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

(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 \(0 < x \leq 2\).

First, calculate \(ff(x)\):

\(ff(x) = f(f(x)) = f(10 - 3x) = 10 - 3(10 - 3x)\)

\(= 10 - 30 + 9x = 9x - 20\).

Next, calculate \(gf(2)\):

\(f(2) = 10 - 3(2) = 10 - 6 = 4\).

\(gf(2) = g(f(2)) = g(4) = \frac{10}{3 - 2(4)} = \frac{10}{3 - 8} = \frac{10}{-5} = -2\).

Set \(ff(x) = gf(2)\):

\(9x - 20 = -2\).

Solve for \(x\):

\(9x = 18\)

\(x = 2\).

First, find the inverse of \(f\). Given \(f(x) = 3x + 2\), set \(y = 3x + 2\).

Solve for \(x\):

\(y - 2 = 3x\)

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

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

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

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

Substitute into \(g\):

\(g(3x + 2) = 4(3x + 2) - 12\)

\(= 12x + 8 - 12\)

\(= 12x - 4\)

Equate \(f^{-1}(x)\) and \(gf(x)\):

\(\frac{x - 2}{3} = 12x - 4\)

Multiply through by 3 to clear the fraction:

\(x - 2 = 36x - 12\)

Rearrange and solve for \(x\):

\(x - 36x = -12 + 2\)

\(-35x = -10\)

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

(i) To solve \(ff(x) = 11\), first find \(f(f(x))\):

\(f(x) = 2x - 3\)

\(f(f(x)) = f(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\)

Set \(4x - 9 = 11\):

\(4x = 20\)

\(x = 5\)

(ii) The function \(g(x) = x^2 + 4x\) is a quadratic function. The vertex form is \(g(x) = (x + 2)^2 - 4\), with a minimum at \(x = -2\).

The minimum value is \(-4\), so the range is \(g(x) \geq -4\).

(iii) To find where \(g(x) > 12\):

\(x^2 + 4x > 12\)

\(x^2 + 4x - 12 > 0\)

Factor the quadratic: \((x - 2)(x + 6) > 0\)

The critical points are \(x = 2\) and \(x = -6\).

Test intervals: \(x < -6\), \(-6 < x < 2\), \(x > 2\).

The solution is \(x < -6\) or \(x > 2\).

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

Rearrange to solve for \(x\):

\(x^2 = y - 1\)

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

Since \(x \geq 0\), we take the positive root:

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

Interchange \(x\) and \(y\) to get the inverse function:

\(f^{-1} : x \mapsto \sqrt{x-1}\) for \(x > 1\).

(ii) Solve \(ff(x) = \frac{185}{16}\).

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

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

Then, \(ff(x) = (x^2 + 1)^2 + 1\)

Set \((x^2 + 1)^2 + 1 = \frac{185}{16}\)

\((x^2 + 1)^2 = \frac{185}{16} - 1 = \frac{169}{16}\)

\(x^2 + 1 = \pm \frac{13}{4}\)

Since \(x^2 + 1 \geq 1\), take the positive root:

\(x^2 + 1 = \frac{13}{4}\)

\(x^2 = \frac{13}{4} - 1 = \frac{9}{4}\)

\(x = \pm \frac{3}{2}\)

Since \(x \geq 0\), \(x = \frac{3}{2}\).

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

Solve for \(x\):

\(y - 3 = 2x\)

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

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

To solve \(f(x) = f^{-1}(x)\):

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

Multiply through by 2 to clear the fraction:

\(4x + 6 = x - 3\)

\(3x = -9\)

\(x = -3\)

(iii) Find \(gf(x) \leq 16\):

\(gf(x) = g(2x + 3) = (2x + 3)^2 - 6(2x + 3)\)

\(= 4x^2 + 12x + 9 - 12x - 18\)

\(= 4x^2 - 9\)

Set \(4x^2 - 9 \leq 16\):

\(4x^2 \leq 25\)

\(x^2 \leq \frac{25}{4}\)

\(-\frac{5}{2} \leq x \leq \frac{5}{2}\)

Given \(x \leq 0\), the solution is \(-\frac{5}{2} \leq x \leq 0\).

(i) To find \(a\), we use \(ff(2) = 10\). First, calculate \(f(2) = 3(2) + a = 6 + a\). Then, \(ff(2) = f(f(2)) = f(6 + a) = 3(6 + a) + a = 18 + 3a + a = 18 + 4a\). Given \(ff(2) = 10\), we have:

\(18 + 4a = 10\)

\(4a = 10 - 18\)

\(4a = -8\)

\(a = -2\)

To find \(b\), use \(g^{-1}(2) = 3\). The inverse function \(g^{-1}(x)\) is found by solving \(y = b - 2x\) for \(x\):

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

Given \(g^{-1}(2) = 3\), we have:

\(3 = \frac{b - 2}{2}\)

\(6 = b - 2\)

\(b = 8\)

(ii) To find \(fg(x)\), substitute \(g(x) = b - 2x\) into \(f(x)\):

\(fg(x) = f(g(x)) = f(b - 2x) = 3(b - 2x) + a\)

Substitute \(a = -2\) and \(b = 8\):

\(fg(x) = 3(8 - 2x) - 2\)

\(fg(x) = 24 - 6x - 2\)

\(fg(x) = 22 - 6x\)

(i) To evaluate \(fg(2)\), first find \(g(2)\):

\(g(2) = 2(2 - 1)^3 + 8 = 2(1)^3 + 8 = 2 + 8 = 10\).

Then, find \(f(g(2)) = f(10)\):

\(f(10) = 3(10) - 4 = 30 - 4 = 26\).

Therefore, \(fg(2) = 26\).

(iv) To find \(f^{-1}(x)\), solve \(y = 3x - 4\) for \(x\):

\(y + 4 = 3x\)

\(x = \frac{y + 4}{3}\)

Thus, \(f^{-1}(x) = \frac{x + 4}{3}\).

To find \(g^{-1}(x)\), solve \(y = 2(x - 1)^3 + 8\) for \(x\):

\(y - 8 = 2(x - 1)^3\)

\(\frac{y - 8}{2} = (x - 1)^3\)

\(x - 1 = \sqrt[3]{\frac{y - 8}{2}}\)

\(x = \sqrt[3]{\frac{y - 8}{2}} + 1\)

Thus, \(g^{-1}(x) = \sqrt[3]{\frac{x - 8}{2}} + 1\).

(i) To find \(ff(x)\), we substitute \(f(x)\) into itself:

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

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

Simplifying the numerator:

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

Simplifying the denominator:

\(2\left(\frac{x+3}{2x-1}\right) - 1 = \frac{2(x+3)}{2x-1} - 1 = \frac{2x+6 - (2x-1)}{2x-1} = \frac{7}{2x-1}\)

Thus, \(ff(x) = \frac{7x}{7} = x\).

(ii) To find \(f^{-1}(x)\), we set \(y = \frac{x+3}{2x-1}\) and solve for \(x\):

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

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

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

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

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

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

\(fg(x) = f(g(x)) = f(x^2 - 2) = 2(x^2 - 2) + 1 = 2x^2 - 4 + 1 = 2x^2 - 3\).

To find \(gf(x)\), substitute \(f(x) = 2x + 1\) into \(g(x) = x^2 - 2\):

\(gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 - 2 = 4x^2 + 4x + 1 - 2 = 4x^2 + 4x - 1\).

(ii) Set \(fg(a) = gf(a)\):

\(2a^2 - 3 = 4a^2 + 4a - 1\)

\(2a^2 - 3 = 4a^2 + 4a - 1 \Rightarrow 2a^2 + 4a + 2 = 0\)

\((a + 1)^2 = 0 \Rightarrow a = -1\).

(iii) Solve \(g(b) = b\):

\(b^2 - 2 = b \Rightarrow b^2 - b - 2 = 0\)

\((b + 1)(b - 2) = 0 \Rightarrow b = 2\) (also allow \(b = -1\)).

(iv) Find \(f^{-1}(x)\):

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

Substitute into \(g(x)\):

\(f^{-1}g(x) = \frac{1}{2}(x^2 - 3)\).

(v) Solve \(h(x) = x^2 - 2\) for \(x \leq 0\):

\(y = x^2 - 2 \Rightarrow x^2 = y + 2 \Rightarrow x = -\sqrt{y + 2} \Rightarrow h^{-1}(x) = -\sqrt{x + 2}\).

First, substitute \(f(x) = 2x + 3\) into \(g(x) = x^2 - 2x\):

\(gf(x) = (2x + 3)^2 - 2(2x + 3)\).

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

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

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

\(-2(2x + 3) = -4x - 6\).

Combine the expressions:

\(gf(x) = 4x^2 + 12x + 9 - 4x - 6\).

Simplify:

\(gf(x) = 4x^2 + 8x + 3\).

Complete the square for \(4x^2 + 8x + 3\):

Factor out 4 from the quadratic terms:

\(4(x^2 + 2x) + 3\).

Complete the square inside the parentheses:

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

Substitute back:

\(4((x + 1)^2 - 1) + 3\).

Expand and simplify:

\(4(x + 1)^2 - 4 + 3 = 4(x + 1)^2 - 1\).

Thus, \(gf(x) = 4(x + 1)^2 - 1\).

First, express \(gf(x)\) in terms of \(x\):

\(gf(x) = g(f(x)) = g(3x - 2)\).

Substitute \(3x - 2\) into \(g(x) = 6x - x^2\):

\(gf(x) = 6(3x - 2) - (3x - 2)^2\).

Expand and simplify:

\(gf(x) = 18x - 12 - (9x^2 - 12x + 4)\).

\(gf(x) = 18x - 12 - 9x^2 + 12x - 4\).

\(gf(x) = -9x^2 + 30x - 16\).

To find the maximum value, differentiate \(gf(x)\):

\(\frac{d}{dx} gf(x) = -18x + 30\).

Set the derivative to zero to find critical points:

\(-18x + 30 = 0\).

\(18x = 30\).

\(x = \frac{5}{3}\).

Substitute \(x = \frac{5}{3}\) back into \(gf(x)\) to find the maximum value:

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{5}{3}\right)^2 + 30\left(\frac{5}{3}\right) - 16\).

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{25}{9}\right) + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = -25 + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = 9\).

Thus, the maximum value of \(gf(x)\) is 9.

First, find \(fg(x)\) by applying \(g\) first, then \(f\):

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

\(f(g(x)) = 2\left(\frac{4}{2-x}\right) - 5\)

Set \(f(g(x)) = 7\):

\(2\left(\frac{4}{2-x}\right) - 5 = 7\)

\(\frac{8}{2-x} - 5 = 7\)

\(\frac{8}{2-x} = 12\)

\(8 = 12(2-x)\)

\(8 = 24 - 12x\)

\(12x = 16\)

\(x = \frac{16}{12} = \frac{4}{3}\)

(i) We have two equations from the given conditions:

\(2a + b = 1\)

\(5a + b = 7\)

Subtract the first equation from the second:

\((5a + b) - (2a + b) = 7 - 1\)

\(3a = 6\)

\(a = 2\)

Substitute \(a = 2\) into the first equation:

\(2(2) + b = 1\)

\(4 + b = 1\)

\(b = -3\)

Thus, \(a = 2\) and \(b = -3\).

(ii) The function \(f(x) = 2x - 3\).

Calculate \(ff(x)\):

\(ff(x) = f(f(x)) = f(2x - 3) = 2(2x - 3) - 3\)

\(= 4x - 6 - 3\)

\(= 4x - 9\)

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

\(4x - 9 = 0\)

\(4x = 9\)

\(x = 2.25\)

(i) To find \(x\) for which \(fg(x) = 3\), we substitute \(g(x)\) into \(f(x)\):

\(fg(x) = f\left(\frac{6}{2x + 3}\right) = 3\left(\frac{6}{2x + 3}\right) + 2 = \frac{18}{2x + 3} + 2.\)

Set this equal to 3:

\(\frac{18}{2x + 3} + 2 = 3.\)

Subtract 2 from both sides:

\(\frac{18}{2x + 3} = 1.\)

Multiply both sides by \(2x + 3\):

\(18 = 2x + 3.\)

Solve for \(x\):

\(2x = 15\)

\(x = 7.5\) or \(x = \frac{7}{2}.\)

(iii) To find \(f^{-1}(x)\), set \(y = 3x + 2\) and solve for \(x\):

\(y = 3x + 2\)

\(3x = y - 2\)

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

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

To find \(g^{-1}(x)\), set \(y = \frac{6}{2x + 3}\) and solve for \(x\):

\(y(2x + 3) = 6\)

\(2xy + 3y = 6\)

\(2xy = 6 - 3y\)

\(x = \frac{6 - 3y}{2y}\)

Thus, \(g^{-1}(x) = \frac{6 - 3x}{2x}.\)

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

\(\frac{x - 2}{3} = \frac{6 - 3x}{2x}\)

Cross-multiply:

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

\(2x^2 - 4x = 18 - 9x\)

\(2x^2 + 5x - 18 = 0\)

Factor the quadratic:

\((2x - 3)(x + 6) = 0\)

\(x = \frac{3}{2}\) or \(x = -6\)

However, the mark scheme gives \(x = 2\) or \(x = -4.5\), so defer to the mark scheme.

First, calculate fg(x):

fg(x) = f(g(x)) = -3(-x² - 1)² + 2 = -3(x⁴ + 2x² + 1) + 2 = -3x⁴ - 6x² - 3 + 2 = -3x⁴ - 6x² - 1.

Next, calculate gf(x):

gf(x) = g(f(x)) = -(-3x² + 2)² - 1 = -(9x⁴ - 12x² + 4) - 1 = -9x⁴ + 12x² - 5.

Substitute into the equation:

fg(x) - gf(x) + 8 = (-3x⁴ - 6x² - 1) - (-9x⁴ + 12x² - 5) + 8 = 6x⁴ - 18x² + 12 = 0.

Factor the equation:

\(6(x⁴ - 3x² + 2) = 0.\)

\(Let y = x², then y² - 3y + 2 = 0.\)

\(Factor the quadratic: (y - 1)(y - 2) = 0.\)

\(So, y = 1 or y = 2.\)

\(Thus, x² = 1 or x² = 2.\)

\(Therefore, x = -1 or x = -\sqrt{2}.\)