WAEC Further Mathematics Past Questions & Answers - Page 2

$P(A)=\frac{1}{6},P(T)=\frac{1}{9}$

Probability that only one of them will hit the target = $P(A)\times P( \bar T ) + P( \bar A )\times P(T)$

Where $P( \bar T )$ is the probability that Tunde will not hit the target and $P( \bar A )$ is the probability that Atta will not hit the target

$P( \bar T )=1-\frac{1}{9}=\frac{8}{9}$

$P( \bar A )=1-\frac{1}{6}=\frac{5}{6}$

Pr(only one) =$(\frac{1}{6}\times\frac{8}{9}) + (\frac{5}{6} \times \frac{1}{9}) =\frac{4}{27} + \frac{5}{54}$

$\therefore$ pr (only one) = $\frac{13}{54}$

$f :x→\frac{x + 2}{x - 3},x ≠ 3, f = ?$

Let $f :x=y$

$y=\frac{x + 2}{x - 3}$

$=x+2=y(x-3)$

$=x-xy=-3y-2$

$=x(1-y)=-3y-2$

$=x=\frac{-3y - 2}{1 - y}=\frac{-(3y + 2)}{- (y - 1)}$

$=x=\frac{3y + 2}{y - 1}$

$∴f ^{-1} : x=\frac{3x + 2}{x - 1},x ≠ 1$

$P(X⋃Y)=\frac{5}{8}$

$P(X⋂Y)=P(X)\times P(Y)$

Since X and Y are independent events, the probability of their union (X ⋃ Y) can be calculated as:

$P(X⋃Y)=P(X)+P(Y)-P(X⋂Y)$

$=\frac{5}{8}=\frac{1}{8}+P(Y)-\frac{1}{8}\times P(Y)$

$=\frac{5}{8}-\frac{1}{8}=P(Y)-\frac{1}{8}\times P(Y)$

$=\frac{1}{2}=P(Y)(1-\frac{1}{8})$

$=\frac{1}{2}=P(Y)(\frac{7}{8})$

$=P(Y)=\frac{1}{2}÷\frac{7}{8}$

$∴P(Y)=\frac{1}{2}x\frac{8}{7}=\frac{4}{7}$

$y^2 + xy = 5$

By implicit differentiation

$=2y\frac{dy}{dx}+y+x\frac{dy}{dx}=0$

$=2y\frac{dy}{dx}+x\frac{dy}{dx}=-y$

Factor out $\frac{dy}{dx}$

$=\frac{dy}{dx}(2y+x)=-y$

$∴\frac{dy}{dx}=\frac{-y}{2y + x}$

$P : (x, y) → (2x + y, -2y)$

$p\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} 2x & y\\0 &-2y\end{bmatrix}$

$\therefore p = \begin{bmatrix} 2 & 1\\0 &-2\end{bmatrix}$

$\therefore p^2 = \begin{bmatrix} 2&1\\0&-2\end{bmatrix}$ $\begin{bmatrix} 2&1\\0&-2\end{bmatrix}$ = $\begin{bmatrix} 4&0\\0&4\end{bmatrix}$