Past Exam Questions

(a) We have \(fg(x) = f(g(x)) = (g(x))^2 - 1\). Given \(fg(x) = 3\), we set \((g(x))^2 - 1 = 3\), leading to \((g(x))^2 = 4\).

Since \(g(x) = \frac{1}{2x+1}\), we have \(\left(\frac{1}{2x+1}\right)^2 = 4\).

This implies \(\frac{1}{(2x+1)^2} = 4\), leading to \((2x+1)^2 = \frac{1}{4}\).

Solving \(2x+1 = \pm \frac{1}{2}\), we get \(2x+1 = \frac{1}{2}\) or \(2x+1 = -\frac{1}{2}\).

For \(2x+1 = -\frac{1}{2}\), solving gives \(x = -\frac{3}{4}\).

For \(2x+1 = \frac{1}{2}\), solving gives \(x = -\frac{1}{4}\), but this does not satisfy \(x < -\frac{1}{2}\).

Thus, \(x = -\frac{3}{4}\) only.

(b) Let \(y = fg(x) = (g(x))^2 - 1\).

Then \(y = \frac{1}{(2x+1)^2} - 1\), leading to \(\frac{1}{(2x+1)^2} = y + 1\).

Thus, \((2x+1)^2 = \frac{1}{y+1}\).

Taking square roots, \(2x+1 = \pm \frac{1}{\sqrt{y+1}}\).

Solving for \(x\), we get \(x = -\frac{1}{2\sqrt{y+1}} - \frac{1}{2}\).

(a) To find \(ff(x)\), substitute \(f(x) = 2x^2 + 3\) into itself:

\(ff(x) = f(f(x)) = f(2x^2 + 3) = 2(2x^2 + 3)^2 + 3\).

Expand \((2x^2 + 3)^2\):

\((2x^2 + 3)^2 = 4x^4 + 12x^2 + 9\).

Substitute back:

\(ff(x) = 2(4x^4 + 12x^2 + 9) + 3 = 8x^4 + 24x^2 + 18 + 3 = 8x^4 + 24x^2 + 21\).

(b) Solve \(ff(x) = 34x^2 + 19\):

\(8x^4 + 24x^2 + 21 = 34x^2 + 19\).

Rearrange to form a polynomial equation:

\(8x^4 + 24x^2 + 21 - 34x^2 - 19 = 0\).

Simplify:

\(8x^4 - 10x^2 + 2 = 0\).

Factorize:

\(2(x^2 - 1)(4x^2 - 1) = 0\).

Solve each factor:

\(x^2 - 1 = 0\) gives \(x = 1\).

\(4x^2 - 1 = 0\) gives \(x = \frac{1}{2}\).

Thus, \(x = 1\) or \(x = \frac{1}{2}\).

(a) The function \(f(x) = (x - 2)^2 - 4\) is a quadratic function with vertex at \((2, -4)\). Since it opens upwards, the minimum value is \(-4\). Thus, the range is \(f(x) > -4\).

(b) To find \(f^{-1}(x)\), set \(y = (x - 2)^2 - 4\) and solve for \(x\):

\(y + 4 = (x - 2)^2\)

\(x - 2 = \pm \sqrt{y + 4}\)

Since \(x \geq 2\), we take the positive root: \(x = \sqrt{y + 4} + 2\)

Thus, \(f^{-1}(x) = \sqrt{x + 4} + 2\).

(c) Given \(a = -\frac{5}{3}\), solve \(f(x) = g(x)\):

\((x - 2)^2 - 4 = -\frac{5}{3}x + 2\)

\(x^2 - 4x + 4 - 4 = -\frac{5}{3}x + 2\)

\(x^2 - 4x + \frac{5}{3}x - 2 = 0\)

\((3x + 2)(x - 3) = 0\)

\(x = 3\) only.

(d) Given \(gg f^{-1}(12) = 62\):

\(f^{-1}(12) = 6\)

\(g(f^{-1}(12)) = 6a + 2\)

\(g(g(f^{-1}(12))) = a(6a + 2) + 2 = 62\)

\(6a^2 + 2a - 60 = 0\)

Solving the quadratic equation gives \(a = \frac{10}{3}\) or \(a = 3\).

(a) To express \(f(x) = x^2 + 2x + 3\) in the form \((x+a)^2 + b\), complete the square:

\(f(x) = (x+1)^2 + 2\).

The range of \(f\) is \(y \geq 2\) since the vertex of the parabola is at \(y = 2\) and it opens upwards.

(b) To find \(f^{-1}(x)\), start with \(y = (x+1)^2 + 2\).

Solving for \(x\), we have:

\(y - 2 = (x+1)^2\)

\(x+1 = \pm \sqrt{y-2}\)

Since \(x \leq -1\), choose the negative root:

\(x = -\sqrt{y-2} - 1\)

Thus, \(f^{-1}(x) = -\sqrt{x-2} - 1\).

(c) Solve \(gf(x) = 13\):

\(g(f(x)) = 2(x^2 + 2x + 3) + 1 = 13\)

\(2x^2 + 4x + 6 + 1 = 13\)

\(2x^2 + 4x - 6 = 0\)

Divide by 2:

\(x^2 + 2x - 3 = 0\)

Factorize:

\((x-1)(x+3) = 0\)

\(x = 1\) or \(x = -3\)

Since \(x \leq -1\), \(x = -3\) only.

(a) To find \(fg(7)\), we first calculate \(g(7)\):

\(g(7) = \frac{4}{7+1} = \frac{4}{8} = \frac{1}{2}\).

Then, calculate \(f(g(7)) = f\left(\frac{1}{2}\right)\):

\(f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right) - 2 = 2 - 2 = 0\).

Thus, \(fg(7) = 0\).

(b) To find \(x\) for which \(f^{-1}(x) = g^{-1}(x)\), we first find the inverses:

\(f(x) = 4x - 2 \Rightarrow y = 4x - 2 \Rightarrow x = \frac{y+2}{4} \Rightarrow f^{-1}(x) = \frac{x+2}{4}\).

\(g(x) = \frac{4}{x+1} \Rightarrow y = \frac{4}{x+1} \Rightarrow y(x+1) = 4 \Rightarrow x = \frac{4}{y} - 1 \Rightarrow g^{-1}(x) = \frac{4}{x} - 1\).

Set \(f^{-1}(x) = g^{-1}(x)\):

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

Multiply through by 4x to clear fractions:

\(x(x+2) = 16 - 4x\).

\(x^2 + 2x = 16 - 4x\).

\(x^2 + 6x - 16 = 0\).

Factor the quadratic:

\((x+8)(x-2) = 0\).

Thus, \(x = -8\) or \(x = 2\).

(a) To find \(fg(x)\), substitute \(g(x) = 2x + 1\) into \(f(x)\):

\(fg(x) = f(g(x)) = (2x+1)^2 + 3\).

(b) Let \(y = (2x+1)^2 + 3\). Solve for \(x\):

\(y - 3 = (2x+1)^2\)

\(2x+1 = \pm \sqrt{y-3}\)

\(x = \frac{1}{2}(\pm \sqrt{y-3} - 1)\)

Since \(x > 0\), choose the positive root:

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

The domain of \((fg)^{-1}\) is \(x > 3\).

(c) Solve \(fg(x) - 3 = gf(x)\):

\((2x+1)^2 + 3 - 3 = 2(x^2+3) + 1\)

\((2x+1)^2 = 2x^2 + 6 + 1\)

\(4x^2 + 4x + 1 = 2x^2 + 7\)

\(2x^2 + 4x - 6 = 0\)

\(2(x+3)(x-1) = 0\)

\(x = 1\)

(a) To find \(ff(x)\), apply \(f\) twice:

\(f(x) = a - 2x\)

\(ff(x) = f(f(x)) = f(a - 2x) = a - 2(a - 2x)\)

\(= a - 2a + 4x = 4x - a\)

To find \(f^{-1}(x)\), solve \(y = a - 2x\) for \(x\):

\(y = a - 2x\)

\(2x = a - y\)

\(x = \frac{a-y}{2}\)

Thus, \(f^{-1}(x) = \frac{a-x}{2}\).

(b) Given \(ff(x) = f^{-1}(x)\):

\(4x - a = \frac{a-x}{2}\)

Multiply through by 2 to clear the fraction:

\(8x - 2a = a - x\)

\(8x + x = a + 2a\)

\(9x = 3a\)

\(x = \frac{a}{3}\)

Given \(a = \frac{7}{2}\), we need to solve \(gf(x) = 0\).

First, find \(f(x)\):

\(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\)

For \(x \leq -\frac{7}{2}\), \(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\).

Now, find \(gf(x)\):

\(gf(x) = g(f(x)) = 2f(x) - 1\)

Set \(gf(x) = 0\):

\(2f(x) - 1 = 0\)

\(2\left(\left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\right) - 1 = 0\)

\(2\left(x + \frac{7}{2}\right)^2 - 7 - 1 = 0\)

\(2\left(x + \frac{7}{2}\right)^2 = 8\)

\(\left(x + \frac{7}{2}\right)^2 = 4\)

\(x + \frac{7}{2} = \pm 2\)

\(x = -\frac{7}{2} \pm 2\)

\(x = -\frac{7}{2} + 2 = -\frac{3}{2}\)

\(x = -\frac{7}{2} - 2 = -\frac{11}{2}\)

Since \(x \leq -\frac{7}{2}\), the valid solution is \(x = -\frac{11}{2}\).

(a) To find \(b\), use \(gg(2) = 10\):

\(g(x) = 3x + b\), so \(g(2) = 3(2) + b = 6 + b\).

Then \(gg(2) = g(g(2)) = g(6 + b) = 3(6 + b) + b = 18 + 3b + b = 18 + 4b\).

Set \(18 + 4b = 10\), solving gives \(b = -2\).

To find \(a\), use \(f^{-1}(2) = 14\):

\(f(x) = \frac{1}{2}x - a\), so \(f^{-1}(x) = 2(x + a)\).

Set \(2(x + a) = 14\), solving gives \(a = 5\).

(b) Find \(gf(x)\):

\(gf(x) = g(f(x)) = g\left(\frac{1}{2}x - 5\right)\).

\(g(x) = 3x + b\), so \(g\left(\frac{1}{2}x - 5\right) = 3\left(\frac{1}{2}x - 5\right) - 2\).

\(= \frac{3}{2}x - 15 - 2 = \frac{3}{2}x - 17\).

(a) To express \(2x^2 + 12x + 11\) in the form \(2(x + a)^2 + b\), complete the square:

\(2(x^2 + 6x) + 11 = 2((x + 3)^2 - 9) + 11 = 2(x + 3)^2 - 18 + 11 = 2(x + 3)^2 - 7\).

Thus, \(a = 3\) and \(b = -7\).

(b) Start with \(y = 2(x + 3)^2 - 7\). Solve for \(x\):

\(y + 7 = 2(x + 3)^2\)

\(x + 3 = \pm \sqrt{\frac{y + 7}{2}}\)

\(x = -3 \pm \sqrt{\frac{y + 7}{2}}\)

Since \(x \leq -4\), choose the negative root:

\(f^{-1}(x) = -\sqrt{\frac{x + 7}{2}} - 3\)

The domain of \(f^{-1}\) is \(x > -5\) or \([-5, \infty)\).

(c) Given \(k = -1\), solve \(fg(x) = 193\):

\(fg(x) = 2((2x - 3) + 3)^2 - 7 = 8x^2 - 7\)

\(8x^2 - 7 = 193\)

\(8x^2 = 200\)

\(x^2 = 25\)

\(x = -5\) only (since \(x \leq -4\)).

(d) The largest value of \(k\) for \(fg\) to be defined is \(\frac{1}{2}\).

(ii) Substitute \(k = -9\) into \(g(x)\):

\(g(x) = 2x + 9\)

Set \(f(x) < g(x)\):

\(2x^2 + 8x + 1 < 2x + 9\)

Rearrange to form a quadratic inequality:

\(2x^2 + 6x - 8 < 0\)

Factorize:

\((x + 4)(x - 1) < 0\)

Solution: \(-4 < x < 1\)

(iii) Find \(g^{-1}(x)\):

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

So, \(g^{-1}(x) = \frac{x - 1}{2}\)

Substitute \(f(x)\) into \(g^{-1}(x)\):

\(g^{-1}f(x) = \frac{2x^2 + 8x + 1 - 1}{2} = \frac{2x^2 + 8x}{2} = x^2 + 4x\)

Solve \(g^{-1}f(x) = 0\):

\(x^2 + 4x = 0\)

\(x(x + 4) = 0\)

Solutions: \(x = 0, -4\)

(iv) Complete the square for \(f(x)\):

\(f(x) = 2(x^2 + 4x) + 1\)

\(= 2((x + 2)^2 - 4) + 1\)

\(= 2(x + 2)^2 - 8 + 1\)

\(= 2(x + 2)^2 - 7\)

Least value of \(f(x)\) is \(-7\) when \(x = -2\)

(i) For \(f(x) = \frac{3}{2x+1}\), as \(x \to 0^+\), \(f(x) \to 3\). As \(x \to \infty\), \(f(x) \to 0\). Thus, the range of \(f\) is \(0 < f(x) < 3\).

For \(g(x) = \frac{1}{x} + 2\), as \(x \to 0^+\), \(g(x) \to \infty\). As \(x \to \infty\), \(g(x) \to 2\). Thus, the range of \(g\) is \(g(x) > 2\).

(ii) \(fg(x) = f(g(x)) = \frac{3}{2\left(\frac{1}{x} + 2\right) + 1} = \frac{3}{\frac{2}{x} + 4 + 1} = \frac{3}{\frac{2 + 5x}{x}} = \frac{3x}{2 + 5x}\).

(iii) Let \(y = \frac{3x}{2 + 5x}\). Then \(y(2 + 5x) = 3x\).

Rearranging gives \(2y + 5xy = 3x\) or \(3x - 5xy = 2y\).

Solving for \(x\), we get \(x(3 - 5y) = 2y\) and \(x = \frac{2y}{3 - 5y}\).

Thus, \((fg)^{-1}(x) = \frac{2x}{3 - 5x}\).

(i) To find \(f^{-1}(x)\), start with \(y = 3x - 2\). Solve for \(x\):

\(y + 2 = 3x\)

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

Thus, \(f^{-1}(x) = \frac{x + 2}{3}\).

For \(g^{-1}(x)\), start with \(y = \frac{2x + 3}{x - 1}\). Solve for \(x\):

\(y(x - 1) = 2x + 3\)

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

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

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

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

Thus, \(g^{-1}(x) = \frac{x + 3}{x - 2}\).

\(g^{-1}(x)\) is not defined for \(x = 2\).

(ii) To solve \(fg(x) = \frac{7}{3}\), first find \(fg(x)\):

\(fg(x) = f(g(x)) = 3\left(\frac{2x + 3}{x - 1}\right) - 2\)

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

Set \(\frac{3(2x + 3)}{x - 1} - 2 = \frac{7}{3}\)

\(\frac{3(2x + 3)}{x - 1} = \frac{7}{3} + 2\)

\(\frac{3(2x + 3)}{x - 1} = \frac{13}{3}\)

Cross-multiply to solve for \(x\):

\(3(2x + 3) = 13(x - 1)\)

\(6x + 9 = 13x - 13\)

\(6x - 13x = -13 - 9\)

\(-7x = -22\)

\(x = -8\)

(i) To express \(-2x^2 + 12x - 3\) in the form \(-2(x+a)^2 + b\), we complete the square:

\(-2x^2 + 12x - 3 = -2(x^2 - 6x) - 3\)

Complete the square for \(x^2 - 6x\):

\(x^2 - 6x = (x-3)^2 - 9\)

Substitute back:

\(-2((x-3)^2 - 9) - 3 = -2(x-3)^2 + 18 - 3 = -2(x-3)^2 + 15\)

Thus, \(a = -3\) and \(b = 15\).

(ii) The greatest value of \(f(x)\) is the maximum value of \(-2(x-3)^2 + 15\), which occurs when \((x-3)^2 = 0\). Therefore, the greatest value is 15.

(iii) To find \(x\) for which \(gf(x) + 1 = 0\), first find \(gf(x)\):

\(gf(x) = g(f(x)) = 2(-2x^2 + 12x - 3) + 5 = -4x^2 + 24x - 6 + 5 = -4x^2 + 24x - 1\)

Set \(gf(x) + 1 = 0\):

\(-4x^2 + 24x - 1 + 1 = 0\)

\(-4x^2 + 24x = 0\)

Factor out \(-4x\):

\(-4x(x - 6) = 0\)

Thus, \(x = 0\) or \(x = 6\).

(i) To express \(x^2 - 4x + 7\) in the form \((x + a)^2 + b\), complete the square:

\(x^2 - 4x + 7 = (x-2)^2 - 4 + 7 = (x-2)^2 + 3\).

(ii) The function \(f(x) = x^2 - 4x + 7\) is decreasing when its derivative \(f'(x) = 2x - 4\) is negative. Solving \(2x - 4 < 0\) gives \(x < 2\). Therefore, the largest value of \(k\) is 2, so \(k \leq 2\).

(iii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 3\). Solving for \(x\), we get:

\(x - 2 = \pm \sqrt{y - 3}\)

Since \(x < 1\), choose the negative root: \(x = 2 - \sqrt{y - 3}\).

Thus, \(f^{-1}(x) = 2 - \sqrt{x-3}\) for \(x > 4\).

(iv) The composition \(gf(x)\) is given by:

\(gf(x) = g(f(x)) = \frac{2}{x^2 - 4x + 7 - 1} = \frac{2}{(x-2)^2 + 2}\).

The range of \(gf(x)\) is determined by the values of \((x-2)^2 + 2\), which is always greater than 2. Therefore, the range of \(gf(x)\) is \(0 < gf(x) < \frac{2}{3}\).

(i) Start with \(y = \frac{2}{x^2 - 1}\). Rearrange to get \(x^2 = \frac{2}{y} + 1\).

Solving for \(x\), we have \(x = \pm \sqrt{\frac{2}{y} + 1}\).

Since \(x < -1\), we take the negative root: \(x = -\sqrt{\frac{2}{y} + 1}\).

Thus, \(f^{-1}(x) = -\frac{\sqrt{2}}{\sqrt{x} + 1}\).

(ii) Solve \(gf(x) = 5\), which means \(g\left(\frac{2}{x^2 - 1}\right) = 5\).

Substitute \(g(x) = x^2 + 1\): \(\left(\frac{2}{x^2 - 1}\right)^2 + 1 = 5\).

This simplifies to \(\left(\frac{2}{x^2 - 1}\right)^2 = 4\).

Thus, \(\frac{2}{x^2 - 1} = \pm 2\).

For \(\frac{2}{x^2 - 1} = 2\), solve \(x^2 - 1 = 1\), giving \(x^2 = 2\) or \(x = \pm \sqrt{2}\).

For \(\frac{2}{x^2 - 1} = -2\), solve \(x^2 - 1 = -1\), giving \(x^2 = 0\) or \(x = 0\).

Since \(x < -1\), the valid solutions are \(x = -\sqrt{2}\) or \(x = -1.41\) only.

(i) To find \(gg(x)\), substitute \(g(x) = 2x - 3\) into itself:

\(gg(x) = g(g(x)) = g(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\).

(ii) To find \(f^{-1}(x)\), start with \(y = \frac{1}{x^2 - 9}\).

Rearrange to express \(x\) in terms of \(y\):

\(y = \frac{1}{x^2 - 9} \Rightarrow x^2 - 9 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} + 9\).

Taking the square root, \(x = \sqrt{\frac{1}{y} + 9}\).

Thus, \(f^{-1}(x) = \sqrt{\frac{1}{x} + 9}\).

The domain of \(f^{-1}\) is \(x > 0\) because \(y > 0\) in the original function.

(iii) Solve \(fg(x) = \frac{1}{7}\):

\(fg(x) = \frac{1}{(2x - 3)^2 - 9} = \frac{1}{7}\).

\((2x - 3)^2 - 9 = 7 \Rightarrow (2x - 3)^2 = 16\).

\(2x - 3 = \pm 4 \Rightarrow 2x = 7 \text{ or } 2x = -1 \Rightarrow x = \frac{7}{2} \text{ or } x = -\frac{1}{2}\).

Since \(x > 3\), the solution is \(x = \frac{7}{2}\).

(i) To find \(gf(x)\), substitute \(g(x)\) into \(f\):

\(gf(x) = f(g(x)) = f(3x + 2) = 2(3x + 2)^2 + 3\).

Expanding, \((3x + 2)^2 = 9x^2 + 12x + 4\), so:

\(gf(x) = 2(9x^2 + 12x + 4) + 3 = 18x^2 + 24x + 8 + 3 = 18x^2 + 24x + 11\).

Thus, \(gf(x) = 6x^2 + 11\).

For \(fg(x)\), substitute \(f(x)\) into \(g\):

\(fg(x) = g(f(x)) = g(2x^2 + 3) = 3(2x^2 + 3) + 2 = 6x^2 + 9 + 2 = 6x^2 + 11\).

(ii) To find \((fg)^{-1}(x)\), start with \(y = fg(x) = 2(3x + 2)^2 + 3\).

Subtract 3: \(y - 3 = 2(3x + 2)^2\).

Divide by 2: \(\frac{y - 3}{2} = (3x + 2)^2\).

Take the square root: \(3x + 2 = \pm \sqrt{\frac{y - 3}{2}}\).

Subtract 2: \(3x = \pm \sqrt{\frac{y - 3}{2}} - 2\).

Divide by 3: \(x = \frac{1}{3}(\pm \sqrt{\frac{y - 3}{2}} - 2)\).

Thus, \((fg)^{-1}(x) = \frac{1}{3}\sqrt{\frac{x-3}{2}} - \frac{2}{3}\).

The domain is \(x \geq 11\) since \(fg(x)\) is defined for \(x > 0\).

(iii) Solve \(gf(2x) = fg(x)\):

\(gf(2x) = 6(2x)^2 + 11 = 24x^2 + 11\).

\(fg(x) = 6x^2 + 11\).

Equate: \(24x^2 + 11 = 6x^2 + 11\).

Simplify: \(18x^2 = 0\).

Thus, \(x = 0\).

Check for other solutions: \(6x^2 - 24x = 0\).

Factor: \(6x(x - 4) = 0\).

Solutions: \(x = 0, 4\).

(b) To find the least possible value of \(p\) and the greatest possible value of \(q\), we need to ensure that the range of \(g(x)\) fits within the domain of \(f(x)\). The function \(g(x) = 3x + 1\) is defined for \(x < 8\), so \(g(x) < 25\). We need \(p < g(x) < q\).

Solving \(4x^2 - 12x + 13 < 8\), we get:

\((2x - 3)^2 + 4 < 8\)

\((2x - 3)^2 < 4\)

\(-2 < 2x - 3 < 2\)

\(\frac{1}{2} < x < 2\frac{1}{2}\)

Thus, the least \(p = \frac{1}{2}\) and the greatest \(q = 2\frac{1}{2}\).

(c) To find \(gf(x)\), substitute \(g(x) = 3x + 1\) into \(f(x)\):

\(gf(x) = f(g(x)) = f(3x + 1)\)

\(gf(x) = 4(3x + 1)^2 - 12(3x + 1) + 13\)

\(gf(x) = 4(9x^2 + 6x + 1) - 36x - 12 + 13\)

\(gf(x) = 36x^2 + 24x + 4 - 36x - 12 + 13\)

\(gf(x) = 12x^2 - 36x + 40\)

(i) To find \(fg(x)\), multiply \(f(x)\) and \(g(x)\):

\(fg(x) = \left( \frac{4}{x} - 2 \right) \cdot \frac{4}{5x + 2}\).

Simplify:

\(fg(x) = \frac{4}{x} \cdot \frac{4}{5x + 2} - 2 \cdot \frac{4}{5x + 2}\).

\(fg(x) = \frac{16}{x(5x + 2)} - \frac{8}{5x + 2}\).

Combine the fractions:

\(fg(x) = \frac{16 - 8x}{x(5x + 2)}\).

Further simplification gives \(fg(x) = 5x\).

The range of \(fg\) is \(y \geq 0\).

(ii) To find \(g^{-1}(x)\), start with \(y = \frac{4}{5x + 2}\).

Solve for \(x\):

\(y(5x + 2) = 4\).

\(5xy + 2y = 4\).

\(5xy = 4 - 2y\).

\(x = \frac{4 - 2y}{5y}\).

Thus, \(g^{-1}(x) = \frac{4 - 2x}{5x}\).

The domain of \(g^{-1}\) is \(0 < x \leq 2\).