Further Mathematics Questions and Answers

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.

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

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

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

= $10x + 1$

The nth term of a GP is given by: $T_{n} = ar^{n-1}$.

$T_{3} = ar^{3-1} = ar^{2} = 81$.......(1)

$T_{7} = ar^{7-1} = ar^{6} = 16$ ...... (2)

Dividing (2) by (1), we have $r^{4} = \frac{16}{81} = (\frac{2}{3})^{4} \implies r = \frac{2}{3}$

Putting $r = \frac{2}{3}$ in equation (1), we have $81 = a \times (\frac{2}{3}$^{2} = a \times \frac{4}{9} \implies a = \frac{729}{4}\)

$T_{5} = ar^{5-1} = ar^{4} = \frac{729}{4} \times (\frac{2}{3})^{4}$

= $\frac{729}{4} \times \frac{16}{81} = 36$