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