Further Mathematics Questions and Answers

\(f(x) = 3x^{3} + 8x^{2} + 6x + k\)

\(f(2) = 3(2^{3}) + 8(2^{2}) + 6(2) + k = 1\)

\(\implies 24 + 32 + 12 + k = 1\)

\(68 + k = 1  \therefore k = 1 - 68 = -67\)

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

\((3 * b) = \frac{3+b}{2}\)

\(x \circ y = \frac{x^{2}}{y}\)

\((\frac{3+b}{2}) \circ 48 = \frac{(\frac{3+b}{2})^{2}}{48} = \frac{1}{3}\)

\(\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}\)

\((3 + b)^{2} = \frac{48 \times 4}{3} = 64\)

\(b^{2} + 6b + 9 = 64 \implies b^{2} + 6b + 9  - 64 = 0\)

\(b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0\)

\(b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (} b + 11) = 0\)

Since b > 0, b - 5 = 0 

b = 5.

When you have two lines, \(y_{1}, y_{2}\), perpendicular to each other, the product of their slopes = -1.

\(3x + 4y + 6 = 0 \implies 4y = -6 - 3x\)

\(\therefore y = \frac{-6}{4} - \frac{3}{4}x\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{-3}{4}\)

Also, \(4x - by + 3 = 0  \implies by = 4x + 3\)

\(y = \frac{4}{b}x + \frac{3}{b}\) 

\(\frac{\mathrm d y}{\mathrm d x} = \frac{4}{b}\)

\(\frac{-3}{4} \times \frac{4}{b} = -1 \implies \frac{4}{b} = \frac{4}{3}\)

\(b = 3\)

\((1 + \sin \theta)(1 - \sin \theta) = 1 - \sin \theta + \sin \theta - \sin^{2} \theta\)

\(= 1 - \sin^{2} \theta\)

Recall, \(\cos^{2} \theta + \sin^{2} \theta = 1\)

\(\therefore 1 - \sin^{2} \theta = \cos^{2} \theta\).

\(\frac{1}{5^{-y}} = 25(5^{4-2y})\)

\(\implies 5^{y} = (5^{2})(5^{4-2y})\)

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

Comparing bases, we have

\(y = 6 - 2y\)

\(3y = 6 \implies y = 2\)