2021 - JAMB Mathematics Past Questions and Answers - page 4

(m\(^2\) + 2) (m - 1)

Sum of the 4 angles of a quadrilateral = 360°

(5x-30) + (3x + 61) + (60-x) + (4x+ 60) = 360°

11x + 151 = 360°

11x = 360 - 151 = 209

x = 209/11 = 19°

Each angle:

5x - 30 =  65°

4x+ 60 = 136°

60 - x =41°

3x + 61 = 118°

The smallest of the angles is 41° 

Given the sequence 2, 6, 12, 20...

The nth term = n\(^2\) + n

Check: n = 1, u1 = 2

n = 2, u2 = 4 + 2 = 6

n = 3, u3 = 9 + 3 = 12

n = 4, u4 = 16 + 4 = 20

Identity(e) : a \(\ast\) e = a

m \(\ast\) e = m...(i)

m \(\ast\) e = me + m + e

m \(\ast\) e = m 

m = me + m + e

m - m = e(m + 1)

e = \(\frac{0}{m + 1}\)

e = 0

81a\(^4\) - 16b\(^4\) = (9a\(^2\))\(^2\) - (4b\(^2\))\(^2\)

= (9a\(^2\) + 4b\(^2\))(9a\(^2\) - 4b\(^2\))

9a\(^2\) - 4b\(^2\) = (3a - 2b)(3a + 2b)

Given log\(_9\)x = 1.5,

9\(^{1.5}\) = x

9^\(\frac{3}{2}\) = x

(√9)\(^3\) = 3
x = 27

-2 < 2x - 6   AND  2x - 6 < 4

-2 + 6 <2x  AND  2x < 4 + 6

4 <2x  AND  2x < 10

: 2 <x  AND x <5

2 < x < 5 

Hence,

3, 4

Sinθ = \(\frac{6}{12}\)

Sinθ = \(\frac{1}{2}\)

θ =  Sin\(^0.5\)
θ = 30°

Bearing of Z from X, (270 - 30)° = 240°

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

The number of women in the group = 6+4+7+(1+2+2+3) 

= 25

Given (x + 2) and (x - 1), i.e. x = -2 or +1

At x = -2

L(-2) + 2k(-2)\(^2\) + 24 = 0

f(-2) = -2L + 8k = -24...(i)

At x = 1

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

f(1):L + 2k = -24...(ii)

Substitute L = -24 - 2k in equations (i)

-2(-24 - 2k) + 8k = -24

+48 + 4k + 8k = -24

12k = -24 - 48 = -72

k = \(frac{-72}{12}\)

k = -6

Where L = -24 - 2k

L = -24 - 2(-6)

L = -24 + 12

L = -12

K = -6, L = -12