1994 - JAMB Mathematics Past Questions and Answers - page 2

a²x - b²y - b²x + a²y = a²x - b²x - b²y + a²y Rearrange

= x(a² - b²) + y(a² - b²)

= (x + y)(a² - b²)

= (x + y)(a + b)(a + b)

Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero

∴ (x - 1) = 0

x = 1

Substitute in the polynomial the value x = 1

= p(1)³ + q(1)² + 11(1) - 6 = 0

p + q + 5 = 0 .....(i)

Also since x - 3 is a factor, ∴ x - 3 = 0

x = 3

Substitute p(3)³ + q(3)² + 11(3) - 6 = 0

27p + 9q = -27 ......(2)

Combine eqns. (i) and (ii)

Multiply equation (i) by 9 to eliminate q

9p + 9q = -45

Subt. (\frac{27p + 9q = -27}{-18p = -18})

∴ p = 1

(\frac{a}{a - x}) = (\frac{b}{x - b})

(\frac{1}{1 - x}) = (\frac{3}{x - 3})

∴ 3(1 - x) = x - 3

3 - 3x = x - 3

Rearrange 6 = 4x; x = (\frac {6}{4})

= (\frac{3}{2})

(\frac{1}{r - 1}) + (\frac{2}{r + 1}) = (\frac{3}{r})

Multiply through by r(r -1) which is the LCM

= (r)(r + 1) + 2(r)(r - 1)

= 3(r - 1)(r + 1)

= r² + r + 2r² - 2r

3r² - 3 = 3r²

r = 3r² - 3

-r = -3

∴ r = 3

(\frac{x - 3}{(1 - x)(x + 2)}) = (\frac{p}{1 - x}) + (\frac{Q}{x + 2})

Multiply both sides by LCM i.e. (1 - x(x + 2))

∴ x - 3 = p(x + 2) + Q(1 - x)

When x = +1

(+1) - 3 = p(+1 + 2) + Q(1 - 1)

-2 = 3p + 0(Q)

3p = -2

∴ p = (\frac{-2}{3})

(\frac{1}{x}) > 2 = (\frac{x}{x^2}) > 2

x > 2x²

= 2x² < x

= 2x² - x < 0

= x(2x - 10 < 0

Case 1(+, -) = x > 0, 2x - 1 < 0

x > 0, x < (\frac{1}{2}) (solution)

Case 2(-, 4) = x < 0, 2x - 1 > 0

x < 0, x , (\frac{1}{2}) = 0

In an AP, Tn = a + (n - 1)d

T₆ = a + 5d = 11

The first term = a = 1

∴ T₆ = 1 + 5d = 11

5d = 11 - 1

5d = 10

∴ d = 2

log₁₀ r + log₁₀ r² + log₁₀ r⁴ + log₁₀ r⁸ + log₁₀ r¹⁶ + log₁₀ r³² = 63

log₁₀r⁶³ = 63

⁶³ = 10⁶³

∴ r = 10

(\frac{(n + 1)(n + 2)}{2})

If n = 1, the expression becomes 3

n = 2, the expression becomes 6

n = 4, the expression becomes 15

n = 5, the expression becomes 21

a (\oplus) b = ab

The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a (\oplus) b = ab; b (\oplus) a = ba = ab

The number 1 is the identity element under multiplication