WAEC Further Mathematics Past Questions & Answers - Page 129

641.

A car is moving at 120\(kmh^{-1}\). Find its speed in \(ms^{-1}\).

A.

33.3\(ms^{-1}\)

B.

66.6\(ms^{-1}\)

C.

99.9\(ms^{-1}\)

D.

120.0\(ms^{-1}\)

Correct answer is A

\(120 kmh^{-1} = \frac{120 \times 1000}{3600} = \frac{100}{3} = 33.3ms^{-1}\)

642.

A particle starts from rest and moves through a distance \(S = 12t^{2} - 2t^{3}\) metres in time t seconds. Find its acceleration in 1 second.

A.

24\(ms^{-2}\)

B.

18\(ms^{-2}\)

C.

12\(ms^{-2}\)

D.

10\(ms^{-2}\)

Correct answer is C

\(\frac{\mathrm d s(t)}{\mathrm d t} = v(t)\) and \(\frac{\mathrm d v(t)}{\mathrm d t} = a(t)\)

\(\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}\)

\(\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)\)

\(a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}\)

643.

Find the constant term in the binomial expansion \((2x^{2} + \frac{1}{x})^{9}\)

A.

84

B.

168

C.

336

D.

672

Correct answer is D

Let the power of \(2x^{2}\) be t and the power of \(\frac{1}{x} \equiv x^{-1}\) = 9 - t.

\((2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}\)

Dealing with x alone, we have

\((x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0\)

\(3t - 9 = 0 \therefore t = 3\)

The binomial expansion is then,

\(^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}\)

= 84 x 8

= 672

644.

Find the angle between forces of magnitude 7N and 4N if their resultant has a magnitude of 9N.

A.

39.45°

B.

73.40°

C.

75.34°

D.

106.60°

Correct answer is B

\(F_{1} = 7i + 0j\)

\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)

\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)

\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)

\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)

Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)

\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)

\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)

\(\theta = \cos^{-1} 0.2857  = 73.40°\)

645.

A body of mass 28g, initially at rest is acted upon by a force, F Newtons. If it attains a velocity of \(5.4ms^{-1}\) in 18 seconds, find the value of F.

A.

0.0082N

B.

0.0084N

C.

0.082N

D.

0.084N

Correct answer is B

\(F = mass \times acceleration\) but \(accl = \frac{v - u}{t}\)

\(\therefore F = m(\frac{v - u}{t})\)

\(Mass = 28g = 0.028kg\)

\(v = 5.4 ms^{-1}; u = 0; t = 18secs\)

\(\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N\)