Further Mathematics Questions and Answers

\(f(x) = p + qx\)

\(f(1) = p + q(1) \implies p + q = 7 .... (1)\)

\(f(5) = p + 5q = 19 .....(2)\)

Solving for p and q using simultaneous equation, p = 4, q = 3

\(f(3) = 4 + 3(3) = 13\)

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding this, we have \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the given equation, \(3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0\) (making the coefficients of \(x^{2}\) and \(y^{2}\) = 1 , we get that

\(-2a = 2 \implies a = -1\)

\(2b = 4 \implies b = 2\)

\(r^{2} - a^{2} - b^{2} = -2\)

\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3\)

\(r = \sqrt{3}\)

\(3x^{2} + 4x + 1 > 0 \)

\(3x^{2} + 3x + x + 1 > 0\)

\(3x(x + 1) + 1(x + 1) > 0\)

\((3x + 1)(x + 1) > 0\)

\(3x + 1 > 0 \implies 3x > -1 \)

\(x > -\frac{1}{3}\)

\(x + 1 > 0 \implies x > -1\)

\(x > -1, x > -\frac{1}{3}\)

\(\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{\frac{-1}{2}} \)

\(\frac{2}{3} - (\frac{9}{4})^{\frac{1}{2}}\)

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

= \(\frac{-5}{6}\)

\(f(x) = \frac{3x + 1}{x^{2} - 1}\)

\(f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1\)