1992 - JAMB Mathematics Past Questions and Answers - page 4

\(\frac{dm}{dt}\) = 4 - 0.6t

\(\int\)dm = \(\int\)(4 - 0.6t)dt

m = 4t - 0.3t² + c, when t = 0, m = 2g

∴ c = 2

m = 4t - 0.3t² + 2, when t = 5 minutes

m = 4(5) - 0.3(5)² + 2 = 20 - 7.5

= 14.5

f(x) = x³ - 12x + 11

\(\frac{df(x)}{dx)}\) = 3x² - 12 = 0

∴ 3x² - 12 = 0 \(\to\) x²m = 4

x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15

= f(-2) = (-8) + 24 + 11

= 35 - 8 = 27

∴ maximum value = 27

\(\frac{dv}{dt}\) = \(\pi\)(20 - t²)cm²S^-1

\(\int\)dv = \(\pi\)(20 - t²)dt

V = \(\pi\) \(\int\)(20 - t²)dt

V = \(\pi\)(20 \(\frac{t}{3}\) - t³) + c

when c = 0, V = (20t - \(\frac{t^3}{3}\))

after t = 2 seconds

V = \(\pi\)(40 - \(\frac{8}{3}\)

= \(\pi\)\(\frac{120 - 8}{3}\)

= \(\frac{112}{3}\)

= 37.33\(\pi\)

let I = \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}}\) 2 cos 2x dx

= 2(sin 2x)^{\(\frac{\pi}{4}\)} (sin 2x)^{\(\frac{\pi}{4}\)}_{\(\frac{\pi}{12}\)}

(2)\(\frac{\pi}{12}\)

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

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

H will represent \(\frac{108}{720}\) x \(\frac{360^o}{1}\) = 54^o

Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)

= \(\frac{5.2}{1}\)

= \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\)

= \(\frac{5.2}{1}\)

= \(\frac{84 + 4y}{15 + y}\)

= 5.2(15 + y)

= 84 + 4y

= 5.2 x 15 + 5.2y

= 84 + 4y

= 78 + 5.2y

= 84 = 4y

= 5.2y - 4y

= 84 - 78

1.2y = 6

y = \(\frac{6}{1.2}\)

= \(\frac{60}{12}\)

= 5

(x) = \(\frac{5 + 6 + 7}{3}\)

= \(\frac{18}{3}\)

= 6

\(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\hline 5 & 5 - 6 & 1\ 6 & 6 - 6 & 0 \ 7 & 7 - 6 & 1\ \hline & & 2\end{array}\)

S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x)

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