Further Mathematics Questions and Answers

\((x + \frac{2}{x^2})^5\)

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

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

\(\frac{1}{12}[81 - 16]\)

\(\frac{65}{12}\)