2008 - JAMB Mathematics Past Questions and Answers - page 2

n(M) only = 30-10 = 20

n(E) only = 20-10 = 10

n(M∩E) = 10

∴M∪E = 20+10+10

= 40

(L=\frac{4}{3}M\sqrt{PQ}\

=\frac{3}{4M} \times L = \sqrt{PQ}\

=\left(\frac{3L}{4M}\right)^2=(\sqrt{PQ})^2\

=\frac{9L^2}{16M^2}=PQ\

=Q=\frac{9L^2}{16M^2 P})

2x² - kx - 12 is divisible by x-4

implies x is a factor ∴ x = 4
f(4) implies 2(4)² - k(4) - 12 = 0

32 - 4k - 12 = 0

-4k + 20 = 0

-4k = -20

k = 5

(4x+3y)² - (3x-2y)²

(4x+3y+3x-2y)(4x+3y-(3x-2y))

(4x+3y+3x-2y)(4x+3y-3x+2y)

(x+5y)(7x+y)

(x – 3) ∝ y²

X-3 = Ky²

K = X-3 / y²

= ^5-2/_2²

= ²/₄

= ¹/₂

∴X-3 = ¹/₂y²

X-3 = ¹/₂(6)²

X-3 = ¹/₂ x ³⁰/₁

X-3 = 18

X = 21

P ∝ ¹/_q

P = ^k/_q

K = q²P

= 4²8

∴P = ¹²⁸/_q

32 = ¹²⁸/_q

q² = ¹²⁸/₃₂

q² = 4

q = √4 = +/-2

4X - 7 (\leq) 3X and 3X - 4 (\leq) 4X

4X - 3X (\leq) 7 and 3X - 4X (\leq) 4

X (\leq) 7 and -X (\leq) 4 = X (\geq) -4

Range -4 (\leq) x (\leq) 7

x² - 5x + 6 = 0

(X-2)(X-3) = 0

X-2 = 0 implies X = 2

X-3 = 0 implies X = 3

∴ x ≤ 2, x ≥ 3

U₅ = 24, n = 5 and U₁₁ = 96, n = 11

U_n = a + (n-1)d

24 = a + (5-1)d imply 24 = a+4d .....eqn1

96 = a + (11-1)d imply 96 = a+10d ...eqn2

eqn1 - eqn2 -72 = -6d

d = ⁷²/₆ = 12

but 24 = a+4d

24 = a + 4(12)

24 = a + 48

a = 24-48

a = -24

X * Y = 2X - 3Y + 2

2*3 = 2(2) - 3(3) + 2

=4-9+2

= -3

But -3 does not belong to positive integer