WAEC SSCE - Further Mathematics - 2022 (Objective)

∫0−1 ${\int }_{-1}^0$ (x + 1)(x - 2) dx

= ∫0−1 ${\int }_{-1}^0$ x2−x−2 ${\mathrm{<br>}}^{}$dx

Integrated x2−x−2 ${\mathrm{<br>}}^{}$ = x33−x22−2 $\frac{{\mathrm{<br>}}^{}}{}$

=

(0

-

0

-

0)

-

(

-13

-

12

+

2)

 

=

0

-

(

76

)

=

-76

4p2+4C2−4p3 $\mathrm{<br>}$

npr=n![n−r]!andnCr=n![n−r]!r! $\mathrm{<br>}$

= 4![4−2]!+4![4−2]!2!−4![4−3]!=4!2!+4!2!2!−4!1! $\frac{4!}{[4-2]!}+\frac{4!}{[4-2]!2!}-\frac{4!}{[4-3]!}=\frac{4!}{2!}+\frac{4!}{2!2!}-\frac{4!}{1!}$

= 4∗3∗2!2!+4∗3∗2!2!2!−4∗3∗2∗11! $\frac{4\ast 3\ast 2!}{2!}+\frac{4\ast 3\ast 2!}{2!2!}-\frac{4\ast 3\ast 2\ast 1}{1!}$

12 + 6 - 24 = -6

The problem provides us with two points, (x, 4) and (-x, 3), and tells us that the distance between them is 7 units. We need to find the value of x that satisfies this condition. To find the distance between two points, we can use the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) where (x1, y1) and (x2, y2) are the coordinates of the two points, and d is the distance between them. Using this formula, we can calculate the distance between the given points: d = sqrt((-x - x)^2 + (3 - 4)^2) d = sqrt(4x^2 + 1) We are given that this distance is equal to 7 units: sqrt(4x^2 + 1) = 7 To solve for x, we need to isolate it on one side of the equation. To do this, we will square both sides of the equation: 4x^2 + 1 = 49 Now we can solve for x: 4x^2 = 48 x^2 = 12 x = sqrt(12) = 2sqrt(3) Therefore, the value of x that satisfies the given conditions is 2sqrt(3), which is option D.

P = { 20, 25, 30, 35}, Q = {21, 24, 27, 30, 33}, R = {21, 23, 25, 27, 29, 31, 33, 35}

(P⋃Q)∩R = {20, 21, 24, 25, 27, 30, 33, 35} ∩ {21, 23, 25, 27, 29, 31, 33, 35}

= {21, 25, 27, 33, 35}

Un = kn2 ${}^2$ + pn,

U1 ${}_1$ = -1,

U5 ${}_5$ = 15,

when n = 1
U1 = k(1)2 ${}^2$+ p(1) = -1
k + p = -1 --------eqn1
when n = 5
U5 ${}_5$ = k(5)2 ${}^2$+ p(5) = 15
25k + 5p = 15 --------eqn2
multiply eqn1 by 5 and eqn2 by 1
5k + 5p = -5 -------eqn3
25k + 5p = 15 -------eqn4
eqn4 - eqn3
20k = 20
k = 1
sub for k in eqn1
1 + p = -1
p = -1 -1 = -2

1∫0x2(x3+2)3 ${\mathrm{<br>}}_{}^{}{3}^{}$dx

let u=x3+2,du=3x2dx $\mathrm{<br>}$

when x = 1,  u = 3

when x = 0,  u = 2

dx = du3x2 $\frac{\mathrm{<br>}}{}$

3∫2 $3_2^{\int }$ x2[u]33x2 $\frac{{\mathrm{<br>}}^{}}{}$

3∫2 $3_2^{\int }$ u33 $\frac{{\mathrm{<br>}}^{}}{}$ du 

= u43∗4 $\frac{{\mathrm{<br>}}^{}}{}$2 ${}_{\mathrm{<br>}}$3

112[u4] $\frac{\mathrm{<br>}}{}$ ${}_2$3

112[34−24] $\frac{1}{12}[3^4-2^4]$

112[81−16] $\frac{1}{12}[81-16]$

6512

(3√6√5+√543√5 $\frac{3\surd 6}{\surd 5}+\frac{\surd 54}{3\surd 5}$)−1 ${}^{-1}$

= √5(3√5)3√6+3√6 $\frac{\surd 5(3\surd 5)}{3\surd 6+3\surd 6}$

= 3∗56√6=52√6 $\frac{3\ast 5}{6\surd 6}=\frac{5}{2\surd 6}$

= 5∗2√62√6+2√6=10√64∗6 $\frac{5\ast 2\surd 6}{2\surd 6+2\surd 6}=\frac{10\surd 6}{4\ast 6}$

= 5√612

x3 ${}^{\mathrm{<br>}}$y2 ${}^{\mathrm{<br>}}$ in (x-2y)5 ${}^{\mathrm{<br>}}$

n = 5, r = 3, p = x, q = -2y

 

5C3 ${}_3$ * x3 ${}^{\mathrm{<br>\; <span\; style="font-weight:\; var(--bs-body-font-weight);"> -2y</span>}}$2 ${}^{\mathrm{<br>}}$

5C3 ${}_3$ = 5![5−3]!3!

5∗4∗3!2!3! $\frac{\mathrm{<br>}}{}$ →  5∗42 $\frac{\mathrm{<br>}}{}$

5C3 ${}_{\mathrm{<br>}}$ = 10

: 5C3 ${}_{\mathrm{<br>}}$ * x3 ${}^{\mathrm{<br>}}$ -2y2 ${}^{\mathrm{<br>}}$ = 10 *  x3 ${}^{\mathrm{<br>}}$ 4y2 ${}^{\mathrm{<br>}}$

 40x3 ${}^{\mathrm{<br>}}$y2 ${}^{\mathrm{<br>}}$
the coefficient is 40

GP : 36, P, q4 $\frac{�}{4}$, q, ... p + q = ?

Recall,

common

ratio,

r

=

TnTn-1

=

T2T1

=

T3T2

=

T4T3

 

∴

P36

=

94

÷

p

;

p2 ${}^2$

=

94

x

36

;

p2 ${}^2$

=

81

 

p

=

9

∴

r

=

T2T1

=

936

=

14

 

Also

r

=

T4T3

=

q

÷

94

∴ 14 $\frac{1}{4}$ = q ÷ 94 $\frac{9}{4}$ ;

94 $\frac{9}{4}$ = 4q

16q

=

9

,

q

=

916

∴

p

+

q

=

9

+

916

=

9

916

F = m * a

d = t2 ${}^{\mathrm{<br>}}$ + 3t.

a = d2ddt2 $\frac{{\mathrm{<br>}}^{}}{}$

d[d]dt = 2t + 3

d2ddt2 = 2m/s2 ${}^{\mathrm{<br>}}$

a = 2m/s2 ${}^{\mathrm{<br>}}$

F = m * a

F = 3 × 2 = 6N 

x2+y2+−2x−6y+5=0 ${\mathrm{<br>}}^{}$

When differentiated:

x2+y2+−2x−6y+5=0 ${\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"> \to \; 2x\; +\; 2y\; -\; 2\; -\; 6\; =\; 0</span>}}^{}$

where x=3 and y=2

2[3] + 2[2] - 8 = 0

6 + 4 - 8 = 2

Explanation

log10(3x+1)+log104=log10(9x+2) $\mathrm{<br>}$

log104(3x+1)=log10(9x+2) $\mathrm{<br>}$

4(3x+1) = 9x + 2

12x -4 = 9x + 2

12x - 9x = 2 + 4

3x = 6

x = 2

Converting the forces to their rectangular forms
F = (10N, 060º)
Fx = 10cos60 = 5i
fy = 10sin60 = 8.66j
F = 5i + 8.66j
P = (15N, 120º)
Px = 15cos120 = -7.5i
py = 15sin120 = 12.99j
P = -7.5i + 12.99j
Q = (12N, 200º)
Qx = 12cos200 = -11.28i
Qy = 12sin200 = -4.1j
Q = -11.28i -4.1j
The resultant force = F + P + Q
R = 5i + 8.66j + (-7.5i + 12.99j) + (-11.28i -4.1j)
R = -13.78i + 17.55j

dy/dx = 2x - 6

y = ∫ 2x - 6

y = 2x
2−6+c $\frac{2}{}$y = x2 ${}^{\mathrm{<br>}}$ - 6x + c
passes through (1,2)
2 = 12 ${}^{\mathrm{<br>}}$ - 6(1) + c
2 = 1 - 6 + c
c = 7
y = x2 ${}^{\mathrm{<br>}}$ - 6x + c

y = x2 ${}^{\mathrm{<br>}}$ -  6x + 7

m1u1 + m2u2 = (m1 + m2)v

m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s

18(4) + 6(-10) = (18+6)v

72 - 60 = 24v
12 = 24v
v = 12 $\frac{\mathrm{<br>}}{}$ m/s

32x−2−28(3x−2)+3=0 ${\mathrm{<br>}}^{}$

32x32−28.3x32+3=0 $\frac{{\mathrm{<br>}}^{}}{}$

32x9−28.3x9+3=0 $\frac{{\mathrm{<br>}}^{}}{}$

let p = 3x ${}^{\mathrm{<br>}}$

p29−28p9+3=0 $\frac{{\mathrm{<br>}}^{}}{}$

multiply through by 9
p2 ${}^2$ - 28p + 27 = 0
p2 ${}^2$ - p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3x ${}^{\mathrm{<br>}}$
3x ${}^{\mathrm{<br>}}$ = 1
3x ${}^{\mathrm{<br>}}$ = 30 ${}^0$
x = 0
when p = 27
3x = 27
3x = 33
x = 3

9∗3n+1−3n+23n+1−3n $\frac{\mathrm{<br>}}{}$

= 3n∗3n∗31∗−32∗323n∗31−3n $\frac{{\mathrm{<br>}}^{}}{}$

= 3n(32∗31)3n(31−1) $\frac{{\mathrm{<br>}}^{}}{}$

= 27−93−1 $\frac{27-9}{3-1}$

= 182 $\frac{18}{2}$

= 9

The semi-interquartile range is a measure of variability in a dataset. To calculate it, we first need to find the median of the dataset, which is the middle value when the data is arranged in order from lowest to highest. Then, we need to find the quartiles, which are the values that divide the dataset into four equal parts. The semi-interquartile range is half of the difference between the upper quartile and the lower quartile. The upper quartile is the value that separates the highest 25% of the data from the lowest 75%, while the lower quartile is the value that separates the lowest 25% of the data from the highest 75%. Looking at the given options, the formula that corresponds to the semi-interquartile range is: 1/2 (Upper Quartile - Lower Quartile) Therefore, the correct answer is option d) "1/2 (Upper Quartile - Lower Quartile)".

x∆y = x+y−xy4−−−−−−−−−√

4∆3 = 4+3−4∗34−−−−−−−−−√ $\sqrt{4+3-\frac{4\ast 3}{4}}$ 

= 4+3−3−−−−−−−−√ $\sqrt{4+3-3}$

= 4–√ $\sqrt{4}$

= 2.

Recall:

xa

+

yb

=1

Where 'a' and 'b' are the x and y intercept respectively.
 

x-3

+

y2

=1

2x-3y = -6
2x - 3y + 6 = 0 -------(1)
multiply through by -1
-2x + 3y - 6 = 0

from the equation of motion

u = 5m/s, a = 2m/s2 ${}^{\mathrm{<br>}}$, t = 4s 

s = ut + 12at2 $\frac{\mathrm{<br>}}{}$

s = 5*4 + 122∗42 $\frac{\mathrm{<br>}}{}$

s =  20 + 16

s = 36m

6C2=6![6−2]![2!] ${}_2<br>$

6∗5∗4!4!∗2! $\frac{\mathrm{<br>}}{}$

= 6∗52 $\frac{\mathrm{<br>}}{}$

= 15

5x3+x2x $\frac{\mathrm{<br>}}{}$ → 5x3x+x2x $\frac{\mathrm{<br>}}{}$

5x2 ${}^{\mathrm{<br>}}$ + x 

Then dy/dx = 10x + 1

x - 1 = 2
x = 3
f(2) = (3)3 ${}^{\mathrm{<br>}}$ + 3(3)2 ${}^{\mathrm{<br>}}$ + 4(3) - 5
f(2) = 27 + 27 + 12 - 5

= 61

To find the values of x for which 3f(x) = g(x), we need to substitute the given functions f(x) and g(x) and then solve for x. Substituting the function f(x) = 2x^2 + 3x - 7, we get: 3f(x) = 3(2x^2 + 3x - 7) = 6x^2 + 9x - 21 Substituting the function g(x) = 5x^2 + 7x - 6, we get: g(x) = 5x^2 + 7x - 6 Now we can set these two equations equal to each other and solve for x: 6x^2 + 9x - 21 = 5x^2 + 7x - 6 Subtracting 5x^2 and 7x from both sides, we get: x^2 + 2x - 15 = 0 Factoring this quadratic equation, we get: (x + 5)(x - 3) = 0 Therefore, the values of x that satisfy the equation 3f(x) = g(x) are x = -5 and x = 3. Hence, the answer is: x = -5 or 3.

This is a question about vector addition. To find →PR, we need to add →PQ and →RQ, but in reverse order (→PR = →PQ + →RQ). Given →PQ = -2i + 5j and →RQ = -i - 7j, we can add them as follows: →PR = →PQ + →RQ →PR = (-2i + 5j) + (-i - 7j) →PR = -2i - i + 5j - 7j →PR = -3i - 2j Therefore, the answer is "-3i - 2j". Option (B) "-3i + 12j", (C) "-i + 12j", and (D) "i - 12j" are incorrect.

To find →PQ, we need to subtract the coordinates of point P from those of point Q, giving us: →PQ = Q - P = (2-(-4), 3-(-5)) = (6, 8) The magnitude, or length, of →PQ is found using the distance formula: k = √(6² + 8²) = √100 = 10 To find the bearing θ, we use trigonometry. The tangent of θ is the ratio of the opposite side (the change in y-coordinates) to the adjacent side (the change in x-coordinates): tan θ = 8/6 = 4/3 Using a calculator, we can find that θ is approximately 53.13º. However, we need to adjust this value depending on which quadrant →PQ lies in. Since both x and y are positive, →PQ lies in the first quadrant, so we don't need to make any adjustments. Therefore: θ = 53.13º So the vector →PQ can be expressed in the form (k,θ) as: (10 units, 53.13º) Therefore, the correct answer is (a) (10 units, 053º).

2x2 ${}^2$ + 7x - 15 ≥ 0

2x2 ${}^2$ -3x + 10x - 15 ≥ 0
x(2x - 3) + 5(2x - 3) ≥ 0
(x+5)(2x-3) ≥ 0
the points on x-axis where the graph ≥ 0

x ≤ -5 or x ≥ 32 $\frac{\mathrm{<br>}}{}$

 

1 - x2 ${}^{\mathrm{<br>}}$ ≥ 0
-x2 ${}^{\mathrm{<br>}}$ ≥ -1
x2 ${}^{\mathrm{<br>}}$ ≤ 1
√x2 ${}^{\mathrm{<br>}}$ ≤ 1
|x| ≤ 1
-1 ≤ x ≤ 1

To find the mean height of all the students, you need to add up the heights of all the students and divide by the total number of students. First, you need to find the total number of students, which is the sum of 20, 16 and 14 students, which is 50 students in total. Then, you need to find the total height of all the students by multiplying the number of students in each group by their respective mean height and adding them all up. Total height of all students = (20 x 1.67) + (16 x 1.50) + (14 x 1.40) = 33.4 + 24 + 19.6 = 77 meters Finally, to find the mean height of all the students, you need to divide the total height of all the students by the total number of students: Mean height of all students = Total height of all students / Total number of students = 77 / 50 = 1.54 meters Therefore, the answer is 1.54m.

U1 = x - 4
U2 = x + 2
U3 = 3x + 1

u2u1=u3u2 $\frac{{\mathrm{<br>}}_{}}{}$

x+2x−4=3x+1x+2 $\frac{\mathrm{<br>}}{}$

(x+2)(x+2) = (x-4)(3x+1)
x2 ${}^2$ + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x2 ${}^2$ - 15x - 8 =0
2x2 ${}^2$ + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0

x = (−12,8 $\frac{-1}{2},8$)

5+4+x+9+2x+1 = 40
19+3x = 40
3x = 21
x = 7

Distance(km)	3	4	5	6	7	8
Frequency	5	4	7	9	14	1

The probability that the hunter covered at least 6km, means the hunter covered either 6km or 7km, or 8km.

24 hunters covered at least 6km

2440

=

35

x2 ${}^2$ + mx - n = 0

a = 1, b = m, c = -n

α + β = −ba $\frac{<br>}{}$ = −m1 $\frac{<br>}{}$ = -m

αβ = ca $\frac{\mathrm{<br>}}{}$ = −n1 $\frac{<br>}{}$ = -n

the roots are = 1α $\frac{1}{\alpha }$ and 1β $\frac{1}{\beta }$

sum of the roots = 1α $\frac{1}{\alpha }$ + 1β $\frac{1}{\beta }$

1α $\frac{\mathrm{<br>}}{}$ + 1β $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><span\; style="font-weight:\; var(--bs-body-font-weight);"> = </span></span>}}{}$α+βαβ $\frac{<br>}{}$

α + β = -m
αβ = -n

α+βαβ $\frac{<br>}{}$ 

product of the roots = 1α $\frac{\mathrm{<br>}}{}$ * 1β $\frac{\mathrm{<br>}}{}$

1α $\frac{\mathrm{<br>}}{}$ + 1β $\frac{\mathrm{<br>}}{}$ = 1αβ $\frac{\mathrm{<br>}}{}$ → 

x2 ${}^{\mathrm{<br>}}$ - (sum of roots)x + (product of roots)
x2 ${}^{\mathrm{<br>}}$ - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx2 ${}^{\mathrm{<br>}}$ - mx - 1 = 0