Correct Answer: A. determine how income is to shared

Explanation

The level of sectoral performance is important because it can indicate the strength of the economy and the potential for future growth. By measuring the level of sectoral performance, governments can identify areas where the economy is doing well and areas where it needs improvement.

Correct Answer: D. relegation of morality

Correct Answer: C. Adah's desire for personal fulfilment

Correct Answer: A. of unknown and uncertain qualities

Correct Answer: B. the colonialists

Correct Answer: B. voluntary

Correct Answer: A. environment

Correct Answer: C. potentiometer

Correct Answer: A. Malaysia

Explanation

(a) Let \(y = \frac{x^{2} + 1}{(x + 1)^{2}}\)

Let \(u = x^{2} + 1 ; v = (x + 1)^{2}\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x ; \frac{\mathrm d v}{\mathrm d x} = 2(x + 1)\)

Using the quotient rule,

\(\frac{\mathrm d y}{\mathrm d x} = \frac{v \frac{\mathrm d u}{\mathrm d x} - u \frac{\mathrm d v}{\mathrm d x}}{v^{2}}\)

= \(\frac{(x + 1)^{2} (2x) - (x^{2} + 1)(2(x + 1))}{((x + 1)^{2})^{2}}\)

= \(\frac{(x + 1)[(x + 1)(2x) - (x^{2} + 1)(2)}{(x + 1)^{4}}\)

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

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

(b)(i) \(\begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\)

= \(1(9 + 1) - 2(6 - 1) - 1(2 + 3)\)

= \(10 - 10 - 5 = -5\)

(ii) \(x + 2y - z = 4\)

\(2x + 3y - z = 2\)

\(-x + y + 3z = -1\)

In matrix form, this is

\(\begin{pmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\)

Cofactors :

\(A_{11} = +(3 \times 3 + 1) = 10\)

\(A_{12} = -(2 \times 3 - 1) = -5\)

\(A_{13} = +(2 \times 1 + 3) = 6\)

\(A_{21} = -(2 \times 3 + 1) = -7\)

\(A_{22} = +(1 \times 3 - 1) = 2\)

\(A_{23} = -(1 \times 1 + 1 \times 2) = -3\)

\(A_{31} = +(2 \times -1 + 3) = 1\)

\(A_{32} = -(-1 + 2) = -1\)

\(A_{33} = +(1 \times 3 - 2 \times 2) = -1\)

\(C = \begin{pmatrix} 10 & -5 & 5 \\ -7 & 2 & -3 \\ 1 & -1 & -1 \end{pmatrix}\)

adj A = \(C^{T} = \begin{pmatrix} 10 & -7 & 1 \\ -5 & 2 & -1 \\ 5 & -3 & -1 \end{pmatrix}\)

= \(\frac{1}{-5} \begin{pmatrix} 10 & -7 & 1 \\ -5 & 2 & -1 \\ 5 & -3 & -1 \end{pmatrix}\)

= \(\begin{pmatrix} -2 & \frac{7}{5} & \frac{-1}{5} \\ 1 & \frac{-2}{5} & \frac{1}{5} \\ -1 & \frac{3}{5} & \frac{1}{5} \end{pmatrix}\)

\(\therefore x = A^{-1} . b \)

= \(\begin{pmatrix} -2 & \frac{7}{5} & \frac{-1}{5} \\ 1 & \frac{-2}{5} & \frac{1}{5} \\ -1 & \frac{3}{5} & \frac{1}{5} \end{pmatrix} \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\)

= \(\begin{pmatrix} - 8 + \frac{14}{5} + \frac{1}{5} \\ 4 - \frac{4}{5} + \frac{1}{5} \\ -4 + \frac{6}{5} - \frac{1}{5} \end{pmatrix}\)

= \(\begin{pmatrix} -5 \\ 3 \\ -3 \end{pmatrix}\)

x = -5 ; y = 3 ; z = -3.