WAEC SSCE - Further Mathematics - 2022

In how many ways can six persons be paired?

Awọn alaye Idahun

6C2=6![6−2]![2!] ${}_2<br>$

6∗5∗4!4!∗2! $\frac{\mathrm{<br>}}{}$

= 6∗52 $\frac{\mathrm{<br>}}{}$

= 15

Evaluate1∫0x2(x3+2)3

Awọn alaye Idahun

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

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.

Awọn alaye Idahun

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 

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

Awọn alaye Idahun

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)".

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?

Awọn alaye Idahun

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.

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

Awọn alaye Idahun

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

If 36, p,94 $\frac{\mathrm{<br>}}{}$ and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.

Awọn alaye Idahun

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

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

94 $\frac{9}{4}$ = 4q

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

Awọn alaye Idahun

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.

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

Awọn alaye Idahun

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

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.

Awọn alaye Idahun

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.

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.

Awọn alaye Idahun

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

A straight line makes intercepts of -3 and 2 on the x and y axes respectively. Find the equation of the line.

Awọn alaye Idahun

Recall:

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

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

A particle moving with a velocity of 5m/s accelerates at 2m/s2 ${}^2$. Find the distance it covers in 4 seconds.

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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 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

Awọn alaye Idahun

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.

Solve: 32x−2−28(3x−2)+3=0

Awọn alaye Idahun

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

Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

Awọn alaye Idahun

Find the coefficient of x3 ${}^3$y2 ${}^2$ in the binomial expansion of (x-2y)5

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

5x2 ${}^{\mathrm{<br>}}$ + x 

Then dy/dx = 10x + 1

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$)

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

Given that 8x+mx2−3x−4≡5x+1+3x−4

Awọn alaye Idahun

8x+mx2−3x−4≡5x+1+3x−4 $\frac{\mathrm{<br>}}{}$

8x+mx2−3x−4 $\frac{\mathrm{<br>}}{}$ ≡ 5(x−1)+3(x+4)x2−3x−4 $\frac{\mathrm{<br>}}{}$

multiplying both sides by x2-3x-4
8x+m ≡ 5(x-4)+3(x+1)
8x + m ≡ 5x - 20 + 3x + 3
8x - 5x - 3x + m = -20 + 3
m = -17

The functions f:x → 2x2 ${}^2$ + 3x -7 and g:x →5x2 ${}^2$ + 7x - 6 are defined on the set of real numbers, R. Find the values of x for which 3f(x) = g(x).

Awọn alaye Idahun

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.

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

Awọn alaye Idahun

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.

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.

Awọn alaye Idahun

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

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

If a hunter is selected at random, find the probability that the hunter covered at least 6km.

Awọn alaye Idahun

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

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

Find the range of values of x for which 2x2 ${}^2$ + 7x - 15 ≥ 0.

Awọn alaye Idahun

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

Awọn alaye Idahun

∣∣∣21−34∣∣∣∣∣∣−6k∣∣∣∣∣∣3−26∣∣∣=15 $|$