1992 - JAMB Mathematics Past Questions and Answers - page 2

Given S = ut + (\frac{1}{2} at^2)

S = ut + (\frac{1}{2} at^2)

∴ 2S = 2ut + at²

= at² + 2ut - 2s = 0

t = (\frac{-2u \pm 4u^2 + 2as}{2a})

= -2u (\pi) (\frac{\sqrt{u^2 4u^2 + 2as}}{2a})

= (\frac{1}{a}) (-u + (\sqrt{U^2 - 2as}))

x + 1 is a factor of x³ + 3x² + kx + 4

Let f(x) = x³ + 3x² + kx + 4

∴ f(-1) = (-1)³ + 3(-1)² + k(-1) + 4 = 0

-1 + 3 - k + 4 = 0

∴ k = 6

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

(\frac{A}{x - 1}) + (\frac{B}{x + 2})

A(x + 2) + B(x - 1) = 3

when x = 1, 3A = 3 (\to) a = 1

when x = -2, -3B = 3 (\to) B = -1

3 = 1 - 1

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

To solve -11 (\leq) 4 - 3x (\leq) 28

-11 (\leq) 4 - 3x also 4 -3x (\leq) 28

15 (\leq) -3x (\leq) 24 = 15 (\geq) 3x - 3x (\geq) -24

-5 (\geq) x, x (\geq) -8

i.e. x (\leq) 5

∴ -8 (\leq) x (\leq) 5

3 + 2 + (\frac{4}{3}) + (\frac{8}{9}) + (\frac{16}{17}) + .....

a = 3

r = (\frac{2}{3})

s (\alpha) = (\frac{a}{1 - r}) = (\frac{3}{1 - \frac{2}{3}})

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

= 3 x 3

= 9

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

a = 2, d = 3 and n = 11

To find S_{n/sub> = (\frac{n}{2}) [2a + (n - 1) (\delta)]}

= (\frac{11}{2}) [2(2) + (11 - 1) 3]

= (\frac{11}{2})n [4 + 10(3)]

= (\frac{11}{2})(34)

= 11 x 17

= 187

</sub>

m (\ast) n = mn + m + n

m (\ast) e = me + m + e, e (\ast) m = e

∴ me + m + e, m(e + 1)e - e = 0

e + 1 = 0

∴ e = -1