2013 - WAEC Mathematics Past Questions and Answers - page 3
p = (y : 2y (\geq) 6)
2y (\leq) 6
y (\leq \frac{6}{2})
y = (\leq) 3
and Q = (y : y -3 (\geq) 4)
y - 3 (\geq) 4
y (\geq) 4 + 3
y (\geq) 7
therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....}
P(\cap)Q = {3, 4, 5, 6, 7}
6k2 = 5k + 6
6k2 - 5k - 6 = 0
6k2 - 0k + 4k - 6 = 0
3k(2k - 3) + 2(2k - 3) = 0
(3k + 2)(2k - 3) = 0
3k + 2 = 0 or 2k - 3 = 0
3k = -2 or 2k = 3
k = (\frac{-2}{3}) or k = (\frac{3}{2})
k = ((\frac{-2}{3}), k = (\frac{3}{2}))
y (\alpha) \sqrt{x + 1}), y = k\sqrt{x + 1})
6 = k(\sqrt{3 + 1})
6 = k(\sqrt{4})
6 = 2k
k = (\frac{6}{2}) = 3
y = (\sqrt{(x + 1)})
9 = 3(\sqrt{(x + 1)})(divide both side by 3)
(\frac{9}{3}) = (\frac{3\sqrt{x + 1}}{3})
3 = (\sqrt{x + 1})(square both sides)
9 = x + 1
x = 9 - 1
x = 8
y = x2 + 2x + k at point(2,0) x = 2, y = 0
0 = (2)2 + 2(20 + k)
0 = 4 + 4 + k
0 = 8 + k
k = -8
Let the number be y, subtract y from 2 i.e 2 - y
2 - y = 4 < (\frac{1}{5}) x y, 2 - y = 4 < (\frac{y}{5})
2 - y - 4 < (\frac{y}{5})
2 - 4 - y (\frac{x}{5} - 4), multiplying through by 5
5(2 - x) = 5((\frac{x}{5})) - 5(4)
10 - 5x = x - 20
-5x - x = -20 - 10
-6x = -30
x = (\frac{-30}{-6})
= 5
(\frac{2}{x + 3} - \frac{1}{x - 2}) = (\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)})
= (\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6})
= (\frac{x -7}{x^2 + x - 6})
= (\frac{x - 7}{x^2 + x - 6})
Let the interior angle = xo
interior angle = 5xo (sum of int. angle ann exterior)
(angles = angle or straight line)
6x = 180
x = (\frac{180}{6})
x = 30o
no. of sides = (\frac{\text{sum of exterior angles}}{\text{exterior angle}})
= (\frac{360}{30}) = 12
P = x2 + 4x - 2, Q = 2x - 1
Q - p = 2, (2x - 1) - (x2 + 4x - 2) = 2
2x - 1 - x2 - 4x + 2 = 2
-2x - x2 + 1
-x2 - 2x - 1 = 0
x2 + 2x + 1 = 0
x2 + x + x + 1 = 0
x(x + 1) + 1(x + 1) = 0
(x + 1)(x + 1) = 0
x + 1 = 0 or x + 1 = 0
x = -1 or x = -1
x = -1
Volume of pyramid = (\frac{1}{3}) x base area x height
= (\frac{1}{3} \times 12^4 \times 8 \times 14)
= 4 x 8 x 14 = 448m3