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<sub>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