JAMB Mathematics Past Questions & Answers - Page 330

3(x + 2) > 6(x + 3)

3x + 6 > 6x + 18

3x - 6x > 18 - 6

-3x > 12

x < -4

$r \propto \frac{1}{\sqrt{s}}, r \propto \frac{1}{\sqrt{t}}$

$r \propto \frac{1}{\sqrt{s}}$ ..... (1)

$r \propto \frac{1}{\sqrt{t}}$ ..... (2)

Combining (1) and (2), we get

$r = \frac{k}{\sqrt{s} \times \sqrt{t}} = \frac{k}{\sqrt{st}}$

This gives $\sqrt{st} = \frac{k}{r}$

By taking the square of both sides, we get

st = $\frac{k^2}{r^2}$

s = $\frac{k^2}{r^{2}t}$

P $\propto$ mu, p $\propto \frac{1}{q}$

p = muk ................ (1)

p = $\frac{1}{q}k$.... (2)

Combining (1) and (2), we get

P = $\frac{mu}{q}k$

4 = $\frac{m \times u}{1}k$

giving k = $\frac{4}{6} = \frac{2}{3}$

So, P = $\frac{mu}{q} \times \frac{2}{3} = \frac{2mu}{3q}$

Hence, P = $\frac{2 \times 6 \times 4}{3 \times \frac{8}{5}}$

P = $\frac{2 \times 6 \times 4 \times 5}{3 \times 8}$

p = 10

Let f(p) = 6p³ - p² - 47p + 30

Then by the remainder theorem,

(p - 3): f(3) = remainder R,

i.e. f(3) = 6(3)³ - (3)² - 47(3) + 30 = R

162 - 9 - 141 + 30 = R

192 - 150 = R

R = 42

Let f(x) = x² - x - k
Then by the factor theorem,

(x - 4): f(4) = (4)² - (4) - k = 0

16 - 4 - k = 0

12 - k = 0

k = 12