Correct Answer: B. \(\frac{4}{9}\)

Explanation

Let the number of candidates that speak both Igbo and Hausa = x

The number that speaks Igbo only = 56 - x

The number that speaks Hausa only = 50 - x

56 - x + x + 50 - x = 90

106 - x = 90

x = 16

P(Igbo only) = \(\frac{56 - 16}{90} = \frac{4}{9}\)

Explanation

(a)(i) Scale : On x- axis, 2cm = 1 unit

On y- axis, 2cm = 5 units.

(ii) Equation of line of symmetry is x = 1.25.

(iii) Roots of the equation \(ax^{2} + bx + c = 0\) are x = 0.25 and x = 2.25.

(b) Coordinates are D(0, 1), E(1, -2) and G(3, 4).

Substituting for y and x in \(ax^{2} + bx + c = y\)

D(0, 1) : \(1 = a(0^{2}) + b(0) + c \implies c = 1\)

E(1, -2) : \(-2 = a(1^{2}) + b(1) + c \implies -2 = a + b + c\)

\(a + b = -2 - 1 = -3 ... (1)\)

G(3, 4) : \(4 = a(3^{2}) + b(3) + c \implies 4 = 9a + 3b + c\)

\(9a + 3b = 4 - 1 = 3 ... (2)\)

\(\implies 3a + b = 1 ... (2a)\)

(2a) - (1) : \(2a = 4 \implies a = 2\)

\(a + b = -3 \implies 2 + b = -3\)

\(b = -3 - 2 = -5\)

\(\therefore\) The equation is \(y = 2x^{2} - 5x + 1\)

(c) The greatest value of y = 33.5.

Correct Answer: A. {21, 91, 221}

Explanation

multiples of 5 less than 20 = 5, 10 and 15

= [ 5, 10 and 15 ]x\(^2\) - x + 1

x\(^2\) - x + 1

when x = 5

5\(^2\) - 5 + 1; 25 - 5 + 1 --> 21

when x = 10

10\(^2\) - 10 + 1

100 - 9 = 91

when x = 15

15\(^2\) - 15 + 1

225 - 15 + 1 = 211

Explanation

(a)

(b)

(c)(i) x = -2.9 or 1.9

(ii) \(y + (3 - 2x - 2x^{2}) = 11 - 2x - 2x^{2}\)

\(y = 11 - 2x - 2x^{2} - 3 + 2x + 2x^{2}\)

\(y = 8\)

\(\therefore x = \text{-1.8 or 0.8}\)

(iii) \(y = 11 - 2x - 2x^{2}\)

\(\frac{\mathrm d y}{\mathrm d x} = -2 - 4x\)

Gradient at x = 1 : \(-2 - 4(1) = -2 - 4 = -6\)

Correct Answer: D. 1 + \(\frac{\sqrt{2}}{2}\)

Correct Answer: (b) 604.8x 10³ secs

Explanation

\(\frac{log_2 ^8 + log_2 ^ {16} - 4 log_2^ 2}{log_4^{16}}\)

= \(\frac{3log_2 ^2 + 4 log_2^ 2 - 4 log_2 2}{2 log_4^4}\)

= \(\frac{3 + 4 - 4}{2}\)

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

(b)

T\(_1\) + T\(_2\) = 6

a + a + d = 6

2a + d = 6.....(i)

a = a

a + 2d = ar, a + 6d = ar\(^2\)

\(\frac{a + 2d}{a} = \frac{a + 6d}{a + 2d}\)

(a + 2d)\(^2\) = \(\frac{a + 6d}{a + 2d}\)

(a + 2d)\(^2\) = a(a + 6d)

a\(^2\) + 4d\(^2\) + 4ad = a\(^2\) + 6ad

4d\(^2\) = 6ad - 4ad

4d\(^2\) = 2ad

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

a = 2d ......(ii)

from;

2a + d = 6

2(2d) + d = 6

4d + d = 6

5d = 6

d = \(\frac{6}{5}\)

= 1\(\frac{1}{5}\)

from; a = 2d

a = \(\frac{2 \times 6}{5}\)

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

= 2\(\frac{2}{5}\)

First term = 2\(\frac{2}{5}\)

Common difference = 1\(\frac{4}{5}\)

Explanation

(a) Distance along arc PQ = \(\frac{\theta}{360°} \times 2 \pi r\)

where \(r = R \cos \theta\)

\(|PQ| = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 6400 \times \cos 40°\)

\(|PQ| = \frac{1}{6} \times 2 \times \frac{22}{7} \times 6400 \times 0.7660\)

= \(5134.74 km \approxeq 5135km\) (to the nearest whole number)

(b) Difference in latitude between Q and T = 40° - 28° = 12°

\(\therefore |QT| \text{along the line of longitude} = \frac{12}{360} \times 2 \times \frac{22}{7} \times 6400\)

\(\frac{1}{30} \times 2 \times \frac{22}{7} \times 6400 = 1340.95km \approxeq 1341km\)

(c) Average speed = \(\frac{\text{Total distance covered}}{\text{Total time taken}}\)

= \(\frac{5134.74 + 1340.95}{12} = 539.64km/hr\)

\(\approxeq 540km/hr\) (to 3 significant figures)

Correct Answer: B. 1

Correct Answer: C. \(2a^{3} - \frac{2}{a^{3}}\)

Explanation

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

\(h(a) = a^{3} - \frac{1}{a^{3}}\)

\(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\)

\(h(a) - h(\frac{1}{a}) = (a^{3} - \frac{1}{a^{3}}) - (\frac{1}{a^{3}} - a^{3}) = 2a^{3} - \frac{2}{a^{3}}\)