2012 - JAMB Mathematics Past Questions and Answers - page 2

Hence f(x) = 2x³ - 11x² + 8x - 1

f(-3) = 2(-3)³ - 11(-3)² + 8(-3) - 1

= 2(-27) - 11(9) + 8(-3) - 1

= -54 - 99 - 24 - 1

= -178

x² - y² = 4 .... (1)

x + y = 2 .... (2)

Simplify eqn (1)

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

From eqn (2)

x + y = 2 so substitute it into simplified eqn (1), we have

2 (x - y) = 4

therefore,

x - y = 2 ... (1)

x + y = 2

---------

2x = 4

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x = 2, when y = 0

y (\alpha \sqrt{n})

y = k(\sqrt{n})

when y = 4, n = 4

4 = k(\sqrt{4})

4 = 2k

k = 2

Therefore,

y = 2(\sqrt{n})

y = 2(\sqrt{\frac{16}{9}})

y = 2((\frac{4}{3}))

y = (\frac{8}{3})

U (\alpha \frac{1}{V^3})

U = (\frac{k}{V^3})

k = UV³

k = 81 x 2³ = 81 x 8

When V = 3,

U = (\frac{k}{V^3})

U = (\frac{81 \times 8}{3^3})

U = (\frac{81 \times 8}{27}) = 24

(\frac{1}{5}y + \frac{1}{5} < \frac{1}{2}y + \frac{2}{5})

Collect like terms

(\frac{y}{5} - \frac{y}{2} < \frac{2}{5} - \frac{1}{5})

(\frac{2y - 5y}{10} < \frac{2 - 1}{5})

(\frac{-3y}{10} < \frac{1}{5})

(y > \frac{-2}{3})

(m - 3)(m - 4) < 0

(m - 3) < 0 ; (m - 4) < 0

m < 3 ; m < 4

3 < m < 4

n² - 6n - 4

For the 3rd term,

3² - 6(3) - 4

9 - 18 -4 = -13

For the 4th term,

4² - 6(4) - 4

16 - 24 - 4 = -12

Sum of both terms

-13 - 12 = -25

S_r = (\frac{a}{1 - r})

(-\frac{1}{10}) = (\frac{1}{8} \times \frac{1}{1 - r})

(-\frac{1}{10}) = (\frac{1}{8(1 - r)})

(-\frac{1}{10}) = (\frac{1}{8 - 8r})

cross multiply...

-1(8 - 8r) = -10

-8 + 8r = -10

8r = -2

r = -1/4

p * q = pq + p - q

First execute for 3 * 4

==> 3(4) + 3 - 4 = 12 + 3 - 4 = 11

Now we execute for 2 * 11

==> 2(11) + 2 - 11 = 22 + 2 - 11 = 13

m (\propto) n = (\frac{mn}{2} - a)

Identify = e = 2

a (\propto) a^-1 = e

a (\propto) a^-1 = 2

-5 (\propto) a^-1 = 2

(\frac{-5 \times a^{-1}}{2} = 2)

(a^{-1} = \frac{2 \times 2}{-5})

(a^{-1} = -\frac{4}{5})