WAEC SSCE - Further Mathematics - 2022

Maelezo ya Majibu

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.

Simplify 9∗3n+1−3n+23n+1−3n

Maelezo ya Majibu

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 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.

Maelezo ya Majibu

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.

Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

Maelezo ya Majibu

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.

Maelezo ya Majibu

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

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

What is the mode of the distribution?
 

Maelezo ya Majibu

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.

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

Maelezo ya Majibu

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

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.

Maelezo ya Majibu

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}

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.

Maelezo ya Majibu

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

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

94 $\frac{9}{4}$ = 4q

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

Maelezo ya Majibu

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 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

Maelezo ya Majibu

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.

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

Maelezo ya Majibu

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

Maelezo ya Majibu

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 P = (-4, -5) and Q = (2,3), express →PQ in the form (k,θ). where k is the magnitude and θ the bearing.

Maelezo ya Majibu

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).

Find the coefficient of x2 ${}^2$in the binomial expansion of (x+2x2)5

Maelezo ya Majibu

(x+2x2)5 $<br>$

n = 5,  r = 4,  p = x  and q = 2x2 $\frac{\mathrm{<br>}}{{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"><br></span>}}^{}}$

5C4 ${}_4$x4 ${}^4$ (2x2 $\frac{\mathrm{<br>}}{}$)1 = 5C4 ${}_4$ 2x4x2 $\frac{\mathrm{<br>}}{}$

5C4 ${}_4$ 2x2 ${}^2$ = 5![5−4]!4! $\frac{\mathrm{<br>}}{}$ * 2x2 ${}^{\mathrm{<br>}}$

5∗4!4!∗2x2 $\frac{5\ast 4!}{4!}\ast 2{\mathrm{<br>}}^{}$ = 5 * 2x2 ${}^2$ = 10x2 ${}^{\mathrm{<br>}}$

The coefficient is 10.

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

Maelezo ya Majibu

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

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?

Maelezo ya Majibu

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.

Evaluate 4p2+4C2−4p3

Maelezo ya Majibu

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

(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

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

Maelezo ya Majibu

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

Maelezo ya Majibu

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

Maelezo ya Majibu

∣∣∣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

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

Maelezo ya Majibu

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

5x2 ${}^{\mathrm{<br>}}$ + x 

Then dy/dx = 10x + 1

In how many ways can six persons be paired?

Maelezo ya Majibu

6C2=6![6−2]![2!] ${}_2<br>$

6∗5∗4!4!∗2! $\frac{\mathrm{<br>}}{}$

= 6∗52 $\frac{\mathrm{<br>}}{}$

= 15

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

Maelezo ya Majibu

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 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.

Maelezo ya Majibu

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 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).

Maelezo ya Majibu

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.

Express 4π2 radians in degrees.

Maelezo ya Majibu

4π2 $\frac{\mathrm{<br>}}{}$ = 45∗180 $\frac{4}{5}\ast 180$

= 144º

Solve: 32x−2−28(3x−2)+3=0

Maelezo ya Majibu

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