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$\theta$ = $\frac{15}{x}$

= $\frac{15}{17}$