Further Mathematics Questions and Answers

\(a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\); \(b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\)

\(\implies 2 \times a = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) and \(\frac{1}{4} \times b = \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\)

\(\therefore 2a - \frac{1}{4}b = \begin{pmatrix} 4 - \frac{-1}{4} \\ 6 - 1 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)

Sample size = 6 x 6 = 36.

Let A and B be the sum for the boys and girls respectively.

\(\frac{A + B}{10 + 15} = \frac{A + B}{25} = 90\)

\(\implies A + B = 90 \times 25 = 2250\)

Given the average for girls = x, we have \(\frac{B}{15} = x \implies B = 15x)

\(\therefore A + 15x = 2250; A = 2250 - 15x \implies\) average score for boys \(= \frac{2250 - 15x}{10}\)

= \(225 - \frac{3x}{2}\) 

The line of symmetry of the curve is at the minimum point of the curve (ie y' = 0)

\(\frac{ \mathrm d}{ \mathrm d x} \left ( 5-x-2x^{2} \right)\) = -1 - 4x

If y' = 0, we have \(-1 - 4x = 0 \implies 4x = -1\)

\(x = \frac{-1}{4}\)

This can be done either by using quotient rule or by direct division of the equation, then differentiate.

\(\frac{\mathrm d}{\mathrm d x} \left( \frac{5x^{3} + x^{2}}{x} \right)\)     

= \(\frac{\mathrm d}{\mathrm d x} \left ( \frac{5x^{3}}{x} + \frac{x^{2}}{x} \right)\)

= \(\frac{\mathrm d}{\mathrm d x} \left ( 5x^{2} + x \right)\)

= \(10x + 1\)