WAEC SSCE - Further Mathematics - 2022

Question 1 Report

A binary operation ∆ is defined on the set of real numbers R, by x∆y = $\sqrt{x + y - \frac{x y}{4}}$ , where x, yER. Find the value of 4∆3

16

8

4

2

Answer Details

x∆y = $\sqrt{x + y - \frac{x y}{4}}$

4∆3 = $\sqrt{4 + 3 - \frac{4 * 3}{4}}$

= $\sqrt{4 + 3 - 3}$

= $\sqrt{4}$

= 2.

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Binary Operations

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Question 2 Report

Evaluate $1_{0}^{\int} x^{2} (x^{3} + 2)^{3}$

$\frac{56}{12}$

$\frac{65}{12}$

12

65

Answer Details

$1_{0}^{\int} x^{2} (x^{3} + 2)^{3}$ dx

let $u = x^{3} + 2, d u = 3 x^{2} d x$

when x = 1, u = 3

when x = 0, u = 2

dx = $\frac{d u}{3 x^{2}}$

$3_{2}^{\int}$ $\frac{x^{2} [u]^{3}}{3 x^{2}}$

$3_{2}^{\int}$ $\frac{u^{3}}{3}$ du

= $\frac{u^{4}}{3 * 4}$ $_{2}$ 3

$\frac{1}{12} [u^{4}]$ $_{2}$ 3

$\frac{1}{12} [3^{4} - 2^{4}]$

$\frac{1}{12} [81 - 16]$

$\frac{65}{12}$

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Integration

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Question 3 Report

Solve: $3^{2 x - 2} - 28 (3^{x - 2}) + 3 = 0$

x = -2 or x = 1

x = 0 or x = -3

x = 2 or x = 1

x = 0 or x = 3

Answer Details

$3^{2 x - 2} - 28 (3^{x - 2}) + 3 = 0$

$\frac{3^{2 x}}{3^{2}} - \frac{{28.3}^{x}}{3^{2}} + 3 = 0$

$\frac{3^{2 x}}{9} - \frac{{28.3}^{x}}{9} + 3 = 0$

let p = 3 $^{x}$

$\frac{p^{2}}{9} - \frac{28 p}{9} + 3 = 0$

multiply through by 9
p $^{2}$ - 28p + 27 = 0
p $^{2}$ - p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3 $^{x}$
3 $^{x}$ = 1
3 $^{x}$ = 3 $^{0}$
x = 0
when p = 27
3x = 27
3x = 33
x = 3

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Differentiation

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Question 4 Report

The functions f:x → 2x $^{2}$ + 3x -7 and g:x →5x $^{2}$ + 7x - 6 are defined on the set of real numbers, R. Find the values of x for which 3f(x) = g(x).

x = -3 or -5

x = -3 or 5

x = 3 or -5

x = 3 or 5

Functions

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Question 5 Report

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

x

^{2}

- 6x - 5

x

^{2}

- 6x + 5

x

^{2}

- 6x - 3

x

^{2}

- 6x + 7

Answer Details

dy/dx = 2x - 6

y = ∫ 2x - 6

y = $\frac{2 x^{2}}{2} - 6 + c$ y = x $^{2}$ - 6x + c
passes through (1,2)
2 = 1 $^{2}$ - 6(1) + c
2 = 1 - 6 + c
c = 7
y = x $^{2}$ - 6x + c

y = x $^{2}$ - 6x + 7

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Functions

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Question 6 Report

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

What is the mode of the distribution?

Distance(km)	3	4	5	6	7	8
Frequency	5	4	x	9	2x	1

5

6

7

8

Statistics

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Question 7 Report

Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

All strong wrestlers are weightlifters

Some strong wrestlers are not weightlifters

Some weak wrestlers are weightlifters

All weightlifters are wrestlers

Logical Reasoning

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Question 8 Report

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.

17.55i + 13.78j

17.55j - 13.78i

-17.55i + 13.78j

-17.55i - 13.78j

Statics

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Question 9 Report

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

x ≤ -5 or x ≥

\frac{3}{2}

x ≥ -5 or x ≤

\frac{3}{2}

-5 ≤ x ≤

\frac{3}{5}

\frac{3}{5}

≤ x ≤ -5

Answer Details

2x $^{2}$ + 7x - 15 ≥ 0

2x $^{2}$ -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 ≥ $\frac{3}{2}$

View Answer

Question 10 Report

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.

(10 units, 053º)

(9 units, 049º)

(10 units, 037º)

(9 units, 027º)

Vectors

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Question 11 Report

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?

0.06

0.22

0.78

0.80

Probability

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Question 12 Report

( $\frac{3 \sqrt 6}{\sqrt 5} + \frac{\sqrt 54}{3 \sqrt 5}$ ) $^{- 1}$

\frac{5 \sqrt 3}{6}

\frac{3 \sqrt 15}{6}

\frac{5 \sqrt 6}{12}

\frac{5 \sqrt 3}{12}

Answer Details

( $\frac{3 \sqrt 6}{\sqrt 5} + \frac{\sqrt 54}{3 \sqrt 5}$ ) $^{- 1}$

= $\frac{\sqrt 5 (3 \sqrt 5)}{3 \sqrt 6 + 3 \sqrt 6}$

= $\frac{3 * 5}{6 \sqrt 6} = \frac{5}{2 \sqrt 6}$

= $\frac{5 * 2 \sqrt 6}{2 \sqrt 6 + 2 \sqrt 6} = \frac{10 \sqrt 6}{4 * 6}$

= $\frac{5 \sqrt 6}{12}$

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Surds

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Question 13 Report

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

10

40

32

80

Answer Details

$(x + \frac{2}{x^{2}})^{5}$

n = 5, r = 4, p = x and q = $\frac{2}{x^{2}}$

5C $_{4}$ x $^{4}$ ( $\frac{2}{x^{2}}$ )1 = 5C $_{4}$ $\frac{2 x^{4}}{x^{2}}$

5C $_{4}$ 2x $^{2}$ = $\frac{5!}{[5 - 4]! 4!}$ * 2x $^{2}$

$\frac{5 * 4!}{4!} * 2 x^{2}$ = 5 * 2x $^{2}$ = 10x $^{2}$

The coefficient is 10.

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Binomial Theorem

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Question 14 Report

Given that $\frac{8 x + m}{x^{2} - 3 x - 4} \equiv \frac{5}{x + 1} + \frac{3}{x - 4}$

23

17

-17

18

Answer Details

$\frac{8 x + m}{x^{2} - 3 x - 4} \equiv \frac{5}{x + 1} + \frac{3}{x - 4}$

$\frac{8 x + m}{x^{2} - 3 x - 4}$ ≡ $\frac{5 (x - 1) + 3 (x + 4)}{x^{2} - 3 x - 4}$

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

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Binomial Theorem

Polynomial Functions

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Question 15 Report

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

76

^{\circ}

53

^{\circ}

37

^{\circ}

14

^{\circ}

Answer Details

tanθ

=

m1 - m21 + m1m2

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

y

=

12

x

-

3

m2

=

12

tanθ

=

2 -

\frac{1}{2}

1+2(

\frac{1}{2}

)

tanθ = $\frac{3}{2}$ ÷ (1+1)

tanθ = $\frac{3}{2}$ ÷ 2

tanθ

=

34

θ = $t a n^{- 1} (\frac{3}{4})$

θ = 36.87º
θ = 37º

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Co-ordinate Geometry

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Question 16 Report

Evaluate $\int_{- 1}^{0}$ (x + 1)(x - 2) dx

7/6

5/6

-5/6

-7/6

Answer Details

$\int_{- 1}^{0}$ (x + 1)(x - 2) dx

= $\int_{- 1}^{0}$ $x^{2} - x - 2$ dx

Integrated $x^{2} - x - 2$ = $\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2$

=

(0

-

0

-

0)

-

(

-13

-

12

+

2)

=

0

-

(

76

)

=

-76

View Answer

Question 17 Report

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

16m

26m

36m

46m

Answer Details

from the equation of motion

u = 5m/s, a = 2m/s $^{2}$ , t = 4s

s = ut + $\frac{1}{2} a t^{2}$

s = 5*4 + $\frac{1}{2} 2 * 4^{2}$

s = 20 + 16

s = 36m

View Answer

Question 18 Report

Simplify $\frac{9 * 3^{n + 1} - 3^{n + 2}}{3^{n + 1} - 3^{n}}$

3

9

27

81

Answer Details

$\frac{9 * 3^{n + 1} - 3^{n + 2}}{3^{n + 1} - 3^{n}}$

= $\frac{3^{n} * 3^{n} * 3^{1} * - 3^{2} * 3^{2}}{3^{n} * 3^{1} - 3^{n}}$

= $\frac{3^{n} (3^{2} * 3^{1})}{3^{n} (3^{1} - 1)}$

= $\frac{27 - 9}{3 - 1}$

= $\frac{18}{2}$

= 9

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Functions

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Question 19 Report

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.

1.63m

1.54m

1.52m

1.42m

Statistics

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Question 20 Report

Given

| \begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix} | | \begin{matrix} - 6 \\ k \end{matrix} | | \begin{matrix} 3 \\ - 26 \end{matrix} | = 15

Solve for k.

-8

-5

-4

-3

Answer Details

$| \begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix} | | \begin{matrix} - 6 \\ k \end{matrix} | | \begin{matrix} 3 \\ - 26 \end{matrix} | = 15$

$| \begin{matrix} 2 [- 6] & - 3 k \\ 1 [- 6] & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

$| \begin{matrix} - 12 & - 3 k \\ - 6 & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

-12 - 3k = 3

-3k = 3 + 12

k = $\frac{15}{- 3}$

k = -5

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Surds

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Question 21 Report

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

-80

10

40

80

Answer Details

x $^{3}$ y $^{2}$ in (x-2y) $^{5}$

n = 5, r = 3, p = x, q = -2y

5C $_{3}$ * x $^{3}$ $^{2}$

5C $_{3}$ = $\frac{5!}{[5 - 3]! 3!}$

$\frac{5 * 4 * 3!}{2! 3!}$ → $\frac{5 * 4}{2}$

5C $_{3}$ = 10

: 5C $_{3}$ * x $^{3}$ -2y $^{2}$ = 10 * x $^{3}$ 4y $^{2}$

40x $^{3}$ y $^{2}$
the coefficient is 40

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Binomial Theorem

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Question 22 Report

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.

{20, 21, 25, 30, 33}

{21, 25, 27, 33, 35}

{20, 21, 25, 27, 33, 35}

{21, 25, 27, 30, 33, 35}

Sets

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Question 23 Report

A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.

\frac{1}{2}

m/s

\frac{1}{3}

m/s

2m/s

3m/s

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 = $\frac{1}{2}$ m/s

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Matrices And Linear Transformation

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Question 24 Report

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

x < -1 or x > 1

x ≤ -1 or x ≥1

-1 ≤ x ≤ 1

-1 < x < 1

Answer Details

1 - x $^{2}$ ≥ 0
-x $^{2}$ ≥ -1
x $^{2}$ ≤ 1
√x $^{2}$ ≤ 1
|x| ≤ 1
-1 ≤ x ≤ 1

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Surds

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Question 25 Report

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

equation

whose

roots

are

1α

and

1β

.

mnx

^{2}

- n

^{2}

x - m = 0

mx

^{2}

- nx + 1 = 0

nx

^{2}

- mx + 1 = 0

nx

^{2}

- mx - 1 = 0

Answer Details

x $^{2}$ + mx - n = 0

a = 1, b = m, c = -n

α + β = $\frac{- b}{a}$ = $\frac{- m}{1}$ = -m

αβ = $\frac{c}{a}$ = $\frac{- n}{1}$ = -n

the roots are = $\frac{1}{α}$ and $\frac{1}{β}$

sum of the roots = $\frac{1}{α}$ + $\frac{1}{β}$

$\frac{1}{α}$ + $\frac{1}{β}$ $\frac{α + β}{α β}$

α + β = -m
αβ = -n

$\frac{α + β}{α β}$

product of the roots = $\frac{1}{α}$ * $\frac{1}{β}$

$\frac{1}{α}$ + $\frac{1}{β}$ = $\frac{1}{α β}$ → $\frac{1}{- n}$

x $^{2}$ - (sum of roots)x + (product of roots)
x $^{2}$ - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx $^{2}$ - mx - 1 = 0

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Polynomial Functions

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Question 26 Report

If $x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$ , evaluate dy/dx when x=3 and y=2.

2

-2

-4

4

Answer Details

$x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$

When differentiated:

$x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$

where x=3 and y=2

2[3] + 2[2] - 8 = 0

6 + 4 - 8 = 2

View Answer

Question 27 Report

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.

\frac{- 1}{2}, 8

\frac{1}{2}, - 8

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

\frac{1}{2}, 8

Answer Details

U1 = x - 4
U2 = x + 2
U3 = 3x + 1

$\frac{u_{2}}{u_{1}} = \frac{u_{3}}{u_{2}}$

$\frac{x + 2}{x - 4} = \frac{3 x + 1}{x + 2}$

(x+2)(x+2) = (x-4)(3x+1)
x $^{2}$ + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x $^{2}$ - 15x - 8 =0
2x $^{2}$ + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0

x = ( $\frac{- 1}{2}, 8$ )

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Polynomial Functions

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Question 28 Report

If Un = kn $^{2}$ + pn, U $_{1}$ = -1, U $_{5}$ = 15, find the values of k and p.

k = -1, p = 2

k = -1, p = -2

k = 1, p = -2

k = 1, p = 2

Answer Details

Un = kn $^{2}$ + pn,

U $_{1}$ = -1,

U $_{5}$ = 15,

when n = 1
U1 = k(1) $^{2}$ + p(1) = -1
k + p = -1 --------eqn1
when n = 5
U $_{5}$ = k(5) $^{2}$ + 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

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Polynomial Functions

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Question 29 Report

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

2x + 3y + 6 = 0

3x - 2y - 6 = 0

-3x 2y - 6 = 0

-2x + 3y - 6 = 0

Co-ordinate Geometry

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Question 30 Report

If →PQ = -2i + 5j and →RQ = -i - 7j, find →PR

-3i + 12j

-3i - 12j

-i + 12j

i - 12j

Vectors

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Question 31 Report

The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.

4√3

2√6

3√2

2√3

Surds

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Question 32 Report

Evaluate $4 p_{2} + 4 C_{2} - 4 p_{3}$

18

6

-6

-18

Answer Details

$4 p_{2} + 4 C_{2} - 4 p_{3}$

$n p_{r} = \frac{n!}{[n - r]!} a n d n C_{r} = \frac{n!}{[n - r]! r!}$

= $\frac{4!}{[4 - 2]!} + \frac{4!}{[4 - 2]! 2!} - \frac{4!}{[4 - 3]!} = \frac{4!}{2!} + \frac{4!}{2! 2!} - \frac{4!}{1!}$

= $\frac{4 * 3 * 2!}{2!} + \frac{4 * 3 * 2!}{2! 2!} - \frac{4 * 3 * 2 * 1}{1!}$

12 + 6 - 24 = -6

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Sets

Permutation And Combinations

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Question 33 Report

Express $\frac{4 π}{2}$ radians in degrees.

288º

200º

144º

120º

Answer Details

$\frac{4 π}{2}$ = $\frac{4}{5} * 180$

= 144º

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Trigonometry

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Question 34 Report

If 36, p, $\frac{9}{4}$ and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.

9/16

81/16

9

\frac{9}{16}

Answer Details

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

Recall,

common

ratio,

r

=

TnTn-1

=

T2T1

=

T3T2

=

T4T3

∴

P36

=

94

÷

p

;

p

^{2}

=

94

x

36

;

p

^{2}

=

81

p

=

9

∴

r

=

T2T1

=

936

=

14

Also

r

=

T4T3

=

q

÷

94

∴ $\frac{1}{4}$ = q ÷ $\frac{9}{4}$ ;

$\frac{9}{4}$ = 4q

16q

=

9

,

q

=

916

∴

p

+

q

=

9

+

916

=

9

916

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Polynomial Functions

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Question 35 Report

Differentiate $\frac{5 x^{3} + x^{2}}{x}$ , x ≠ 0 with respect to x.

10x + 1

10x + 2

x(15x + 1)

x(15x + 2)

Answer Details

$\frac{5 x^{3} + x^{2}}{x}$ → $\frac{5 x^{3}}{x} + \frac{x^{2}}{x}$

5x $^{2}$ + x

Then dy/dx = 10x + 1

View Answer

Question 36 Report

Solve: 4sin $^{2}$ θ + 1 = 2, where 0º < θ < 180º

60º 0r 120º

30º 0r 150º

30º 0r 120º

60º 0r 150º

Answer Details

4sin $^{2}$ θ + 1 = 2

4sin $^{2}$ θ = 2 - 1

4sin $^{2}$ θ = 1

$\sqrt{s} i n^{2} θ$ = $\sqrt{\frac{1}{4}}$

sinθ = $\frac{1}{2}$

θ = $s i n^{- 1} \frac{1}{2}$

θ = 30º 0r 150º

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Trigonometry

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Question 37 Report

Which of the following is the semi-interquartile range of a distribution?

Mode - Median

Highest score - Lowest score

1/2 (Upper Quartile - Median)

1/2 (Upper Quartile - Lower Quartile)

Statistics

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Question 38 Report

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

(

\frac{- 1}{4}

,

\frac{3}{4}

)

(

\frac{1}{4}

,

\frac{3}{4}

(

\frac{- 1}{2}

,

\frac{3}{2}

)

. (

\frac{- 1}{2}

,

\frac{- 3}{2}

)

Answer Details

2x $^{2}$ + 2y $^{2}$ - x - 3y - 41

standard equation of circle
(x-a) $^{2}$ + (x-b) $^{2}$ = r $^{2}$
General form of equation of a circle.
x $^{2}$ + y $^{2}$ + 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.
2x $^{2}$ + 2y $^{2}$ - x - 3y - 41

= x $^{2}$ + y $^{2}$ + 2gx + 2fy + c
2x $^{2}$ + 2y $^{2}$ - x - 3y - 41 = 0
divide through by 2

g = $\frac{- 1}{4}$ ; 2g = $\frac{- 1}{2}$

f = $\frac{- 3}{4}$ ; 2f = $\frac{- 3}{2}$

a = -g → - $\frac{- 1}{4}$ ; = $\frac{1}{4}$

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is ( $\frac{1}{4}$ , $\frac{3}{4}$ )

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Co-ordinate Geometry

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Question 39 Report

If $l o g_{10} (3 x + 1) + l o g_{10} 4 = l o g_{10} (9 x + 2)$ , find the value of x

\frac{1}{3}

1

2

3

Answer Details

Explanation

$l o g_{10} (3 x + 1) + l o g_{10} 4 = l o g_{10} (9 x + 2)$

$l o g_{10} 4 (3 x + 1) = l o g_{10} (9 x + 2)$

4(3x+1) = 9x + 2

12x -4 = 9x + 2

12x - 9x = 2 + 4

3x = 6

x = 2

View Answer

Question 40 Report

In how many ways can six persons be paired?

5

10

15

20

Answer Details

6C $_{2} = \frac{6!}{[6 - 2]! [2!]}$

$\frac{6 * 5 * 4!}{4! * 2!}$

= $\frac{6 * 5}{2}$

= 15

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Permutation And Combinations

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Question 41 Report

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 = t $^{2}$ + 3t. Find the magnitude of F at time t.

0N

2N

3(2t + 3)N

6N

Answer Details

F = m * a

d = t $^{2}$ + 3t.

a = $\frac{d^{2} d}{d t^{2}}$

$\frac{d [d]}{d t}$ = 2t + 3

$\frac{d^{2} d}{d t^{2}}$ = 2m/s $^{2}$

a = 2m/s $^{2}$

F = m * a

F = 3 × 2 = 6N

View Answer

Question 42 Report

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

(7, -2)

(5, -2)

(-2, 7)

(-7, 2)

Matrices And Linear Transformation

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Question 43 Report

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

Distance(km)	3	4	5	6	7	8
Frequency	5	4	x	9	2x	1

If a hunter is selected at random, find the probability that the hunter covered at least 6km.

\frac{3}{5}

\frac{2}{5}

\frac{3}{8}

\frac{9}{40}

Probability

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Question 44 Report

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

61

25

20

13

Answer Details

x - 1 = 2
x = 3
f(2) = (3) $^{3}$ + 3(3) $^{2}$ + 4(3) - 5
f(2) = 27 + 27 + 12 - 5

= 61

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Functions

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Question 45 Report

Solve 3cos2x - sinx = 0 for 0º≤x≤360º

Answer

3cos2x - sinx = 0
cos 2A = 1 – 2 sin $^{2}$ A
3(1 - 2sin $^{2}$ x) - sinx = 0
3 - 6sin $^{2}$ x -sinx = 0
let sinx = p
3 - 6p $^{2}$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.6287)
x = 38.95º

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Answer Details

3cos2x - sinx = 0
cos 2A = 1 – 2 sin $^{2}$ A
3(1 - 2sin $^{2}$ x) - sinx = 0
3 - 6sin $^{2}$ x -sinx = 0
let sinx = p
3 - 6p $^{2}$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.6287)
x = 38.95º

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Trigonometry

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Question 46 Report

A body of mass of 18kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35º. If the system is in equilibrium, calculate the:

i. value of F

ii. tension in the string

Answer

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = $\frac{180}{0.8192}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

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Answer Details

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = $\frac{180}{0.8192}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

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Statics

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Question 47 Report

The vectors 6i + 8j and 8i - 6j are parallel to ?OP and ?OQ respectively. If the magnitude of ?OP and ?OQ are 80 units and 120 units respectively, express: ?OP and ?OQ in terms of i and j;

ii. |?PQ|, in the form c?k, where c and k are constants.

Answer

Vectors

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Question 48 Report

A particle initially at rest moves in a straight line with an acceleration of (10t - 4t $^{2}$ )m/s $^{2}$
Find the:
a. velocity of the particle after t seconds;

ii. average acceleration of the particle during the 4th second.

b. A load of mass 120kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards at a constant velocity. [Take g = 10m/s $^{2}$ ]

ii. with an acceleration of 3m/s $^{2}$ . [Take g = 10m/s $^{2}$ ]

Answer

a. a = (10t - 4t $^{2}$ )m/s $^{2}$ , t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t $^{2}$ )t
v = (10t $^{2}$ - 4t $^{3}$ )m/s $^{2}$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3) $^{2}$
Δa4 = 10(4-1) - 4(4-3) $^{2}$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s $^{2}$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s $^{2}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

Answer Details

a. a = (10t - 4t $^{2}$ )m/s $^{2}$ , t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t $^{2}$ )t
v = (10t $^{2}$ - 4t $^{3}$ )m/s $^{2}$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3) $^{2}$
Δa4 = 10(4-1) - 4(4-3) $^{2}$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s $^{2}$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s $^{2}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

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Question 49 Report

Two functions f and g are defined on the set of real numbers, R, by

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .Find the domain of (g∘f) $^{- 1}$

Answer

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .

(g∘f) $^{- 1}$

(g∘f) $^{x^{2} + 2}$

g $^{x^{2} + 2}$ = $\frac{1}{(x^{2} + 2) + 2}$

(g∘f) = $\frac{1}{(x^{2} + 4}$

let y = $\frac{1}{(x^{2} + 4}$

y $((x^{2} + 4)$ = 1

$y x^{2} + 4 y = 1$

x $^{2}$ = $\frac{1 - 4 y}{y}$

x = $\sqrt{\frac{1 - 4 y}{y}}$

(g∘f) $^{- 1}$ = $\sqrt{\frac{1 - 4 x}{x}}$

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Answer Details

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .

(g∘f) $^{- 1}$

(g∘f) $^{x^{2} + 2}$

g $^{x^{2} + 2}$ = $\frac{1}{(x^{2} + 2) + 2}$

(g∘f) = $\frac{1}{(x^{2} + 4}$

let y = $\frac{1}{(x^{2} + 4}$

y $((x^{2} + 4)$ = 1

$y x^{2} + 4 y = 1$

x $^{2}$ = $\frac{1 - 4 y}{y}$

x = $\sqrt{\frac{1 - 4 y}{y}}$

(g∘f) $^{- 1}$ = $\sqrt{\frac{1 - 4 x}{x}}$

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Sets

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Question 50 Report

The probability that Abiola will be late to the office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will:

(a) only be late for 3 days.

(b) not be late in the week:

(c) be late throughout the six days.

Answer

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

$\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0467

(c) be late to office = 2/5

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$

p = 0.0041

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Answer Details

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

$\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0467

(c) be late to office = 2/5

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$

p = 0.0041

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Probability

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Question 51 Report

Given that p = (8N,030º) and q = (9N, 150º), find, in component from, the unit vector along(p - q).

Answer

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = $\frac{- 14.7224 i + 0.5 j}{14.7309}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

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Answer Details

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = $\frac{- 14.7224 i + 0.5 j}{14.7309}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

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Vectors

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Question 52 Report

Given that nC $_{4}$ , nC $_{5}$ and nC $_{6}$ are the terms of a linear sequence (A.P), find the :

i. value of n

ii. common differences of the sequence.

Answer

nC $_{4}$ = $\frac{n!}{[n - 4]! 4!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4]!}{[n - 4]! 4!}$

nC $_{5}$ = $\frac{n!}{[n - 5]! 5!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5]!}{[n - 5]! 5!}$

and nC $_{6}$ = $\frac{n!}{[n - 6]! 6!}$ → $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5] [n - 6]!}{[n - 6]! 6!]}$

d = U2 - U1 = U3 - U2

nC $_{5}$ - nC $_{4}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$

nC $_{6}$ - nC $_{5}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

nC $_{5}$ - nC $_{4}$ = nC $_{6}$ - nC $_{5}$

$\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

divide both sides by n(n-1)(n-2)(n-3)

$\frac{n - 9}{5!}$ = $\frac{[n - 4] [n - 11]}{5! 6}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n $^{2}$ -11n - 4n + 44
6n - 54 = n $^{2}$ - 15n + 44
n $^{2}$ - 21n + 98 = 0
n $^{2}$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C $_{5}$ - 7C $_{4}$

d = $\frac{7 * 6 * 5!}{2! * 5!}$ - $\frac{7 * 6 * 5 * 4!}{3! * 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C $_{5}$ - 14C $_{4}$

d = $\frac{14 * 13 * 12 * 11 * 10 * 9!}{9! * 5!}$ - $\frac{14 * 13 * 12 * 11 * 10!}{10! * 4!}$

d = 2002 - 1001
d = 1001

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Answer Details

nC $_{4}$ = $\frac{n!}{[n - 4]! 4!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4]!}{[n - 4]! 4!}$

nC $_{5}$ = $\frac{n!}{[n - 5]! 5!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5]!}{[n - 5]! 5!}$

and nC $_{6}$ = $\frac{n!}{[n - 6]! 6!}$ → $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5] [n - 6]!}{[n - 6]! 6!]}$

d = U2 - U1 = U3 - U2

nC $_{5}$ - nC $_{4}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$

nC $_{6}$ - nC $_{5}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

nC $_{5}$ - nC $_{4}$ = nC $_{6}$ - nC $_{5}$

$\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

divide both sides by n(n-1)(n-2)(n-3)

$\frac{n - 9}{5!}$ = $\frac{[n - 4] [n - 11]}{5! 6}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n $^{2}$ -11n - 4n + 44
6n - 54 = n $^{2}$ - 15n + 44
n $^{2}$ - 21n + 98 = 0
n $^{2}$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C $_{5}$ - 7C $_{4}$

d = $\frac{7 * 6 * 5!}{2! * 5!}$ - $\frac{7 * 6 * 5 * 4!}{3! * 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C $_{5}$ - 14C $_{4}$

d = $\frac{14 * 13 * 12 * 11 * 10 * 9!}{9! * 5!}$ - $\frac{14 * 13 * 12 * 11 * 10!}{10! * 4!}$

d = 2002 - 1001
d = 1001

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Question 53 Report

Solve $2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Answer

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Let p = 2 $^{y}$
$2^{2 y} (2^{1}) - 5 (2^{y})$ + 2 = 0
2p $^{2}$ - 5p + 2 = 0
2p $^{2}$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or $\frac{1}{2}$

p = 2 $^{y}$
when p = 2
2 $^{y}$ = 2
y = 1

when p = $\frac{1}{2}$

2 $^{y}$ = $\frac{1}{2}$

2 $^{y}$ = 2 $^{- 1}$

y = -1

y = -1 or 1

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Answer Details

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Let p = 2 $^{y}$
$2^{2 y} (2^{1}) - 5 (2^{y})$ + 2 = 0
2p $^{2}$ - 5p + 2 = 0
2p $^{2}$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or $\frac{1}{2}$

p = 2 $^{y}$
when p = 2
2 $^{y}$ = 2
y = 1

when p = $\frac{1}{2}$

2 $^{y}$ = $\frac{1}{2}$

2 $^{y}$ = 2 $^{- 1}$

y = -1

y = -1 or 1

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Question 54 Report

A solid rectangular block has a base that measures 3x cm by 2x cm. The height of the block is ycm and its volume is 72cm $^{3}$ .

i. Express y in terms of x.

ii. An expression for the total surface area of the block in terms of x only;

iii. the value of x for which the total surface area has a stationary value.

Answer

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

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Answer Details

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

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Question 55 Report

The table shows the scores obtained by a group of artistes in Vocal (X) and the instrument (Y) musical competition.

Vocal (X)	63	69	72	59	82	91	95	68
Instrument (Y)	58	61	67	51	53	79	92	57

Calculate the spearman's rank correlation coefficient between the scores.

Answer

p = $1 - \frac{6 Σ (d_{i})^{2}}{n (n^{2} - 1}$

p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation

X	R $_{x}$	Y	R $_{y}$	D $_{I}$ (R $_{x}$ - R $_{y}$ )	D $_{i}$ $^{2}$
63	2	58	4	-2	4
69	4	61	5	-1	1
72	5	67	6	-1	1
59	1	51	1	0	0
82	6	53	2	4	16
91	7	79	7	0	0
95	8	92	8	0	0
68	3	57	3	0	0

P = $1 - \frac{6 (4 + 1 + 1 + 0 + 16 + 0 + 0 + 0)}{8 (8^{2} - 1)}$

p = $1 - \frac{6 (22)}{8 (63)}$

p = $1 - \frac{132}{504}$

p = $1 - \frac{11}{42}$

p = $\frac{11}{42}$

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Answer Details

p = $1 - \frac{6 Σ (d_{i})^{2}}{n (n^{2} - 1}$

p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation

X	R $_{x}$	Y	R $_{y}$	D $_{I}$ (R $_{x}$ - R $_{y}$ )	D $_{i}$ $^{2}$
63	2	58	4	-2	4
69	4	61	5	-1	1
72	5	67	6	-1	1
59	1	51	1	0	0
82	6	53	2	4	16
91	7	79	7	0	0
95	8	92	8	0	0
68	3	57	3	0	0

P = $1 - \frac{6 (4 + 1 + 1 + 0 + 16 + 0 + 0 + 0)}{8 (8^{2} - 1)}$

p = $1 - \frac{6 (22)}{8 (63)}$

p = $1 - \frac{132}{504}$

p = $1 - \frac{11}{42}$

p = $\frac{11}{42}$

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Statistics

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Question 56 Report

Evaluate: lim $_{x \to - 2}$ $\frac{x^{3} + 8}{x + 2}$ .

Answer

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

Read lesson note on Differentiation (WAEC)

Answer Details

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

Read lesson note on Differentiation (WAEC)

Differentiation

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Question 57 Report

A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers.

Two fruits are drawn one after the other without replacement. Calculate the probability that:

i. the first is an orange and the second is an avocado pear.

ii.both are of same fruit;

iii. at least one is an apple

Answer

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = $\frac{3}{12}$ = $\frac{1}{4}$

probability that apple is drawn = $\frac{4}{12}$ = $\frac{1}{3}$

probability that avocado pear is drawn = $\frac{5}{12}$

probability that the first is orange and the second is an avocado pear = $\frac{3}{12}$ * $\frac{5}{11}$ = $\frac{5}{44}$

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = $\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}$

= $\frac{1}{22} + \frac{1}{11} + \frac{5}{33}$

= $\frac{19}{66}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= $\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11}$

= $\frac{12}{132}$ + $\frac{12}{132}$ + $\frac{20}{132}$ + $\frac{20}{132}$ + $\frac{12}{132}$

= $\frac{76}{132}$

= $\frac{19}{33}$

Read lesson note on Probability (WAEC)

Answer Details

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = $\frac{3}{12}$ = $\frac{1}{4}$

probability that apple is drawn = $\frac{4}{12}$ = $\frac{1}{3}$

probability that avocado pear is drawn = $\frac{5}{12}$

probability that the first is orange and the second is an avocado pear = $\frac{3}{12}$ * $\frac{5}{11}$ = $\frac{5}{44}$

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = $\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}$

= $\frac{1}{22} + \frac{1}{11} + \frac{5}{33}$

= $\frac{19}{66}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= $\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11}$

= $\frac{12}{132}$ + $\frac{12}{132}$ + $\frac{20}{132}$ + $\frac{20}{132}$ + $\frac{12}{132}$

= $\frac{76}{132}$

= $\frac{19}{33}$

Read lesson note on Probability (WAEC)

Probability

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Question 58 Report

A binary operation * is defined on the set T = {-2,-1,1,2} by p*q = p $^{2}$ + 2pq - q $^{2}$ , where p,q ∊ T.
Copy and complete the table.

*	-2	-1	1	2
-2		7		-8
-1		2	-2
1	-7			1
2		-1

Answer

p*q = p $^{2}$ + 2pq - q $^{2}$

when p = -2 and q = -2

p*q = -2 $^{2}$ + 2(-2)(-2) - (-2) $^{2}$
p*q = 4 + 8 - 4 = 8

when p = -2 and q = -1
p*q = -2 $^{2}$ + 2(-2)(-1) - (-1) $^{2}$
p*q = 4 + 4 - 1 = 7

when p = -2 and q = 1
p*q = -2 $^{2}$ + 2(-2)(1) - (1) $^{2}$
p*q = 4 - 4 -1 = -1

when p = -1 and q = -2
p*q = -1 $^{2}$ + 2(-1)(-2) - (-2) $^{2}$
p*q = 1 + 4 - 4 = 1

when p = -1 and q = -1
p*q = -1 $^{2}$ + 2(-1)(-1) - (-1) $^{2}$
p*q = 1 + 2 -1 = 2

when p = -1 and q = 2
p*q = -1 $^{2}$ + 2(-1)(2) - (2) $^{2}$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -2
p*q = 1 $^{2}$ + 2(1)(-2) - (-2) $^{2}$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -1
p*q = 1 $^{2}$ + 2(1)(-1) - (-1) $^{2}$
p*q = 1 - 2 - 1 = -2

when p = 1 and q = 1
p*q = 1 $^{2}$ + 2(1)(1) - (1) $^{2}$
p*q = 1 + 2 - 1 = 2

when p = 1 and q = 2
p*q = 1 $^{2}$ + 2(1)(2) - (2) $^{2}$
p*q = 1 + 4 - 4 = 1

when p = 2 and q = -2
p*q = 2 $^{2}$ + 2(2)(-2) - (-2) $^{2}$
p*q = 4 - 8 - 4 = -8

when p = 2 and q = -1
p*q = 2 $^{2}$ + 2(2)(-1) - (-1) $^{2}$
p*q = 4 - 4 - 1 = -1

when p = 2 and q = 1
p*q = 2 $^{2}$ + 2(2)(1) - (1) $^{2}$
p*q = 4 + 4 - 1 = 7
when p = 2 and q = 2
p*q = 2 $^{2}$ + 2(2)(2) - (2) $^{2}$
p*q = 4 + 8 - 4 = 8

*	-2	-1	1	2
-2	8	7	-1	-8
-1	1	2	-2	-7
1	-7	-2	2	1
2	-8	-1	7	8