WAEC SSCE - Further Mathematics - 2022 (Essay)

a. a = (10t - 4t2 ${}^{\mathrm{<br>}}$)m/s2 ${}^{\mathrm{<br>}}$, t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t2 ${}^2$)t
v = (10t2 ${}^2$ - 4t3 ${}^3$)m/s2 ${}^2$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3)2 ${}^2$
Δa4 = 10(4-1) - 4(4-3)2 ${}^2$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s2 ${}^2$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s2 ${}^{\mathrm{<br>}}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

x3+8x+2 $\frac{{\mathrm{<br>}}^{}}{}$

x3 ${}^{\mathrm{<br>}}$ + 8 = x3 ${}^{\mathrm{<br>}}$ + 23
recall that
x3 ${}^3$ + y3 ${}^3$ = (x+y)(x2 ${}^2$ - xy + y2 ${}^2$)
x = x, y = 2
x3 ${}^3$ + 8 = (x+2)(x3 ${}^3$ - 2(x) + 22)

x3+8x+2=(x+2)(x3−2(x)+22)x+2 $\frac{{\mathrm{<br>}}^{}}{}$ - 2x + 4

x = -2
(-2)2 ${}^{\mathrm{<br>}}$ - 2(2) + 4 = 4 - 4 + 4

= 4

a. To plot a scatter diagram, we need to plot the given X and Y values as individual points on a graph, where X values are taken as horizontal axis and Y values are taken as vertical axis. The scatter diagram shows the relationship between two variables.

b i. To calculate x?, the mean of X, we add up all the X values and divide by the number of values, i.e., x? = (14+16+17+18+22+24+27+28+31+33) / 10 = 23.

Similarly, to calculate ?, the mean of Y, we add up all the Y values and divide by the number of values, i.e., ? = (22+19+15+13+10+12+3+5+3+2) / 10 = 11.

b ii. To calculate x?1, we need to find the X values that are below x? and then calculate their mean. From the X values given, 14, 16, 17, 18, and 22 are below x? = 23. So, x?1 = (14+16+17+18+22) / 5 = 17.4.

Similarly, we need to find the corresponding Y values that are below x? and then calculate their mean. From the Y values given, corresponding Y values for the X values 14, 16, 17, 18, and 22 are 22, 19, 15, 13, and 10 respectively. So, ?1 = (22+19+15+13+10) / 5 = 15.8.

c. To draw the line of best fit, we need to plot the points (x?, ?) and (x?1, ?1) on the scatter diagram and then draw a straight line passing through these two points. The line should be such that it has an equal number of points above and below it.

d i. From the scatter diagram, we can see that there is a negative relationship between X and Y. As X increases, Y decreases.

d ii. To determine the value of Y when X is 20, we can draw a vertical line from the point X = 20 on the X-axis to the line of best fit. From the point where the line intersects the Y-axis, we can read off the value of Y, which is approximately 12.

f:x → x2 ${}^2$ + 2 and g:x → 1x+2.

(g∘f)−1 ${}^{-1}$

(g∘f)x2+2 ${}^{{\mathrm{<br>}}^{}}$

gx2+2 ${}^{{\mathrm{<br>}}^{}}$ = 1(x2+2)+2

(g∘f) = 1(x2+4 $\frac{\mathrm{<br>}}{}$

let y = 1(x2+4 $\frac{\mathrm{<br>}}{}$

y((x2+4) $<br>$ = 1

yx2+4y=1 $\mathrm{<br>}$

x2 ${}^2$ = 1−4yy $\frac{\mathrm{<br>}}{}$

x = 1−4yy−−−−√

(g∘f)−1 ${}^{<br>}$ = 1−4xx−−−−√

p = 1−6Σ(di)2n(n2−1 $1-\frac{6\Sigma ({�}_{�}{)}^2}{�({�}^2-1}$

p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation

P = 1−6(4+1+1+0+16+0+0+0)8(82−1) $\mathrm{<br>}$

p = 1−6(22)8(63) $1<br>$

p =  1−132504 $1\frac{4}{}$

p = 1−1142 $1<br>$

p = 1142 or 0.7381

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = 1800.8192 $\frac{\mathrm{<br>}}{}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

2(2y+1)−5(2y)+2 ${\mathrm{<br>}}^{}$ = 0

Let p = 2y ${}^{\mathrm{<br>}}$
22y(21)−5(2y) ${\mathrm{<br>}}^{}$ + 2 = 0
2p2 ${}^2$ - 5p + 2 = 0
2p2 ${}^2$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or 12 $\frac{1}{2}$

p = 2y ${}^{\mathrm{<br>}}$
when p = 2
2y ${}^{\mathrm{<br>}}$ = 2
y = 1

when p = 12 $\frac{\mathrm{<br>}}{}$

2y ${}^{\mathrm{<br>}}$ = 12 $\frac{\mathrm{<br>}}{}$

2y ${}^{\mathrm{<br>}}$ = 2−1 ${}^{<br>}$

y = -1

y = -1 or 1

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = −14.7224i+0.5j14.7309 $\frac{<br>}{}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = 312 $\frac{\mathrm{<br>}}{}$= 14 $\frac{\mathrm{<br>}}{}$

probability that apple is drawn = 412 $\frac{\mathrm{<br>}}{}$= 13 $\frac{\mathrm{<br>}}{}$

probability that avocado pear is drawn = 512

probability that the first is orange and the second is an avocado pear = 312 $\frac{\mathrm{<br>}}{}$ * 511 $\frac{\mathrm{<br>}}{}$ = 544 $\frac{\mathrm{<br>}}{}$

ii. 

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) =  14∗211+13∗311+512∗411 $\frac{\mathrm{<br>}}{}$

= 122+111+533 $\frac{\mathrm{<br>}}{}$

= 1966 $\frac{\mathrm{<br>}}{}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= 412∗311+311∗412+412∗511+512∗411+412∗311 $\frac{\mathrm{<br>}}{}$

= 12132 $\frac{\mathrm{<br>}}{}$ + 12132 $\frac{\mathrm{<br>}}{}$ + 20132 $\frac{\mathrm{<br>}}{}$ + 20132 $\frac{\mathrm{<br>}}{}$ + 12132 $\frac{\mathrm{<br>}}{}$

= 76132
 

= 1933

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ * 35 $\frac{\mathrm{<br>}}{}$ * 35 $\frac{\mathrm{<br>}}{}$ * 35 $\frac{\mathrm{<br>}}{}$ 

p = 0.0138

(b)   not be late to office is = 1 - 2/5 = 3/5

35 * 35 * 35 * 35 $\frac{\mathrm{<br>}}{}$ * 35 $\frac{\mathrm{<br>}}{}$ * 35 

p = 0.0467

(c) be late to office = 2/5

25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ * 25 $\frac{\mathrm{<br>}}{}$ 

p = 0.0041

The question is asking to express vectors →OP and →OQ in terms of i and j, where 6i + 8j is parallel to →OP and 8i - 6j is parallel to →OQ. We are also given the magnitudes of →OP and →OQ, which are 80 units and 120 units respectively.

To express →OP and →OQ in terms of i and j, we can simply multiply the scalar component of each vector by its respective unit vector. Thus, →OP = 80/10 (6i + 8j) = 48i + 64j, and →OQ = 120/10 (8i - 6j) = 96i - 72j.

To find the magnitude of →PQ, we can use the formula |→PQ| = |→OQ - →OP|. Substituting the values we obtained earlier, we get |→PQ| = |(96i - 72j) - (48i + 64j)| = |48i - 136j|. To simplify this, we can use the Pythagorean theorem, which states that for a right triangle with sides a and b and hypotenuse c, a² + b² = c². Thus, |→PQ| = √(48² + (-136)²) = √20800 = 40√13. Therefore, the magnitude of →PQ is 80√29 units, which is in the form c√k where c = 40 and k = 13.

nC4 ${}_4$ = n![n−4]!4! $\frac{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"> \to </span>}}{}$n[n−1][n−2][n−3][n−4]![n−4]!4! $\frac{\mathrm{<br>}}{}$

nC5 ${}_5$ = n![n−5]!5! $\frac{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"> \to </span>}}{}$n[n−1][n−2][n−3][n−4][n−5]![n−5]!5!]

and nC6 ${}_6$ = n![n−6]!6! $\frac{\mathrm{<br>}}{}$  → n[n−1][n−2][n−3][n−4][n−5][n−6]![n−6]!6!] $\frac{\mathrm{<br>}}{}$

d = U2 - U1 = U3 - U2

nC5 ${}_5$ - nC4 ${}_4$ = n(n−1)(n−2)(n−3)(n−9)5! $\frac{\mathrm{<br>}}{}$

nC6 ${}_6$ - nC5 ${}_5$ = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6 $\frac{\mathrm{<br>}}{}$

nC5 ${}_5$ - nC4 ${}_4$ = nC6 ${}_6$ - nC5 ${}_5$

n(n−1)(n−2)(n−3)(n−9)5! $\frac{\mathrm{<br>}}{}$ = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6 $\frac{\mathrm{<br>}}{}$

divide both sides by n(n-1)(n-2)(n-3) 

n−95!
 = [n−4][n−11]5!6 $\frac{<br>}{}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n2 ${}^2$ -11n - 4n + 44
6n - 54 = n2 ${}^2$ - 15n + 44
n2 ${}^2$ - 21n + 98 = 0
n2 ${}^2$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C5 ${}_5$ - 7C4 ${}_4$

d = 7∗6∗5!2!∗5! $\frac{7\ast 6\ast 5!}{2!\ast 5!}$ - 7∗6∗5∗4!3!∗4! $\frac{7\ast 6\ast 5\ast 4!}{3!\ast 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C5 ${}_5$ - 14C4 ${}_4$

d = 14∗13∗12∗11∗10∗9!9!∗5! $\frac{\mathrm{<br>}}{}$ - 14∗13∗12∗11∗10!10!∗4! $\frac{\mathrm{<br>}}{}$

d = 2002 - 1001
d = 1001