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
\(\cos \theta = 1 \implies \theta = \cos^{-1} 1\)
= \(0°\)
\(\frac{-x}{1 - x}, x \neq 1\)
\(\frac{1}{1 - x}, x \neq 1\)
\(\frac{-1}{1 - x}, x \neq 1\)
\(\frac{x}{1 - x}, x \neq 1\)
Correct answer is A
\(x * y = x + y - xy\)
Let \(x^{-1}\) be the inverse of x, so that
\(x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0\)
\(x + x^{-1} - x(x^{-1}) = 0 \implies x(x^{-1}) - x^{-1} = x\)
\(x^{-1}(x - 1) = x \implies x^{-1} = \frac{x}{x - 1}\)
= \(\frac{x}{-(1 - x)} = \frac{-x}{1 - x}, x \neq 1\)