\(x^{2} + 4x - 5\)
\(x^{2} - 4x + 5\)
\(x^{2} - 1\)
\(x - 1\)
Correct answer is B
\(f(x) = x^{2} + 1\) and \(g(x) = x - 2\)
\(f o g = f(g(x)) = f(x - 2) = (x - 2)^{2} + 1 \)
= \(x^{2} - 4x + 4 + 1 = x^{2} - 4x + 5\)
A car is moving at 120\(kmh^{-1}\). Find its speed in \(ms^{-1}\).
33.3\(ms^{-1}\)
66.6\(ms^{-1}\)
99.9\(ms^{-1}\)
120.0\(ms^{-1}\)
Correct answer is A
\(120 kmh^{-1} = \frac{120 \times 1000}{3600} = \frac{100}{3} = 33.3ms^{-1}\)
24\(ms^{-2}\)
18\(ms^{-2}\)
12\(ms^{-2}\)
10\(ms^{-2}\)
Correct answer is C
\(\frac{\mathrm d s(t)}{\mathrm d t} = v(t)\) and \(\frac{\mathrm d v(t)}{\mathrm d t} = a(t)\)
\(\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}\)
\(\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)\)
\(a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}\)
Find the constant term in the binomial expansion \((2x^{2} + \frac{1}{x})^{9}\)
84
168
336
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
Find the angle between forces of magnitude 7N and 4N if their resultant has a magnitude of 9N.
39.45°
73.40°
75.34°
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, \(\cos^{2}\theta + \sin^{2}\theta = 1\)
\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)
\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)
\(\theta = \cos^{-1} 0.2857 = 73.40°\)