WAEC SSCE - Further Mathematics - 2022

If log10(3x+1)+log104=log10(9x+2)
, find the value of x 

Bayanin Amsa

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

If g(x) = √(1-x2 ${}^2$), find the domain of g(x)

Bayanin Amsa

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

If α and β are roots of x2 ${}^2$ + mx - n = 0, where m and n are constants, form the

Bayanin Amsa

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

A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.

Bayanin Amsa

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

A linear transformation T is defined by T: (x,y) → (3x - y, x + 4y). Find the image of (2, -1) under T.

Bayanin Amsa

The linear transformation T is defined as T: (x, y) → (3x - y, x + 4y). To find the image of (2, -1) under T, we need to apply T to (2, -1) and see what we get. So, T(2, -1) = (3(2) - (-1), 2 + 4(-1)) = (7, -2) Therefore, the image of (2, -1) under T is (7, -2). Answer is correct.

The first, second and third terms of an exponential sequence (G.P) are (x - 4), (x + 2), and (3x + 1) respectively. Find the values of x.

Bayanin Amsa

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 $\frac{-1}{2},8$)

Evaluate ∫0−1 ${<br>}_{\mathrm{<br>}}^{}$ (x + 1)(x - 2) dx

Bayanin Amsa

∫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 $\frac{{\mathrm{<br>}}^{}}{}$

The equation of a circle is given as 2x2 ${}^2$ + 2y2 ${}^2$ - x - 3y - 41 = 0. Find the coordinates of its centre.

Bayanin Amsa

2x2 ${}^{\mathrm{<br>}}$ + 2y2 ${}^{\mathrm{<br>}}$ - x - 3y - 41

standard equation of circle
(x-a)2 ${}^2$ + (x-b)2 ${}^2$ = r2 ${}^2$
General form of equation of a circle.
x2 ${}^2$ + y2 ${}^2$ + 2gx + 2fy + c = 0
a = -g, b = -f., r2 = g2 + f2 - c
the centre of the circle is (a,b)
comparing the equation with the general form of equation of circle.
2x2 ${}^2$ + 2y2 ${}^2$ - x - 3y - 41

= x2 ${}^2$ + y2 ${}^2$ + 2gx + 2fy + c
2x2 ${}^2$ + 2y2 ${}^2$ - x - 3y - 41 = 0
divide through by 2

g = −14 $\frac{<br>}{}$ ; 2g = −12 $\frac{<br>}{}$

f = −34 $\frac{<br>}{}$ ; 2f = −32 $\frac{<br>}{}$

a = -g  → - −14 $\frac{-}{}$ ; = 14 $\frac{\mathrm{<br>}}{}$

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (14 $\frac{\mathrm{<br>}}{}$, 34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$)

Solve: 4sin2 ${}^2$θ + 1 = 2, where 0º < θ < 180º

Bayanin Amsa

4sin2
2 ${}^{}$θ + 1 = 2

4sin2 ${}^{\mathrm{<br>}}$θ  = 2 - 1

4sin2 ${}^2$θ = 1

s√in2θ
 = 14−−√

sinθ = 12 $\frac{\mathrm{<br>}}{}$

θ = sin−112 $\mathrm{<br>}$

θ = 30º 0r 150º

Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6

Bayanin Amsa

y = 2x + 5
m1 = 2
2y = x - 6

tanθ =  32 $\frac{\mathrm{<br>}}{}$ ÷ (1+1)

tanθ = 32 $\frac{\mathrm{<br>}}{}$ ÷ 2

θ = tan−1(34) $\mathrm{<br>}$

θ = 36.87º
θ = 37º

Evaluate1∫0x2(x3+2)3

Bayanin Amsa

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

Given that P = {x: x is a multiple of 5}, Q = {x: x is a multiple of 3} and R = {x: x is an odd number} are subsets of μ = {x: 20 ≤ x ≤ 35}, (P⋃Q)∩R.

Bayanin Amsa

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}

Solve: 32x−2−28(3x−2)+3=0

Bayanin Amsa

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

The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.

Bayanin Amsa

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.

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

Bayanin Amsa

(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

If f(x-1) = x3 ${}^3$ + 3x2 ${}^2$ + 4x - 5, find f(2)

Bayanin Amsa

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

A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t2 ${}^{\mathrm{<br>}}$ + 3t. Find the magnitude of F at time t.

Bayanin Amsa

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 

The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.

Bayanin Amsa

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.

Bayanin Amsa

∣∣∣21−34∣∣∣∣∣∣−6k∣∣∣∣∣∣3−26∣∣∣=15 $|$

∣∣∣2[−6]1[−6]−3k+4k∣∣∣=∣∣∣3−26∣∣∣ $|\begin{array}{cc}\mathrm{<br>}& \end{array}$

∣∣∣−12−6−3k+4k∣∣∣=∣∣∣3−26∣∣∣ $|\begin{array}{cc}<br>& \end{array}$

-12 - 3k = 3

-3k = 3 + 12

k = 15−3 $\frac{\mathrm{<br>}}{}$

k = -5

A binary operation ∆ is defined on the set of real numbers R, by x∆y = x+y−xy4−−−−−−−−−√
, where x, yER. Find the value of 4∆3

Bayanin Amsa

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.

Bayanin Amsa

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.

The gradient of a function at any point (x,y) 2x - 6. If the function passes through (1,2), find the function.

Bayanin Amsa

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

Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

Bayanin Amsa

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

What is the mode of the distribution?
 

Bayanin Amsa

To find the mode of the distribution, we need to identify the value with the highest frequency. In the table, the frequency is given for each distance (in km) covered by the hunters. Let's first find the frequency of distance 5 km covered by the hunters. From the table, we know that the sum of all frequencies should be equal to 40, the total number of hunters. Sum of frequencies = 5 + 4 + x + 9 + 2x + 1 = 40 Simplifying the equation, we get: 3x + 19 = 40 3x = 21 x = 7 Now we know that the frequency of distance 5 km is x = 7. The frequencies for the other distances are: - Frequency of distance 3 km = 5 - Frequency of distance 4 km = 4 - Frequency of distance 6 km = 9 - Frequency of distance 7 km = 2x = 14 - Frequency of distance 8 km = 1 The highest frequency is 14, which corresponds to the distance of 7 km. Therefore, the mode of the distribution is 7 km. Hence, the answer is option (C) 7.

The probability that a student will graduate from college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?

Bayanin Amsa

To find the probability that at least one student will graduate, we can calculate the probability that no student will graduate and then subtract that from 1. The probability that a student will not graduate from college is 1 - 0.4 = 0.6. So, the probability that none of the three selected students will graduate is: 0.6 * 0.6 * 0.6 = 0.216 Therefore, the probability that at least one student will graduate is: 1 - 0.216 = 0.784 So, the answer is 0.78. In summary, the probability that at least one student will graduate from college given that 3 students are selected is 0.784 or 0.78, which is obtained by subtracting the probability that none of the three students will graduate from 1.

Which of the following is the semi-interquartile range of a distribution?

Bayanin Amsa

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)".

If Un = kn2 ${}^2$ + pn, U1 ${}_1$ = -1, U5 ${}_5$ = 15, find the values of k and p.

Bayanin Amsa

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

Differentiate 5x3+x2x $\frac{\mathrm{<br>}}{}$, x ≠ 0 with respect to x.

Bayanin Amsa

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

5x2 ${}^{\mathrm{<br>}}$ + x 

Then dy/dx = 10x + 1

A particle is acted upon by forces F = (10N, 060º), P = (15N, 120º) and Q = (12N, 200º). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.

Bayanin Amsa

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

Given that 8x+mx2−3x−4≡5x+1+3x−4