WAEC Mathematics Past Questions & Answers - Page 64

Since the curve cuts the x-axis at x = -2 and x = 3,

(x + 2)(x - 3) = 0

\(x^2 - 3x + 2x - 6\) = 0

\(x^2 - x - 6\) = 0

Hence, the equation of the curve is

y = \(x^2 - x - 6\)

\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\)

= \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\)

= \(2(1 - 3m)\)

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)

\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)

Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)

y = \(\frac{3}{7}T_2\) and z =  \(\frac{4}{7}T_2\)

Using y = y

\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)

\(\frac{T_1}{T_2}\)  = \(\frac{15}{21}\)

\(T_1\) = 15 and \(T_2\) = 21

Therefore;

x = \(\frac{2}{5}\) x 15 = 6

y = \(\frac{3}{5}\) x 15 = 9

y = \(\frac{3}{7}\)  x 21 = 9 (again)

z = \(\frac{4}{7}\) x 21 = 12

Hence;

\(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\)

\(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\)

= 1\(\frac{1}{2}\)

y + 2x = 4 .....(1)

y - 3x = -1 ......(2)

Subtract (2) from (1)

2x - (-3x) = 4 - (-1)

2x + 3x = 4 + 1

5x = 5

X = \(\frac{5}{5}\)

= 1

Substitute 1 for x in (1);

y + 2(1) = 4

y + 2 = 4

y = 4 - 2 = 2

Hence, (x + y) = (1 + 2)

= 3

F = \(\frac{9}{5}\)C + 32

When F = 98.6

98.6 = \(\frac{9}{5}\)C + 32

98.6 - 32 = \(\frac{9}{5}\)C

66.6= \(\frac{9}{5}\)C

66.6 x 5 = 9C

C = \(\frac{66.6 \times 5}{9}\)

= 37