Further Mathematics Questions and Answers

$x^{2} + 4x + k = (x + r)^{2} + 1$

$x^{2} + 4x + k = x^{2} + 2rx + r^{2} + 1$

Comparing the LHS and RHS equations, we have

$2r = 4 \implies r = 2$

$k = r^{2} + 1 = 2^{2} + 1 = 5$

Given, $x^{2} + x - 2 = 0$, a = 1, b = 1 and c = -2.

$\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1$

$\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2$

$\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\beta^{2} + \alpha^{2}}{(\alpha\beta)^{2}}$

$\beta^{2} + \alpha^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = (-1)^{2} - 2(-2) = 1 + 4 = 5$

$\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{5}{(-2)^{2}} = \frac{5}{4}$.

The gradient of a curve is gotten by differentiating the equation of the curve. Therefore, given the gradient, integrate to get the equation of the curve back.

$\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4x$

$y = \int {(3x^{2} - 4x)} \mathrm {d} x = \frac{3x^{2+1}}{2+1} - \frac{4x^{1+1}}{1+1} + c$

= $x^{3} - 2x^{2} + c$ 

To find c (the constant of integration), when x = -2, y = 0

$0 = (-2^{3}) - 2(-2^{2}) + c$

$0 = -8 - 8 + c \implies c = 16$

$\therefore y = x^{3} - 2x^{2} + 16$

$\overrightarrow{x} . \overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta$

$\overrightarrow{x} . \overrightarrow{y} = (i - 3j) . (6i + j) = 6 - 3 = 3$

$|\overrightarrow{x}| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}$

$|\overrightarrow{y}| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$

$\therefore 3 = (\sqrt{10})(\sqrt{37})\cos \theta$

$\cos\theta = \frac{3}{\sqrt{370}} = 0.1559$

$\theta = \cos^{-1} 0.1559 \approxeq 81°$

Since 2 students must be included, we have to arrange the remaining 3 students from the 6 students left

= $^{6}C_{3} = \frac{6!}{(6-3)!3!}$

= 20 ways