2009 - JAMB Mathematics Past Questions and Answers - page 9

x(4 - x)

4x - x²

\(\frac{dy}{dx}\) = 4 - 2x

\(\frac{dy}{dx}\) = 0

2x = 4

x = \(\frac{4}{2}\)

= 2

S = (t³ - t² - t + 5)cm

\(\frac{ds}{dt}\) = 3t² - 2t - 1

= (3t + 1)(t - 1)

t = -\(\frac{1}{3}\) or t = 1

Substitute value of t

s = (1³ - 1² - 1 + 5)cm

= 4.0 cm

\(\int\) sec²\(\theta\) d \(\theta\) = tan\(\theta\) + k

3x + 170 = 260

3x = 260 - 170

3x = 90 \(\to\) x = 30

therefore, 20 + 2x + 60 + 40

= 180

\(\begin{array}{c|c} x & f & fx\ \hline 5 & 3 & 15\ 8 & 2 & 16 \ 6 & 4 & 24\ k & 1 & k\ & \sum f & \sum fx = 55 + k\end{array}\)

mean \(\bar{x}\) = \(\frac{55 + k}{10}\)

= 5.7

k = 2

deviation x - \(\bar{x}\)

x - 2 + 2x - 2 + (x + 1) - 2 + 3x - 2 = 0

x - 2 + 2x - 2 -2x - 2 + 3x - 2 = 0

= 4x - 8

4x = 8

x = \(\frac{8}{4}\)

= 2

\(\frac{n!}{n - r!}\) = \(\frac{9!}{9 - 3!}\)

= \(\frac{9!}{6!}\)

= 9 x 8 x 7

= 502

\(\frac{\begin{pmatrix} 8 \ 3 \end{pmatrix}}{\begin{pmatrix} 4 \ 1 \end{pmatrix}\begin{pmatrix} 4\ 1\end{pmatrix}}\)(Two boys and a girl) OR \(\frac{\begin{pmatrix} 8 \ 3 \end{pmatrix}}{\begin{pmatrix} 4 \ 1 \end{pmatrix}\begin{pmatrix} 4\ 1\end{pmatrix}}\)(Two girls and a boy in the committee)

\(\frac{336}{24}\) or \(\frac{336}{24}\)

14 + 14

= 28

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

p(fail) = \(\frac{1}{3}\)

p(pass at 5 marks) = \(\frac{1}{3}\) x \(\frac{1}{3}\) x \(\frac{1}{3}\)

= \(\frac{1}{27}\)

p (pass at 5mark) = \(\frac{12}{20}\)

= \(\frac{3}{5}\)