Further Mathematics Questions and Answers

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

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

\(\implies x^{2} + y^{2} - 2ax - 2by = r^{2} - a^{2} - b^{2}\)

Comparing with the given equation: \(3x^{2} + 3y^{2} + 24x - 12y = 15\)

Making the coefficients of \(x^{2}\) and \(y^{2}\) = 1, we have

\(x^{2} + y^{2} + 8x - 4y = 5\)

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

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

\(r^{2} - a^{2} - b^{2} = 5 \implies r^{2} = 5 + (-4)^{2} + (2)^{2} = 5 + 16 + 4 = 25\)

\(\therefore r = 5\)

Given: \(f(x + 1) = x^{3} + 3x^{2} - 4x + 2\). 

\(f(2) = f(x + 1) \implies x + 1 = 2; x = 1\)

\(f(2) = 1^{3} + 3(1)^{2} - 4(1) + 2 = 1 + 3 - 4 + 2 = 2\)

If (x + 1) is a factor, then f(-1) = 0.

\((-1)^{3} + p(-1)^{2} + (-1) + 6 = 0\)

\(-1 + p - 1 + 6 = 0 \implies p + 4 = 0\)

\(p = -4\)

\(\overrightarrow{SQ} = \overrightarrow{SR} + \overrightarrow{RQ}\)

\(\overrightarrow{RQ} = -\overrightarrow{QR} = - (3i + 2j) = -3i - 2j\)

\(\overrightarrow{SQ} = (-5i + 3j) - 3i - 2j = -8i + j\)

\(\log_{10} (\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})\)

\(\frac{1}{3} + \frac{1}{4} = \frac{7}{12}\)

= \(\log_{10} (\frac{7}{12} \times 2^{2} \times \frac{3}{7})\)

= \(\log_{10} 1 = 0\)