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