Find the coefficient of \(x^{3}\) in the expansion of \([\frac{1}{3}(2 + x)]^{6}\)
\(\frac{135}{729}\)
\(\frac{149}{729}\)
\(\frac{152}{729}\)
\(\frac{160}{729}\)
Correct answer is D
\([\frac{1}{3}(2 + x)]^{6} = (\frac{2}{3} + \frac{x}{3})^{6}\)
The coefficient of \(x^{3}\) is
\(^{6}C_{3}(\frac{2}{3})^{3}(\frac{1}{3})^{3}x^{3} = (\frac{6!}{3!3!})(\frac{8}{27})(\frac{1}{27})x^{3}\)
= \(\frac{160}{729}\)
Given n = 3, evaluate \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\)
\(12\)
\(2\frac{1}{2}\)
\(2\)
\(\frac{11}{24}\)
Correct answer is D
n = 3, \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!} = \frac{1}{(3-1)!} - \frac{1}{(3+1)!}\)
= \(\frac{1}{2} - \frac{1}{24} = \frac{12 -1}{24}\)
= \(\frac{11}{24}\)
Simplify \(\frac{^{n}P_{5}}{^{n}C_{5}}\)
80
90
110
120
Correct answer is D
\(\frac{^{n}P_{5}}{^{n}C_{5}} = \frac{n!}{(n-5)!} ÷ \frac{n!}{(n-5)!5!}\)
= \(\frac{n!}{(n-5)!} \times \frac{(n-5)!5!}{n!} = 5! = 120\)
If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.
\(y \geq \frac{1}{x}\)
\(y \leq \frac{1}{x}\)
\(y \leq \frac{1}{x^{2}}\)
\(y \geq \frac{1}{x^{2}}\)
Correct answer is D
\(\log_{10}y + 3\log_{10}x \geq \log_{10}x\)
\(\implies \log_{10}y \geq \log_{10}x - 3 \log_{10}x \)
\(\log_{10}y \geq -2\log_{10}x = \log_{10}y \geq \log_{10}x^{-2}\)
\(\log_{10}y \geq \log_{10}(\frac{1}{x^{2}}) \implies y \geq \frac{1}{x^{2}}\)
Solve for x in the equation \(5^{x} \times 5^{x + 1} = 25\)
\(-2\)
\(\frac{-1}{2}\)
\(\frac{1}{2}\)
\(2\)
Correct answer is C
\(5^{x} \times 5^{x+1} = 25\)
\(5^{x} \times 5^{x+1} = 5^{2}\)
\(5^{x+x+1} = 5^{2}\), equating powers,
\(2x + 1 = 2 \implies 2x = 1\)
\(\therefore x = \frac{1}{2}\)