2021

If (x + 2) and (x – 1) are factors of the expression Lx+2kx2, find the values of L and k.
A. l = -12, k = -6
B. l = -2 , k = 1
C. l = -2 , k = -1
D. l = 0, k = 1

Correct Option: Answer is A
Solution

Given (x + 2) and (x – 1), i.e. x = -2 or +1
when x = -2
L(-2) + 2k(-2)2 + 24 = 0
f(-2) = -2L + 8k = -24…(i)

And x = 1
L(1) + 2k(1) + 24 = 0

f(1):L + 2k = -24…(ii)
Subst, L = -24 – 2k in eqn (i)
-2(-24 – 2k) + 8k = -24
+48 + 4k + 8k = -24
12k = -24 – 48 = -72
k = frac−7212frac−7212
k = -6
where L = -24 – 2k
L = -24 – 2(-6)
L = -24 + 12
L = -12
That is; K = -6 and L = -12

A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group?
A. 25
B. 19
C. 18
D. 17

Correct Option: Answer is A
Solution

Let the three items be M, Y and P.
n{M ∩ Y} only = 4-3 = 1
n{M ∩ P) only = 5-3 = 2
n{ Y ∩ P} only = 2
n{M} only = 12-(1+3+2) = 6
n{Y} only = 10-(1+2+3) = 4
n{P} only = 14-(2+3+2) = 7
n{M∩P∩Y} = 3
Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.

X is due east point of y on a coast. Z is another point on the coast but 6.0km due south of Y. If the distance ZX is 12km, calculate the bearing of Z from X
A. 240°
B. 150°
C. 60°
D. 270°

Correct Option: Answer is A
Solution

Sinθ = 6/12
Sinθ = 1/2
θ = Sin0.5
θ = 30°
Bearing of Z from X, (270 – 30)° = 240°

List all integers satisfying the inequality in -2 < 2x-6 < 4 A. 2,3,4 and 5 B. 2,3 C. 2,5 D. 3,4 Correct Option: Answer is D Solution -2 < 2x - 6 AND 2x - 6 < 4 -2 + 6 <2x AND 2x < 4 + 6 4 <2x AND 2x < 10 : 2

Find x if log9x = 1.5
A. 27
B. 15
C. 3.5
D. 32
Correct Option: Answer is A
Solution

if log9x = 1.5 = 91.5 = x
93/2= x =>(√9)3 = 3
X= 27

Factorize completely 81a4 – 16b4
A. (3a + 2b)(2a – 3b)(9a2 + 4b2)
B. (3a – 2b)(2a – 3b)(4a2 – 9b2)
C. (3a – 2b)(3a + 2b)(9a2 + 4b2)
D. (6a – 2b)(8a – 3b)(4a3 – 9b2)

Correct Option: Answer is C
Solution

81a4 – 16b4 = (9a2)2 – (4b2)2

= (9a2 + 4b2) (9a2 – 4b2)

(9a2 – 4b2)= (3a – 2b)(3a + 2b)

If the binary operation ∗ is defined by m ∗ n = mn + m + n for any real number m and n, find the identity of the elements under this operation
A. e = 1
B. e = -1
C. e = -2
D. e = 0

Correct Option: Answer is D
Solution
Identity(e) : a ∗ e = a
m ∗ e = m…(i)
m ∗ e = me + m + e
Because m ∗ e = m
: m = me + m + e
m – m = e(m + 1)
e = 0/m+1
e = 0

What is the n-th term of the sequence 2, 6, 12, 20…?
A. 4n – 2
B. 2(3n – 1)
C. n2 + n
D. n2 + 3n + 2

Correct Option: Answer is C
Solution
Given that 2, 6, 12, 20…?
the nth term = n2 + n

check: n = 1, u1 = 2

n = 2, u2 = 4 + 2 = 6

n = 3, u3 = 9 + 3 = 12

∴ n = 4, u4 = 16 + 4 = 20

The angles of a quadrilateral are 5x-30, 4x+60, 60-x and 3x+61.find the smallest of these angles
A. 5x – 30
B. 4x + 60
C. 60 – x
D. 3x + 61

Correct Option: Answer is C
Solution

Sum of all 4 angles of a quadrilateral = 360°
(5x-30) + (3x + 60) + (60-x) + (4x+ 50) = 360°
(12x – x) + ( 170 – 30) = 360°
11x + 140 = 360°
11x = 360 – 140 = 220
x = 220/11 = 20°
Each angles is :
5x – 30 = 100 – 30 = 70°
3x+ 60 = 60 + 60 = 120°
60 – x = 60 – 20 = 40°
4x + 50 = 80 + 50 = 130°
Smallest of these angles is 40°

factorize m3 – m2 + 2m – 2
A. (m2 + 1)(m – 2)
B. (m – 1)(m + 1)(m + 2)
C. (m – 2)(m + 1)(m – 1)
D. (m2 + 2)(m – 1)

Correct Option: Answer is D
Solution

factorize m3 – m2 + 2m – 2

do the grouping m3 – m2 + 2m – 2 =( m3 – m2 )+(2m-2)
factor out m2 in the first and 2 in the second group

m2(m-1)+2(m-1) => factor out common term m-1
= (m2+2) (m-1)
=(m2 + 2)(m – 1)