2011 - JAMB Mathematics Past Questions and Answers - page 2

X² + 1 \(\frac{X - 2}{\sqrt{X^3 - 2X^2 + 3n - 3}}\)

= \(\frac {- 6X^3 + n}{-2X^2 + 2X - 3}\)

= \(\frac{(-2X^2 - 2)}{2X - 1}\)

Remainder is 2X - 1

9y² - 16x²

= 3²y² - 4²x²

= (3y - 4x)(3y +4x)

-2x -5y = 3

x + 3y = 0

x = -3y

-2 (-3y) - 5y = -3

6y - 5y = 3

y = 3

but, x = -3y

x = -3(3)

x = -9

therefore, x = -9, y = 3

x \(\alpha\sqrt y\)

x = k\(\sqrt y\)

81 = k\(\sqrt9\)

k = \(\frac{81}{3}\)

= 27

therefore, x = 27\(\sqrt y\)

y = 1\(\frac{7}{9}\) = \(\frac{16}{9}\)

x = 27 x \(\sqrt{\frac{16}{9}}\)

= 27 x \(\frac{4}{3}\)

dividing 27 by 3

= 9 x 4

= 36

T \(\alpha \frac{1}{R^3}\)

T = \(\frac{k}{R^3}\)

k = TR³

= \(\frac{2}{81}\) x 3³

= \(\frac{2}{81}\) x 27

dividing 81 by 27

k = \(\frac{2}{2}\)

therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)

When R = 2

T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)

= \(\frac{1}{12}\)

-6(x + 3) \(\leq\) 4(x - 2)

-6(x +3) \(\leq\) 4(x - 2)

-6x -18 \(\leq\) 4x - 8

-18 + 8 \(\leq\) 4x +6x

-10x \(\leq\) 10x

10x \(\leq\) -10

x \(\leq\) 1

x² + 2x > 15

x² + 2x - 15 > 0

(x² + 5x) - (3x - 15) > 0

x(x + 5) - 3(x + 5) >0

(x - 3)(x + 5) > 0

therefore, x = 3 or -5

then x < -5 or x > 3

i.e. x< 3 or x < -5

3, 6, 9,..., 36.

a = 3, d = 3, i = 36, n = 18

S_n = \(\frac{n}{2}\) [2a + (n - 1)d

S₁₈ = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]

= 9[6 + (17 x 3)]

= 9 [6 + 51] = 9(57)

= 513

T₂ = 4, T₄ = 16

T_x = ar^n-1

T₂ = ar^2-1 = 4 i.e. ar³ = 16, i.e. ar = 4

T₄ = ar^4-1

therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)

r² = 4 and r = 2

but ar = 4

a = \(\frac{4}{r}\) = \(\frac{4}{2}\)

a = 2

Sⁿ = \(\frac{a(r^n - 1)}{r - 1}\)

S⁵ = \(\frac{2(2^5 - 1)}{2 - 1}\)

= \(\frac{2(32 - 1)}{2 - 1}\)

= 2(31)

= 62

N + Y = XY + X + Y

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

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

= \(\frac{1 - 2}{1 - 3}\)

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