WAEC Further Mathematics Past Questions & Answers - Page 107

531.

Find the equation of a circle with centre (2, -3) and radius 2 units.

A.

\(x^{2} + y^{2} - 4x + 6y + 9 = 0\)

B.

\(x^{2} + y^{2} + 4x - 6y - 9 = 0\)

C.

\(x^{2} + y^{2} + 4x + 6y - 9 = 0\)

D.

\(x^{2} + y^{2} + 4x - 6y + 9 = 0\)

Correct answer is A

The equation of a circle with centre coordinate (a, b) and radius r is :

\((x - a)^{2} + (y - b)^{2} = r^{2}\)

Given centre = (2, -3) and radius r = 2 units

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

\(x^{2} - 4x + 4 + y^{2} + 6y + 9 = 4\)

\(x^{2} + y^{2} - 4x + 6y + 4 + 9 - 4 = 0 \implies x^{2} + y^{2} - 4x + 6y + 9 = 0\)

532.

The deviations from the mean of a set of numbers are \((k+3)^{2}, (k+7), -2, \text{k and (} k+2)^{2}\), where k is a constant. Find the value of k.

A.

3

B.

2

C.

-2

D.

-3

Correct answer is D

The sum of deviations from the mean of a set of numbers equals 0.

\((k+3)^{2} + (k+7) + (-2) + k + (k+2)^{2} = 0\)

\((k^2 + 6k + 9) + (k+7) - 2 + k + (k^2 + 4k + 4) = 0\)

\(2k^{2} + 12k + 18 = 0\)

\(2k^{2} + 6k + 6k + 18 = 2k(k + 3) + 6(k + 3) = 0\)

\(k = -3 (twice)\)

533.

Forces 90N and 120N act in the directions 120° and 240° respectively. Find the resultant of these forces.

A.

\(-45(2i + \sqrt{2}j)\)

B.

\(60(\sqrt{3}i + 7j)\)

C.

\(30(7i + \sqrt{3}j)\)

D.

\(-15(7i + \sqrt{3}j)\)

Correct answer is D

\(F = F\cos\theta + F\sin\theta\)

\(\implies 90N = 90\cos 120° + 90\sin 120°\)

\(120N = 120 \cos 240° + 120 \sin 240°\)

\(R = F_{1} + F_{2} \)

= \((90 \cos 120 + 120 \cos 240)i + (90\sin 120 + 120 \sin 240)j\)

= \(90(-0.5) + 120(-0.5))i + (90(\frac{\sqrt{3}}{2}) + (120(-\frac{\sqrt{3}}{2}))j\)

= \(-105i - 15\sqrt{3}j = -15(7i + \sqrt{3}j)\)

534.

If a fair coin is tossed four times, what is the probability of obtaining at least one head?

A.

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

B.

\(\frac{1}{4}\)

C.

\(\frac{13}{16}\)

D.

\(\frac{15}{16}\)

Correct answer is D

P(at least one head) = 1 - P(4 tails)

Let \(p = \frac{1}{2}\) = probability of head and \(q = \frac{1}{2}\) = probability of tail.

\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)

P(4 tails) = \(q^{4} = (\frac{1}{2})^{4} = \frac{1}{16}\)

P(at least one head) = \(1 - \frac{1}{16} = \frac{15}{16}\)

535.

Find the coefficient of \(x^3\) in the binomial expansion of \((3x + 4)^4\) in ascending powers of x

A.

432

B.

194

C.

144

D.

108

Correct answer is A

\((3x + 4)^{4} = ^{4}C_{0}(3x)^{0}(4)^{4} + ^{4}C_{1}(3x)^{1}(4)^{3} + ^{4}C_{2}(3x)^{2}(4)^{2} + ^{4}C_{3}(3x)^{3}(4)^{1} + ^{4}C_{4}(3x)^{4}(4)^{0}\)

\(x^{3} = ^{4}C_{3}(3x)^{3}(4) = \frac{4!}{3!1!} \times 3^{3} \times 4\)

= \(432x^{3}\)