Evaluate \(\int_{1}^{2} [\frac{x^{3} - 1}{x^{2}}] \mathrm {d} x\)
0.5
1.0
1.5
2.0
Correct answer is B
\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)
\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)
= \((\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})\)
= \((2 + \frac{1}{2}) - (\frac{1}{2} + 1)\)
= \(2\frac{1}{2} - 1\frac{1}{2}\)
= \(1.0\)
If \(\frac{^{8}P_{x}}{^{8}C_{x}} = 6\), find the value of x.
1
2
3
6
Correct answer is C
\(\frac{^{8}P_{x}}{^{8}C_{x}} = \frac{8!}{(8 - x)!} ÷ \frac{8!}{(8 - x)! x!} = 6\)
\(\frac{8!}{(8 - x)!} \times \frac{(8 - x)! x!}{8!} = x! = 6\)
\(x = 3\)
Given that \(\log_{3}(x - y) = 1\) and \(\log_{3}(2x + y) = 2\), find the value of x
1
2
3
4
Correct answer is D
\(\log_{3}(x - y) = 1 \implies x - y = 3^{1} = 3 .... (1)\)
\(\log_{3}(2x + y) = 2 \implies 2x + y = 3^{2} = 9 ..... (2)\)
From (1), y = x - 3
From (2), y = 9 - 2x
\(\implies 9 - 2x = x - 3\)
\(9 + 3 = x + 2x = 3x\)
\(x = 4\)
Simplify \((216)^{-\frac{2}{3}} \times (0.16)^{-\frac{3}{2}}\)
\(\frac{125}{288}\)
\(\frac{2}{125}\)
\(\frac{4}{225}\)
\(\frac{2}{225}\)
Correct answer is A
\((216)^{-\frac{2}{3}} = (\frac{1}{216})^{\frac{2}{3}} = (\sqrt[3]{\frac{1}{216}})^{2} = (\frac{1}{6})^{2} = \frac{1}{36}\)
\((0.16)^{-\frac{3}{2}} = (\frac{100}{16})^{\frac{3}{2}} = (\sqrt{100}{16})^{3} = \frac{1000}{64}\)
\(\frac{1}{36} \times \frac{1000}{64} = \frac{125}{288}\)
\(f(x) = x^{3} - \frac{1}{x^{4}} + 2\)
\(f(x) = x^{3} + \frac{1}{x^{4}} + 2\)
\(f(x) = x^{3} - \frac{1}{x^{4}} - 2 \)
\(f(x) = x^{3} + \frac{1}{x^{4}} - 2\)
Correct answer is B
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - \frac{4}{x^5} = 3x^{2} - 4x^{-5}\)
\(y = \int (3x^{2} - 4x^{-5}) \mathrm {d} x \)
\(y = x^{3} + \frac{1}{x^{4}} + c\)
f(1) = 4; \(4 = 1^{3} + \frac{1}{1^{4}} + c \implies 4 = 2 + c\)
\(c = 2\)
\(f(x) = x^{3} + \frac{1}{x^{4}} + 2\)