WAEC Further Mathematics Past Questions & Answers - Page 110

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 \(\sqrt{a \times c} = b\).

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

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