Further Mathematics Questions and Answers

Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Given that \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)

Expanding the equation of a circle, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Comparing this expansion with the given equation, we have

\(2a = 4 \implies a = 2\)

\(-2b = 8 \implies b = -4\)

\(r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9\)

\(r = 3\)

\(Area = \pi r^{2} = \pi \times 3^{2}\)

= \(9\pi\)

\(m . n = |m||n|\cos \theta\)

\((3i + 4j) . (2i - j) = 6 - 4 = 2\)

\(2 = |(3i + 4j)||(2i - j)| \cos \theta\)

\(|3i + 4j| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)

\(|2i - j| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\)

\(2 = 5(\sqrt{5})(\cos \theta)\)

\(\cos \theta = \frac{2}{5\sqrt{5}} = 0.08\sqrt{5}\) 

\(\theta = \cos^{-1} 0.1788 = 79.7°\)

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 125\)

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 5^{3} = 3\log_{5} 5 = 3\)

\(\log_{2} y^{\frac{1}{2}} = 3 \implies y^{\frac{1}{2}} = 2^{3} = 8\)

\(y = 8^{2} = 64\)

\(y = 4x^{3} + kx^{2} - 6x + 4\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)

At P(1, m)

\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)

\(6 + 2k = 0 \implies k = -3\)

\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)

= -1

P = (1, -1)

\(y = 4x^{3} + kx^{2} - 6x + 4\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)

At P(1, m)

\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)

\(6 + 2k = 0 \implies k = -3\)