WAEC SSCE - Further Mathematics - 2022 (Essay)

3cos2x - sinx = 0
cos 2A = 1 – 2 sin2 ${}^2$A
3(1 - 2sin2 ${}^2$x) - sinx = 0
3 - 6sin2 ${}^2$x -sinx = 0
let sinx = p
3 - 6p2 ${}^2$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn: 

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin−1 ${}^{-1}$ p
x = sin−1 ${}^{-1}$(0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin−1 ${}^{-1}$ p
x = sin−1 ${}^{-1}$(0.6287)
x = 38.95º

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.

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

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

p*q = p2 ${}^2$ + 2pq - q2 ${}^2$

when p = -2 and q = -2

p*q = -22 ${}^2$ + 2(-2)(-2) - (-2)2 ${}^2$
p*q = 4 + 8 - 4 = 8

when p = -2 and q = -1
p*q = -22 ${}^2$ + 2(-2)(-1) - (-1)2 ${}^2$
p*q = 4 + 4 - 1 = 7

when p = -2 and q = 1
p*q = -22 ${}^2$ + 2(-2)(1) - (1)2 ${}^2$
p*q = 4 - 4 -1 = -1

when p = -1 and q = -2
p*q = -12 ${}^2$ + 2(-1)(-2) - (-2)2 ${}^2$
p*q = 1 + 4 - 4 = 1

when p = -1 and q = -1
p*q = -12 ${}^2$ + 2(-1)(-1) - (-1)2 ${}^2$
p*q = 1 + 2 -1 = 2

when p = -1 and q = 2
p*q = -12 ${}^2$ + 2(-1)(2) - (2)2 ${}^2$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -2
p*q = 12 ${}^2$ + 2(1)(-2) - (-2)2 ${}^2$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -1
p*q = 12 ${}^2$ + 2(1)(-1) - (-1)2 ${}^2$
p*q = 1 - 2 - 1 = -2

when p = 1 and q = 1
p*q = 12 ${}^2$ + 2(1)(1) - (1)2 ${}^2$
p*q = 1 + 2 - 1 = 2

when p = 1 and q = 2
p*q = 12 ${}^2$ + 2(1)(2) - (2)2 ${}^2$
p*q = 1 + 4 - 4 = 1

when p = 2 and q = -2
p*q = 22 ${}^2$ + 2(2)(-2) - (-2)2 ${}^2$
p*q = 4 - 8 - 4 = -8

when p = 2 and q = -1
p*q = 22 ${}^2$ + 2(2)(-1) - (-1)2 ${}^2$
p*q = 4 - 4 - 1 = -1

when p = 2 and q = 1
p*q = 22 ${}^2$ + 2(2)(1) - (1)2 ${}^2$
p*q = 4 + 4 - 1 = 7
when p = 2 and q = 2
p*q = 22 ${}^2$ + 2(2)(2) - (2)2 ${}^2$
p*q = 4 + 8 - 4 = 8

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

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

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.

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

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