Correct Answer: A. \(\frac{5}{4}\)

Explanation

Given, \(x^{2} + x - 2 = 0\), a = 1, b = 1 and c = -2.

\(\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1\)

\(\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2\)

\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\beta^{2} + \alpha^{2}}{(\alpha\beta)^{2}}\)

\(\beta^{2} + \alpha^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = (-1)^{2} - 2(-2) = 1 + 4 = 5\)

\(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{5}{(-2)^{2}} = \frac{5}{4}\).

Explanation

(a)

(b)

(c) (ii) roots of the equation : \(x^{2} - x - 2 = 0\) are x = -1 or x = 2.

(iii) gradient of the curve at x = -0.5 is 2.1.

Correct Answer: B. 13.4km

Explanation

ZY\(^2\) = 12\(^2\) + 6\(^2\)

ZY\(^2\) = 144 + 36 = 180

ZY = \(\sqrt{180}\)

= \(13.416\)

= 13.4 km

Correct Answer: A. 5

Correct Answer: B. 1.0

Explanation

\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)

\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)

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

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

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

= \(1.0\)

Correct Answer: B. 1 or 2

Explanation

\(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)

= \((3^{x})^{2} - (3^{x}).(3^{2}) = (3^{x}).(3^{1}) - 27\)

Let \(3^{x}\) be B; we have

= \(B^{2} - 9B - 3B + 27 = B^{2} - 12B + 27 = 0\).

Solving the equation, we have B = 3 or 9.

\(3^{x} = 3\) or \(3^{x} = 9\)

\(3^{x} = 3^{1}\) or \(3^{x} = 3^{2}\)

Equating, we have x = 1 or 2.

Explanation

(a) Let the salary be x

Shopping mall = \(\frac{1}{4}\) x \(x = \frac{x}{4}\)

Open market = \(\frac{1}{3} \times x = \frac{x}{3}\)

The remainder x - [\(\frac{x}{4} + \frac{x}{3}\)]

= x - [\(\frac{3x + 4x}{12}\)]

= \(\frac{x}{1} - \frac{7x}{12}\)

= \(\frac{12x - 7x}{12}\)

= \(\frac{5x}{12}\)

mechanic = \(\frac{2}{5} \times \frac{5x}{12} = \frac{x}{6}\)

\(\frac{x}{6} + \frac{x}{3} + \frac{x}{4}\) + 225,00 = x

\(\frac{2x + 4x + 3x}{12}\) + 225,000

\(\frac{9x}{12}\) + 225,000 = x

225,000 = \(\frac{x}{1} = \frac{9x}{12}\)

225,000 = \(\frac{12x - 9x}{12}\)

225,000 = \(\frac{3x}{12}\)

\(\frac{225,000 \times 12}{3}\) = x

N900,000 = x

(i) Her monthly income = N900,000

(ii) Amount in open market

= \(\frac{N900,000}{3}\)

= N300,000

(b) T\(_3\) = 4m - 2n

T\(_9\) = 2m - 8n

a + 2d = 4m - 2n .....(i)

a + 8d = 2m - 8n ......(ii)

a + 8d = 2m - 8n

-(a + 2d = 4m - 2n

\(\overline{6d = 6m - 6n}\)

6d = 6(m - n)

d = m - n

Common difference = m - n

Explanation

(a) 5b, 4g, 3y = 12 balls.

(i) p(all 3 balls have the same colour) = p(all 3 blue or all 3 green or all 3 yellow)

= \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} + \frac{3}{12} \times \frac{2}{11} \times \frac{1}{10}\)

= \(\frac{60}{1320} + \frac{24}{1320} + \frac{6}{1320}\)

= \(\frac{90}{1320} = \frac{3}{44}\)

(ii) Two balls of the same colour :

Sample space for selecting with restriction (2 blue 1 green or 2 blue 1 yellow or 2 green 1 blue or 2 green 1 yellow or 2 yellow 1 blue or 2 yellow 1 green)

Ways : \(^{5}C_{2} \times ^{4}C_{1} + ^{5}C_{2} \times ^{3}C_{1} + ^{4}C_{2} \times ^{5}C_{1} + ^{4}C_{2} \times ^{3}C_{1} + ^{3}C_{2} \times ^{5}C_{1} + ^{3}C_2} \times ^{4}C_{1}\)

= \(40 + 30 + 30 + 18 + 15 + 12 = 145\)

Selecting without restriction 3 balls out of 12 balls : \(^{12}C_{3}\)

= \(\frac{12!}{3! 9!}\)

= \(220\)

Probability = \(\frac{\text{selection with restriction}}{\text{selection without restriction}} = \frac{145}{220}\)

= \(\frac{29}{44}\)

(b)

Spearman's rank correlation cofficient:

= \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)

= \(1 - \frac{6(16)}{7(7^{2} - 1)}\)

= \(1 - \frac{2}{7}\)

= \(0.714\)

Correct Answer: B. 20

Correct Answer: B. 108