Further Mathematics Questions and Answers

\(f(x) = 5x^{2} - 4x + 3\)

\(f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1\)

When x = 1, f(x) = y = \(5(1^{2}) - 4(1) + 3 = 4\)

The coordinates are (1,4)

On the y-intercept, x=0. At this point, y = 5(0) - 2y = -6; y = 3, so the y- intercept has coordinates (0,3).

The equation of the circle is given as \((x-a)^{2} + (y-b)^{2} = r^{2}\), where (a,b) is the centre and r is the radius.

The radius of the circle through the y- intercept = \(\sqrt{(4-0)^{2} + (5-3)^{2}} = \sqrt{20}\)

The equation of the circle is \((x-4)^{2} + (y-5)^{2} = 20\); expanding, we have

\(x^{2} + y^{2} + 8x - 10y + 21 = 0\)

\(\cos(x+y) = \cos x\cos y - \sin x\sin y\)

Given \(\sin\) of an angle implies we have the value of the opposite and hypotenuse of the right-angled triangle. We find the adjacent side using Pythagoras' theorem.

\(Adj^{2} = Hyp^{2} - Opp^{2}\)

For triangle with angle x, \(adj = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12\)

For triangle with angle y, \(adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15\)

\(\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}\)

\(\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}\)

= \(\frac{140}{221}\)

\(B.B^{-1} = 1\), let \(B^{-1} = \begin{pmatrix}  a & b  \\  c & d  \end{pmatrix}\)

\(\begin{pmatrix}  2 & 5  \\  1 & 3  \end{pmatrix}\)\(\begin{pmatrix}  a & b  \\  c & d  \end{pmatrix}\) = \(\begin{pmatrix}  1 & 0  \\  0 & 1  \end{pmatrix}\)

Multiplying \(B \times B^{-1}\), we have the following equations:

\(2a+5c = 1......... (1)\); \(a+3c = 0 ........(2)\)

\(2b+5d = 0 ..........(3)\); \(b+3d = 1..........(4)\)

Solving the equations simultaneously, we have

\(a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix}  3 & -5  \\  -1 & 2  \end{pmatrix}\)

\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}\)

\(\alpha + \beta = \frac{-b}{a}\); \(\alpha\beta = \frac{c}{a}\)

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

From the equation, a = 2, b = -3, c =4

\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)

\(\alpha\beta = \frac{4}{2} = 2\)

\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)

\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)