WAEC SSCE - Further Mathematics - 2023 (Essay)

Antwortdetails

\(\overline{MN}\) = 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

Antwortdetails

(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

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

\(∴t=\frac{14}{2.6}=5.4s\)

Antwortdetails

(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

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

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

\(v = 27i - 10j\)

\(|v| = √(27^2 + (-10)^2)\)

\(|v| = √(729 + 100) = √829\)

\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)

Antwortdetails

(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,

\(∑f_x=0==>P-μN-mgsinθ=0\)

\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)

\(=P-40√3-60=0\)

\(=P=40√3+60\)

\(∴P=129.28N\)

Antwortdetails

(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

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

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

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

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

∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)

Antwortdetails

(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) \(f(x)=5\)

\(=\frac{6 + 4x}{x + 5}=5\)

\(=6+4x=5(x+5)\)

\(=6+4x=5x+25\)

\(=4x-5x=25-6\)

\(=-x=19\)

\(∴x=-19\)

Antwortdetails

(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

\(∴d=\frac{60}{12}=5\)

Antwortdetails

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

Antwortdetails

(a) 

(b) The estimate of the modal height of the seedlings is 1.08

Antwortdetails

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

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

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

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

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

Antwortdetails

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%.

Antwortdetails

(a) 

(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