WAEC SSCE - Further Mathematics - 2022 (Objective)

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

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

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).

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

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

= 144º

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

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

 

tanθ

=

m1 - m21 + m1m2

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

y

=

12

x

-

3

 

m2

=

12

 

tanθ

=

2 - 12 $\frac{1}{2}$1+2(12 $\frac{\mathrm{<br>}}{}$)

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

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

tanθ

=

34

θ = tan−1(34) $\mathrm{<br>}$

θ = 36.87º
θ = 37º

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

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

Recall:

xa

+

yb

=1

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

x-3

+

y2

=1

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

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

Distance(km)	3	4	5	6	7	8
Frequency	5	4	7	9	14	1

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

2440

=

35

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

(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.

6C2=6![6−2]![2!] ${}_2<br>$

6∗5∗4!4!∗2! $\frac{\mathrm{<br>}}{}$

= 6∗52 $\frac{\mathrm{<br>}}{}$

= 15

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

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

∫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 $\frac{{\mathrm{<br>}}^{}}{}$

=

(0

-

0

-

0)

-

(

-13

-

12

+

2)

 

=

0

-

(

76

)

=

-76

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

(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

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

∣∣∣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

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

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

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