WAEC Further Mathematics Past Questions & Answers - Page 139

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}\)

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

= \(\frac{729}{4} \times \frac{16}{81} = 36\)