WAEC SSCE - Further Mathematics - 2022

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).

Bayanin Amsa

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.

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.

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

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

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.

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.

Bayanin Amsa

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

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

94 $\frac{9}{4}$ = 4q

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

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

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

Bayanin Amsa

(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 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

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.

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.

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.

Bayanin Amsa

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).

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 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.

Bayanin Amsa

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

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 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.

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 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.

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

Bayanin Amsa

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

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

Simplify 9∗3n+1−3n+23n+1−3n

Bayanin Amsa

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

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 

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

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

Bayanin Amsa

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

Given that 8x+mx2−3x−4≡5x+1+3x−4

Bayanin Amsa

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

If x2+y2+−2x−6y+5=0 ${\mathrm{<br>}}^{}$, evaluate dy/dx when x=3 and y=2.

Bayanin Amsa

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

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

Evaluate 4p2+4C2−4p3

Bayanin Amsa

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

In how many ways can six persons be paired?

Bayanin Amsa

6C2=6![6−2]![2!] ${}_2<br>$