Past Exam Questions

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 \(0 < x \leq 2\).

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