WAEC Further Mathematics Past Questions & Answers - Page 14

$(x + \frac{2}{x^2})^5$

n = 5,  r = 4,  p = x  and q = $\frac{2}{x^2}$  

5C$_4$x$^4$ ($\frac{2}{x^2}$)1 = 5C$_4$ $\frac{2x^4}{x^2}$ 

5C$_4$ 2x$^2$ = $\frac{5!}{[5-4]!4!}$ * 2x$^2$

$\frac{5*4!}{4!} * 2x^2$ = 5 * 2x$^2$ = 10x$^2$

The coefficient is 10.

$4p_2 + 4C_2 - 4p_3$

$np_r = \frac{n!}{[n-r]!} and  nC_r = \frac{n!}{[n-r]!r!}$

= $\frac{4!}{[4-2]!} + \frac{4!}{[4-2]!2!} - \frac{4!}{[4-3]!} = \frac{4!}{2!} + \frac{4!}{2!2!} - \frac{4!}{1!}$

= $\frac{4*3*2!}{2!} + \frac{4*3*2!}{2!2!} - \frac{4*3*2*1}{1!}$

12 + 6 - 24 = -6

Let (x1, y1) be the image of the point (x, y) under the given transformations.
x1 = 3x - y
y1 = x + 4y
$\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y = \end{vmatrix}  \begin{vmatrix} x_1 \\ y_1 \end{vmatrix}$

$\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} 2 \\ 1 = \end{vmatrix}  \begin{vmatrix} 7 \\ -2 \end{vmatrix}$

$\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix}  \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15$

$\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}$

$\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}$

-12 - 3k = 3

-3k = 3 + 12

k = $\frac{15}{-3}$

k = -5

${1_0^∫} x^2(x^3+2)^3$dx

let $u = x^3 + 2, du = 3x^2dx$

when x = 1,  u = 3

when x = 0,  u = 2

dx = $\frac{du}{3x^2}$

${3_2^∫}$ $\frac{x^2[u]^3}{3x^2}$

${3_2^∫}$ $\frac{u^3}{3}$ du 

= $\frac{u^4}{3*4}$$_2$3

$\frac{1}{12} [u^4]$$_2$3

$\frac{1}{12} [3^4 - 2^4]$

$\frac{1}{12}[81 - 16]$

$\frac{65}{12}$