a = 5, b = 14
a = -5, b = 14
a = 5, b = -14
a = -5, b = -14
Correct answer is B
The terms of the sequence can be written as : \(u_{r} = ar + b\) in this case, being that they have a regular common difference for each of the r terms.
We can rewrite the sequence as \(a + b, 2a + b, 3a + b,...\) where a is the common difference of the sequence and b is a given constant gotten by solving
\(a + b = 9\) or \(2a + b = 4\) or any other one.
The common difference here is 4 - 9 = -1 - 4 = -5.
\(-5 + b = 9 \implies b = 9 + 5 = 14\)
\(\therefore\) The equation can be written as \(u_{r} = -5r + 14\).
Find the coefficient of \(x^{3}\) in the binomial expansion of \((x - \frac{3}{x^{2}})^{9}\).
324
252
-252
-324
Correct answer is A
\(x - \frac{3}{x^{2}} = x - 3x^{-2}\)
Let the power on x be t, so that the power on \(x^{-2}\) = 9 - t
\((x)^{t}(x^{-2})^{9 - t} = x^{3} \implies t - 18 + 2t = 3\)
\(3t = 3 + 18 = 21 \therefore t = 7\)
To obtain the coefficient of \(x^{3}\), we have
\(^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})\)
= \(\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}\)
Solve \(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)
4.75
4.00
1.75
1.00
Correct answer is B
\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)
Recall, \( 1 = \log_{2}2\), so
\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)
= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)
\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)
\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)
-4
-2
2
4
Correct answer is D
An equation can be written as \(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\)
Making the coefficient of \(x^{2}\) = 1 in the given equation, we have
\(x^{2} + \frac{5}{2}x + \frac{n}{2} = 0\)
Comparing, we have \(\alpha\beta = \frac{n}{2} = 2 \implies n = 4\)
Resolve \(\frac{3x - 1}{(x - 2)^{2}}, x \neq 2\) into partial fractions.
\(\frac{x}{2(x - 2)} - \frac{5}{(x - 2)^{2}}\)
\(\frac{5}{(x - 2)} + \frac{x}{2(x - 2)^{2}}\)
\(\frac{1}{2(x - 2)} + \frac{5x}{2(x- 2)^{2}}\)
\(\frac{-1}{2(x - 2)} + \frac{8x}{2(x - 2)^{2}}\)
Correct answer is C
\(\frac{3x - 1}{(x - 2)^{2}} = \frac{A}{(x - 2)} + \frac{Bx}{(x - 2)^{2}}\)
\(\frac{3x - 1}{(x - 2)^{2}} = \frac{A(x - 2) + Bx}{(x - 2)^{2}}\)
Comparing, we have
\(3x - 1 = Ax - 2A + Bx \implies -2A = -1; A + B = 3\)
\(\therefore A = \frac{1}{2}; B = \frac{5}{2}\)
= \(\frac{1}{2(x - 2)} + \frac{5x}{2(x - 2)^{2}}\)