JAMB Mathematics Past Questions & Answers - Page 261

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

H will represent \(\frac{108}{630}\) x \(\frac{360^o}{1}\) ≈ 62°

\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)

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

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

= \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\)

= \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\)

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

\(\frac{dv}{dt}\) = \(\pi\)(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\(\pi\)

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