WAEC SSCE - Further Mathematics - 2023 (Essay)

Answer Details

\(\frac{dV}{dt}=3\frac{1}{2}cm^3s^{-1}=3.5cm^3s^{-1}\)

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

\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)

\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)

\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)

\(\therefore \frac{dL}{dt}0.032cms^{-1}\)

Answer Details

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%.

 

Answer Details

(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 = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)

Answer Details

(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.

Answer Details

(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)\(=\frac{x + 5}{x^2 + 5}\)

Answer Details

\(^9C_x = 4[^7C_{x - 1}]\)

\(\frac{9!}{x!(9 - x)!}=4*\frac{7!}{(x - 1)!(7 - (x - 1)!)}\)

\(\frac{9!}{x!(9 - x)!} = 4*\frac{7!}{(x - 1)!(8 - x)!}\)

\(=\frac{9 \times 8 \times 7!}{x(x - 1)!(9 - x)(8 - x)!}=4*\frac{7!}{(x - 1)!(8 - x)!}\)

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

Answer Details

\(\overline{MN}\) = 8i + 3j

\(\overline{MO}\) = 14i - 5j

Consider ∆MON

\(\overline{MN}\) + \(\overline{NO}\) = \(\overline{MO}\)

\(\overline{NO}\) = \(\overline{MO}\) - \(\overline{MN}\)

\(\overline{NO}\) = 14i - 5j - (8i + 3j) = 6i - 8j

Since P is the midpoint of \(\overline{NO}\), then

\(\overline{NP} =\frac{1}{2}( \overline{NO} )\)

\(=\frac{1}{2}(6i - 8j) = 3i - 4j\)

Consider ∆MNP

\(\overline{MN}\) + \(\overline{NP}\) = \(\overline{MP}\)

(8i + 3j) + (3i - 4j) = \(\overline{MP}\)

\(\overline{MP}\) = 8i + 3i + 3j - 4j

∴ \(\overline{MP}\) = 11i - j

Answer Details

(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

 

Answer Details

(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

⇒\(t=\frac{v - u}{a}=\frac{20 - 6}{2.6}\)

\(∴t=\frac{14}{2.6}=5.4s\)

Answer Details

(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

=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)

= \((x + 2)^2 + (y - 5)^2 = 9\)

= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)

= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)

∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)

Answer Details

(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

\(v=(3i+2j)+8(3i-\frac{3}{2} j)\)

\(v = 3i + 2j + 24i - 12j\)

\(v = 27i - 10j\)

\(|v| = √(27^2 + (-10)^2)\)

\(|v| = √(729 + 100) = √829\)

\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)

Answer Details

(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

\(∴d=\frac{60}{12}=5\)