WAEC SSCE - Further Mathematics - 2022 (Objective)

Answer Details

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

Answer Details

∣∣∣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

Answer Details

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 

Answer Details

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

Answer Details

∫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 $\frac{{\mathrm{<br>}}^{}}{}$

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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.

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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.

Answer Details

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

Answer Details

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).

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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

Answer Details

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)".

Answer Details

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

Answer Details

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

= 144º

Answer Details

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

5x2 ${}^{\mathrm{<br>}}$ + x 

Then dy/dx = 10x + 1

Answer Details

(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.

Answer Details

6C2=6![6−2]![2!] ${}_2<br>$

6∗5∗4!4!∗2! $\frac{\mathrm{<br>}}{}$

= 6∗52 $\frac{\mathrm{<br>}}{}$

= 15

Answer Details

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}

Answer Details

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

Answer Details

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