2009 - JAMB Mathematics Past Questions and Answers - page 2

S∩T∩W = S(This means that all the elements of s are in T and also in W.)

S∪T∪W = W (imply W is a universal set for S and T) T∩W = S imply all the elements of S are in T and W)

S ⊂ T ⊂ W

∴ I, II, and III

(P=\sqrt{\frac{rs^3}{t}}=

P^2 =\frac{rs^3}{t}\

tp^2 = rs^3\

r = \frac{p^2 t}{s^3})

If ⁴/₃ and -³/₅ are roots of a polynomial

Imply x = ⁴/₃ and - ³/₅

3x = 4 and 5x = -3

∴3x-4 = 0 and 5x+3 = 0 are factors

(3x-4)(5x+3) = 0 product of the factors

15x² + 9x – 20x – 12 = 0 By expansion

15x² - 11x – 12 = 0

x = -2 and x = ¹/₄

x = -2 and 4x = 1

x+2 and 4x-1

(x+2)(4x-1) = 0

4x² - x + 8x -2 = 0

4x² + 7x – 2 = 0 but y intercept is positive. Multiply the equation by -1

-4x² - 7x + 2 = 0

∴y = 2 – 7x – 4x²

(W ∝ U\

W = KU\

K = \frac{W}{U}\

K = \frac{5}{3}\

W = \frac{5}{3}U\

\frac{2}{7} = \frac{5}{3}U\

U = \frac{2}{7} \times \frac{3}{5}\

U = \frac{6}{35})

x² > 0

(x+1)(x-1) > 0

x > -1 or -1

x < -1 or x > 1

3x - 7 (\leq) 0 and x + 5 > 0

3x (\leq) 7 and x > -5

x (\leq) (\frac{7}{3})

∴ Range -5 < x (\leq) (\frac{7}{3})

a = 5, d = 6, n = n

Sn = ⁿ/₂(2a + (n-1)d)

= ⁿ/₂(2(5) + (n-1)6)

= ⁿ/₂(10 + 6n-6)

= ⁿ/₂(6n+4)

= ⁶ⁿ²/₂ + ⁴ⁿ/₂

= 3² + 2n

= n(3n + 2)

(a=1, r=\frac{9}{10}\

S_n = \frac{a}{1-r}\

S_n = \frac{1}{1-\frac{9}{10}}\

=1\div \frac{1}{10}\

=1\times \frac{10}{1}\

10)

m * n = n - (m+2)

= -5 - (3+2)

= -5-5

= -10