WAEC SSCE - Further Mathematics - 2022

Express 4π2 radians in degrees.

Answer Details

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

= 144º

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?

Answer Details

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.

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

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

Evaluate 4p2+4C2−4p3

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

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.

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

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

Answer Details

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

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

What is the mode of the distribution?
 

Answer Details

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.

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).

Answer Details

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.

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

Answer Details

(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

Answer Details

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.

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

Answer Details

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

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

Answer Details

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.

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.

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}

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

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

Evaluate1∫0x2(x3+2)3

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

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

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

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

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.

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

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

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.

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 

Solve: 32x−2−28(3x−2)+3=0

Answer Details

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

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

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

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

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

= 61

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.

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

In how many ways can six persons be paired?

Answer Details

6C2=6![6−2]![2!] ${}_2<br>$

6∗5∗4!4!∗2! $\frac{\mathrm{<br>}}{}$

= 6∗52 $\frac{\mathrm{<br>}}{}$

= 15

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

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.

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.

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).

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