1991 - JAMB Mathematics Past Questions and Answers - page 3

g(x) = x² + 3x + 4

= g(x + 1) = (x + 1)^2 + 3(x + 1) + 4

= x² + 1 + 2x + 3x + 3 + 4

= x² + 5x + 8

g(x + 1) - g(x) = x² + 5x + 8 - (x² + 3x + 4)

= x² + 5x + 8 - x² + 3x + 4

= 2x + 4

= 2(x + 2)

m³ - 2m² - m + 2

Let f(m) = m³ - 2m² - m² + 2

= f(1)

= 1 - 2 - 2 + 2 = 0

∴ m - 1 is factor (\frac{m^3 - 2m^2 - m^2 + 2}{m - 1})

= m² - m - 2

= (m - 1)m² - m - 2

= (m - 1)(m + 1)(m - 2)

1 - (a - b)² = [1 + (a - b)][1 - a + b]

= (1 + a - b)(1 - a + b)

rs + tr - pt - ps = rs - ps - tr - pt

= (r - p)s + (r - p)t

= (r - p)(s + t),

hence r - p is a factor

(\frac{x}{2}) + (\frac{x}{3}) + (\frac{x}{4})< 1

= (\frac{6x + 4x + 3x < 12}{12})

i.e. 13 x < 12 = x < (\frac{12}{13})

3n² = 12n

= 3n² - 12n = 0

= 3n(n - 4) = 0

∴ n = 4

Given 27, 9, 3,......this is a G.P

r = (\frac{9}{27})

= (\frac{1}{3})

T = ar^{n - 1}

= 27(\frac{1}{3}) ^{n - 1}

(x - 2) (x - 3) = 12

x² - 3x - 2x + 6 = 12

x² - 5x - 6 = 0

(x - 1)(x - 6) = 0

x = -1 or 6