1994 - JAMB Mathematics Past Questions and Answers - page 4

Sin(-690^o) = Sin(-360 -3300

sin - 360 = sin 0

∴ sin(-690^o) = sin(330^o)

Negative angles are measured in clockwise direction

The acute angle equivalent of sin(-330^o) = sin(30^o)

sin(-330^o) = sin(30^o)

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

= 0.5

y = 3t³ + 2t² - 7t + 3

\(\frac{dy}{dt}\) = 9t² + 4t - 7

When t = -1

\(\frac{dy}{dt}\) = 9(-1)² + 4(-1) - 7

= 9 - 4 -7

= 9 - 11

= -2

Equation of curve;

y = 2x² - 2x + 3

gradient of curve;

\(\frac{dy}{dx}\) = differential coefficient

\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2

∴ \(\frac{dy}{dx}\) = 2

4x - 2 = 2

4x = 4

∴ x = 1

When x = 1, y = 2(1)² - 2(1) + 3

= 2 - 2 + 3

= 5 - 2

= 3

coordinate of the point where the curve; y = 2x² - 2x + 3 has gradient equal to 2 is (1, 3)

\(\int^{1}_{-1}\)(2x + 1)²dx

= \(\int^{1}_{-1}\)(4x² + 4x + 1)dx

= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3}\) + 2x² + c]

= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]

= \(\frac{8}{3}\) + 3 + -1 - C

= \(\frac{8}{3}\) + 2

= \(\frac{14}{3}\)

= \(\frac{42}{3}\)

x = \(\frac{\sum x}{N}\) = \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} x & x - x & x - x \ \hline 4 & -2 & 2\ 5 & 1 & 1\ 9 & 3 & 3\ \hline & & 6\end{array}\) M.D = \(\frac{|x - x|}{N}\) = \(\frac{6}{3}\) = 2

Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h N = Sum of frequencies L = lower class boundary of median class f = sum of all frequencies below L fm = frequency of modal class and h = class width of median class Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5 = 11 + (\(\frac{25 - 21}{20}\))5 = 11 + (\(\frac{(4)}{20}\)) 11 + 1 = 12

\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \ \hline 1 & 2 & 2 & -2 & 4 & 8\ 2 & 1 & 2 & -1 & 1 & 1\ 3 & 2 & 6 & 0 & 0 & 0\ 4 & 1 & 4 & 1 & 1 & 1\ 2 & 2 & 10 & 2 & 4 & 8\ \hline & 8 & 24 & & & 18 \end{array}\)

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

= \(\frac{24}{8}\)

= 3

Variance (6²) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)

= \(\frac{18}{8}\)

= \(\frac{9}{4}\)

40 = 20 - x + x + 35 - x 40 = 55 - x x = 55 - 40 = 15 ∴ P(both) \(\frac{15}{40}\) = \(\frac{3}{8}\)

y = x³ - 27, y = -27 \(\to\) (0, -27)

when y = 0, x = 3 (3, 0)