1997 - JAMB Mathematics Past Questions and Answers - page 2

y = X² - 3x + 2, (\frac{dy}{dx}) = 2x - 3

at turning pt, (\frac{dy}{dx}) = 0

∴ 2x - 3 = 0

∴ x = (\frac{3}{2})

(\frac{d^2y}{dx^2}) = (\frac{d}{dx})((\frac{d}{dx}))

= 270

∴ y_min = 2(\frac{3}{2}) - 3(\frac{3}{2}) + 2

= (\frac{9}{4}) - (\frac{9}{2}) + 2

= -(\frac{1}{4})

t = (\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}})

t² = (\frac{v}{\frac{1}{f} + \frac{1}{g}})

= (\frac{vfg}{ftg})

(\frac{1}{f} + \frac{1}{g}) = (\frac{v}{t^2})

= (g + f)t² = vfg

gt² = vfg - ft²

gt² = f(vg - t²)

f = (\frac{gt^2}{gv-t^2})

4x² - 18xy + g = g (\to) ((\frac{18y}{4}))²

= (\frac{18y^2}{4})

5 + 2r = k(r + 1) + L(r - 2)

but r - 2 = 0 and r = 2

9 = 3k

k = 3

(\frac{f(x)}{g(x)}) < 1

∴ (\frac{2x +4}{6x + 7}) < 1

= 2x + 4 < 6x + 7

= 6x + 7 > 2x + 4

= 6x - 2x > 4 - 7

= 4x > -3

∴ x > -(\frac{3}{4})

12x² < x + 1

12 - x - 1 < 0

12x² - 4x + 3x - 1 < 0

4x(3x - 1) + (3x - 1) < 0

Case 1 (+, -)

4x + 1 > 0, 3x - 1 < 0

x > -(\frac{1}{4})x < (\frac{1}{3})

Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0

-(\frac{1}{4}) < x < - (\frac{1}{3})

Tn = 3^{1 - n}

S₃ = 3^{1 - 1} + 3^{1 - 2} + 3^{1 - 3}

= 1 + (\frac{1}{3}) + (\frac{1}{9})

= (\frac{13}{9})

m (\ast) n = mn - n - 1, m (\oplus) n = mn + n - 2

3 (\oplus) (4 (\ast) 5) = 3 (\oplus) (4 x 5 - 5 - 1) = 3 (\oplus) 14

3 (\oplus) 14 = 3 x 14 + 14 - 2

= 54

x (\ast) y = x + y - xy

(x (\ast) 2) + (x (\ast) 3) = 68

= x + 2 - 2x + x + 3 - 3x

= 86

3x = 63

x = -21