Find the third term in the expansion of \((a - b)^{6}\) in ascending powers of b.
\(-15a^{4}b^{2}\)
\(15a^{4}b^{2}\)
\(-15a^{3}b^{3}\)
\(15a^{3}b^{3}\)
Correct answer is B
\((a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...\)
Third term = \(^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)\)
= \(15a^{4}b^{2}\)
If \(f(x) = x^{2}\) and \(g(x) = \sin x\), find g o f.
\(\sin^{2} x\)
\(\sin x^{2}\)
\((\sin x)x^{2}\)
\(x \sin x\)
Correct answer is B
\(f(x) = x^{2}, g(x) = \sin x\)
\(g \circ f = g(x^{2}) = \sin x^{2}\)
Express \(\log \frac{1}{8} + \log \frac{1}{2}\) in terms of \(\log 2\)
3 log 2
4 log 2
-3 log 2
-4 log 2
Correct answer is D
\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)
= \(\log 2^{-3} + \log 2^{-1}\)
= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)
Given that \(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\), solve for n
-6.00
-1.20
0.83
1.20
Correct answer is D
\(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\)
\(\implies a^{\frac{5}{6} + \frac{-1}{n}} = a^{0}\)
Equating bases, we have
\(\frac{5}{6} - \frac{1}{n} = 0\)
\(\frac{5n - 6}{6n} = 0\)
\(5n - 6 = 0 \implies 5n = 6\)
\(n = \frac{6}{5} = 1.20\)
Solve: \(\sin \theta = \tan \theta\)
200°
90°
60°
0°
Correct answer is D
\(\sin \theta = \tan \theta \implies \frac{\sin \theta}{1} = \frac{\sin \theta}{\cos \theta}\)
Equating, we have
\(\cos \theta = 1 \implies \theta = \cos^{-1} 1\)
= \(0°\)