Further Mathematics Questions and Answers

Let $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T^{-1}$

$T . T^{-1} = I$

$\begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$\implies -2a - 5c = 1$

$-2b - 5d = 0 \implies b = \frac{-5d}{2}$

$3a + 8c = 0 \implies a = \frac{-8c}{3}$

$3b + 8d = 1$

$-2(\frac{-8c}{3}) - 5c = \frac{16c}{3} - 5c = \frac{c}{3} = 1 \implies c = 3$

$3(\frac{-5d}{2}) + 8d = \frac{-15d}{2} + 8d = \frac{d}{2} = 1 \implies d = 2$

$b = \frac{-5 \times 2}{2} = -5$

$a = \frac{-8 \times 3}{3} = -8$

$\therefore  T^{-1} = \begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}$

$y = \sqrt[3]{3x^{3} + 1}  = (3x^{3} + 1)^{\frac{1}{3}}$

Let u = $3x^{3} + 1$; y = $u^{\frac{1}{3}}$

$\frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})$

$\frac{\mathrm d y}{\mathrm d u} = \frac{1}{3}u^{\frac{-2}{3}}$

$\frac{\mathrm d u}{\mathrm d x} = 9x^{2}$

$\frac{\mathrm d y}{\mathrm d x} = (\frac{1}{3}(3x^{3} + 1)^{\frac{-2}{3}})(9x^{2})$

= $\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}$ 

$\frac{x + P}{(x-1)(x-3)} = \frac{Q}{x-1} + \frac{2}{x-3}$

$\frac{x + P}{(x-1)(x-3)} = \frac{Q(x-3) + 2(x-1)}{(x-1)(x-3)}$

Comparing LHS and RHS of the equation, we have

$x + P = Qx - 3Q + 2x -2$

$P = -3Q - 2$

$Q + 2 = 1 \implies Q = 1 - 2 = -1$

$P = -3(-1) - 2 = 3 - 2 = 1$

$P + Q = 1 + (-1) = 0$

$p(blue) = \frac{\text{no of blue balls}}{\text{total no of balls}}$

= $\frac{2}{3} = \frac{k}{k + 5}$

$3k = 2(k + 5)  \implies 3k = 2k + 10$

$3k - 2k = k = 10$

Given an exponential sequence, say $a, b, c,...$, as consecutive terms, then $\sqrt{a \times c} = b$.

$\therefore 2, (k+1), 8 \implies \sqrt{2 \times 8} = k + 1$

$k + 1 = \pm{4} \implies k = \text{-5 or 3}$