40 | 57 | 25 | 11 | 8 | 1 |
Draw a cumulative frequency curve for the distribution.
(ci) Use the curve to estimate the:
number of candidates who scored marks between 24 and 58 ;
(cii) Use the curve to estimate the:
lowest mark for distinction, if 12% of the candidates passed with distinction.
(a)
| Marks | Class Boundaries | Frequency | Cumulative Frequency |
| 11 - 20 | 10.5 - 20.5 | 5 | 5 |
| 21 - 30 | 20.5 - 30.5 | 39 | 5+39=44 |
| 31 - 40 | 30.5 - 40.5 | 14 | 44+14=58 |
| 41 - 50 | 40.5 - 50.5 | 40 | 58+40=98 |
| 51 - 60 | 50.5 - 60.5 | 57 | 98+57=155 |
| 61 - 70 | 60.5 - 70.5 | 25 | 155+25=180 |
| 71 - 80 | 70.5 - 80.5 | 11 | 180+11=191 |
| 81 - 90 | 80.5 - 90.5 | 8 | 191+8=199 |
| 91 - 100 | 90.5 - 100.5 | 1 | 199+1=200 |
(b)
(ci) The number of candidates who scored marks between 24 and 58 = 146 - 14 = 132
(cii) if 12% of the candidates passed with distinction then 88% did not pass with distinction
⇒\(88% of 200 = \frac{88}{100}\times 200=176\)
From the cumulative frequency curve, 176 corresponds to 69 marks
∴ The lowest mark for distinction = 69 marks
Akọwa Nkọwa
(a)
| Marks | Class Boundaries | Frequency | Cumulative Frequency |
| 11 - 20 | 10.5 - 20.5 | 5 | 5 |
| 21 - 30 | 20.5 - 30.5 | 39 | 5+39=44 |
| 31 - 40 | 30.5 - 40.5 | 14 | 44+14=58 |
| 41 - 50 | 40.5 - 50.5 | 40 | 58+40=98 |
| 51 - 60 | 50.5 - 60.5 | 57 | 98+57=155 |
| 61 - 70 | 60.5 - 70.5 | 25 | 155+25=180 |
| 71 - 80 | 70.5 - 80.5 | 11 | 180+11=191 |
| 81 - 90 | 80.5 - 90.5 | 8 | 191+8=199 |
| 91 - 100 | 90.5 - 100.5 | 1 | 199+1=200 |
(b)
(ci) The number of candidates who scored marks between 24 and 58 = 146 - 14 = 132
(cii) if 12% of the candidates passed with distinction then 88% did not pass with distinction
⇒\(88% of 200 = \frac{88}{100}\times 200=176\)
From the cumulative frequency curve, 176 corresponds to 69 marks
∴ The lowest mark for distinction = 69 marks
Ajụjụ 11 Ripọtì
(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
\(v=(3i+2j)+8(3i-\frac{3}{2} j)\)
\(v = 3i + 2j + 24i - 12j\)
\(v = 27i - 10j\)
\(|v| = √(27^2 + (-10)^2)\)
\(|v| = √(729 + 100) = √829\)
\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)
Akọwa Nkọwa
(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
\(v=(3i+2j)+8(3i-\frac{3}{2} j)\)
\(v = 3i + 2j + 24i - 12j\)
\(v = 27i - 10j\)
\(|v| = √(27^2 + (-10)^2)\)
\(|v| = √(729 + 100) = √829\)
\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)
Ajụjụ 12 Ripọtì
(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.
Akọwa Nkọwa
(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.
