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})