WAEC Mathematics Past Questions & Answers - Page 100

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

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

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 = \(\frac{3}{2}\))

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}

2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2)

\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\)

\(\log_x \frac{27}{8} = 3\)

\(\frac{27}{8} = x^3\)

\(x^3 = \frac{27}{8}\)

x = \(\sqrt{\frac{27}{8}}\)

= (\(\frac{27}{8}\))^{\(\frac{1}{3}\)}

= \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\)

= \(\frac{3}{2}\)