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
---------

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

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

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

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