WAEC SSCE - Further Mathematics - 2022

Vraag 1 Verslag

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

Bekijk antwoord

Vraag 2 Verslag

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}

Antwoorddetails

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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Vraag 3 Verslag

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

Antwoorddetails

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

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

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

-12 - 3k = 3

-3k = 3 + 12

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

k = -5

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Surds

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Vraag 4 Verslag

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

61

25

20

13

Antwoorddetails

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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Vraag 5 Verslag

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

Antwoorddetails

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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Vraag 6 Verslag

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

2

-2

-4

4

Antwoorddetails

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

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

Bekijk antwoord

Vraag 7 Verslag

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

288º

200º

144º

120º

Antwoorddetails

$\frac{4 π}{2}$ = $\frac{4}{5} * 180$

= 144º

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Trigonometry

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Vraag 8 Verslag

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

Bekijk antwoord

Vraag 9 Verslag

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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Vraag 10 Verslag

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

23

17

-17

18

Antwoorddetails

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

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

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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Vraag 11 Verslag

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

Antwoorddetails

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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Vraag 12 Verslag

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

Antwoorddetails

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

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

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

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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Vraag 13 Verslag

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

Antwoorddetails

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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Vraag 14 Verslag

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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Vraag 15 Verslag

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

Antwoorddetails

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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Vraag 16 Verslag

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

\frac{5 \sqrt 3}{6}

\frac{3 \sqrt 15}{6}

\frac{5 \sqrt 6}{12}

\frac{5 \sqrt 3}{12}

Antwoorddetails

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

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

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

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

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

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Surds

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Vraag 17 Verslag

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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Vraag 18 Verslag

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

18

6

-6

-18

Antwoorddetails

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

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

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

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

12 + 6 - 24 = -6

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Sets

Permutation And Combinations

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Vraag 19 Verslag

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

Bekijk antwoord

Vraag 20 Verslag

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

10

40

32

80

Antwoorddetails

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

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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Vraag 21 Verslag

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

7/6

5/6

-5/6

-7/6

Antwoorddetails

$\int_{- 1}^{0}$ (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

Bekijk antwoord

Vraag 22 Verslag

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

Antwoorddetails

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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Vraag 23 Verslag

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

60º 0r 120º

30º 0r 150º

30º 0r 120º

60º 0r 150º

Antwoorddetails

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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Vraag 24 Verslag

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

$\frac{56}{12}$

$\frac{65}{12}$

12

65

Antwoorddetails

$1_{0}^{\int} x^{2} (x^{3} + 2)^{3}$ 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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Vraag 25 Verslag

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

Antwoorddetails

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

Bekijk antwoord

Vraag 26 Verslag

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

Antwoorddetails

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

Bekijk antwoord

Vraag 27 Verslag

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}

Antwoorddetails

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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Vraag 28 Verslag

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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Vraag 29 Verslag

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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Vraag 30 Verslag

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}

)

Antwoorddetails

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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Vraag 31 Verslag

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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Vraag 32 Verslag

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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Vraag 33 Verslag

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

10x + 1

10x + 2

x(15x + 1)

x(15x + 2)

Antwoorddetails

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

5x $^{2}$ + x

Then dy/dx = 10x + 1

Bekijk antwoord

Vraag 34 Verslag

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

Antwoorddetails

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

Bekijk antwoord

Vraag 35 Verslag

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

-80

10

40

80

Antwoorddetails

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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Vraag 36 Verslag

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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Vraag 37 Verslag

In how many ways can six persons be paired?

5

10

15

20

Antwoorddetails

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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Vraag 38 Verslag

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

-3i + 12j

-3i - 12j

-i + 12j

i - 12j

Vectors

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Vraag 39 Verslag

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

Antwoorddetails

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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Vraag 40 Verslag

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

Bekijk antwoord

Vraag 41 Verslag

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

Antwoorddetails

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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Vraag 42 Verslag

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

Antwoorddetails

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 ≥ $\frac{3}{2}$

Bekijk antwoord

Vraag 43 Verslag

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

3

9

27

81

Antwoorddetails

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

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

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

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

= $\frac{18}{2}$

= 9

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Vraag 44 Verslag

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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Vraag 45 Verslag

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.

Antwoord

Vectors

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Vraag 46 Verslag

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

Antwoord

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

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 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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Vraag 47 Verslag

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

Antwoord

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) =

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

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

= $\frac{76}{132}$

= $\frac{19}{33}$

Lees lesnotitie op Probability (WAEC)

Antwoorddetails

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) =

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

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

= $\frac{76}{132}$

= $\frac{19}{33}$

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Vraag 48 Verslag

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

Antwoord

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

Lees lesnotitie op Statics (WAEC)

Antwoorddetails

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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Vraag 49 Verslag

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) $^{- 1}$

Antwoord

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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Antwoorddetails

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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Vraag 50 Verslag

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

Antwoord

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