JAMB Mathematics Past Questions & Answers - Page 305

P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)

Using the binomial prob. distribution,

(H + T)3 = H3 + 3H2T1 + 3HT2 + T3

Hence the probability that three heads show in a toss of the three coins is H3

= (\(\frac{1}{2}\))3

= \(\frac{1}{8}\)

A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in \(^{5}C_{2}\) x \(^{6}C_{3}\) ways

= \(\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}\)

= \(\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}\)

= \(\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}\)

= \(\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}}\)

= 10 x \(\frac{6 \times 20}{6}\)

= 200

\(\int^{2}_{0}(x^3 + x^2)\)dx = \(\int^{2}_{0}\)(\(\frac{x^4}{4} + {\frac {x^3}{3}}\))

= (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\))

= (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0

= \(\frac{80}{12}

= {\frac {20}{3}}\) or 6\(\frac{2}{3}\)

If y = x sinx, then

Let u = x and v = sinx

\(\frac{du}{dx}\) = 1 and \(\frac{dv}{dx}\) = cosx

Hence by the product rule,

\(\frac{dy}{dx}\) = v \(\frac{du}{dx}\) + u\(\frac{dv}{dx}\)

= (sin x) x 1 + x cosx

= sinx + x cosx

cot\(\theta\) = \(\frac{1}{\cos \theta}\)

= \(\frac{8}{15}\)(given)

tan\(\theta\) = \(\frac{15}{18}\)

By Pythagoras theorem,

x2 = 152 + 82

x2 = 225 + 64 = 289

x = \(\sqrt{289}\)

= 17

Hence sin\(\theta\) = \(\frac{15}{x}\)

= \(\frac{15}{17}\)