Further Mathematics Questions and Answers

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix}  2 & 3 \\ -1 & k \end{vmatrix} = 6\)

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)\)

\(\begin{vmatrix}  2 & 3 \\ -1 & k \end{vmatrix} = (2k + 3)\)

\(\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6\)

\(k^{2} - 2k - 3 = 0\), factorising, we have \(k + 1 = 0\) or \(k - 3 = 0\)

Since k > 0, k = 3.

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 \(u_{r} = -5r + 14\).

\(x - \frac{3}{x^{2}} = x - 3x^{-2}\)

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

\(^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})\)

= \(\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}\)

\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)

Recall, \( 1 = \log_{2}2\), so

\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)

= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)

\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)

\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)

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 \(\alpha\beta = \frac{n}{2} = 2 \implies n = 4\)