Further Mathematics Questions and Answers

\(f(x) = 2x - 1; g(x) = x^{2} + 1\)

\(f(x) = 2x - 1\)

\(f(y) = 2y - 1 \)

\(x = 2y - 1 \implies 2y = x + 1\)

\(y = \frac{x + 1}{2}\)

\(g(3) = 3^{2} + 1 = 10\)

\(f^{-1}(10) = \frac{10 + 1}{2} = \frac{11}{2}\)

\(S_{n} = \frac{n}{2}(2a + (n - 1)d)\) ( for an arithmetic progression)

\(S_{3} = 18 = \frac{3}{2}(2(4) + (3 - 1) d) \)

\(18 = \frac{3}{2} (8 + 2d)\)

\(18 = 12 + 3d \implies 3d = 6\)

\(d = 2\)

\(\therefore T_{1} = 4 \implies T_{2} = 4 + 2 = 6; T_{3} = 6 + 2 = 8\)

Their product = \(4 \times 6 \times 8 = 192\)

No explanation has been provided for this answer.

Given \(ax^{2} + bx + c = 0 (\text{general form of a quadratic equation})\)

We have \(x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\)

\( x^{2} - (-\frac{b}{a})x + \frac{c}{a} = 0\)

\(\implies x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0\)

\(\therefore \text{The equation is} x^{2} - 3x + 2 = 0\) 

Two lines are parallel if and only if their slopes are equal.

\(kx - 5y + 6 = 0 \implies 5y = kx + 6\)

\(y = \frac{k}{5}x + \frac{6}{5}\)

\(Slope = \frac{k}{5}\)

\(mx + ny - 1 = 0 \implies ny = 1 - mx\)

\(y = \frac{1}{n} - \frac{m}{n}x\)

\(Slope = -\frac{m}{n}\)

\(Parallel \implies \frac{k}{5} = -\frac{m}{n}\)

\(-5m = kn \implies 5m + kn = 0\)