Past Exam Questions

(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 \(g(x) = x + 3\) into \(f(x)\):

\(fg(x) = 2 - \frac{5}{(x+3)+2} = 2 - \frac{5}{x+5}\).

To combine into a single fraction, express \(2\) as \(\frac{2(x+5)}{x+5}\):

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

Combine the fractions:

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

Simplify the numerator:

\(2(x+5) - 5 = 2x + 10 - 5 = 2x + 5\).

Thus, \(fg(x) = \frac{2x+5}{x+5}\).

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.

(a) To find \(a\), use \(f(7) = \frac{5}{2}\):

\(1 + \frac{2a}{7-a} = \frac{5}{2}\)

\(\frac{2a}{7-a} = \frac{3}{2}\)

\(4a = 3(7-a)\)

\(4a = 21 - 3a\)

\(7a = 21\)

\(a = 3\)

To find \(b\), use \(gf(5) = 4\):

\(f(5) = 1 + \frac{6}{5-3} = 4\)

\(g(4) = b(4) - 2 = 4\)

\(4b - 2 = 4\)

\(4b = 6\)

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

(b) The domain of \(f^{-1}\) is \(x > 1\) because \(f(x) = 1 + \frac{6}{x-3}\) implies \(x > 3\), and \(f(x) > 1\).

(c) To find \(f^{-1}(x)\):

Start with \(y = 1 + \frac{6}{x-3}\)

\(y - 1 = \frac{6}{x-3}\)

\((y-1)(x-3) = 6\)

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

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

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

\(x = 3 + \frac{6}{y-1}\)

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

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:

\(6(x⁴ - 3x² + 2) = 0.\)

\(Let y = x², then y² - 3y + 2 = 0.\)

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

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

\(Thus, x² = 1 or x² = 2.\)

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

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

\(gf(x) = g(f(x)) = g\left(x + \frac{1}{x}\right) = a\left(x + \frac{1}{x}\right) + 1\).

(b) Given \(gf(2) = 11\), substitute \(x = 2\) into \(gf(x)\):

\(gf(2) = a\left(2 + \frac{1}{2}\right) + 1 = 11\).

\(a\left(2.5\right) + 1 = 11\)

\(2.5a + 1 = 11\)

\(2.5a = 10\)

\(a = 4\).

(c) The function \(f(x) = x + \frac{1}{x}\) is not one-one because it has a minimum point at \(x = 1\), meaning it is not strictly increasing or decreasing. Therefore, \(f\) does not have an inverse.

(d) To find \(g^{-1}(x)\), solve \(y = ax + 1\) for \(x\):

\(y - 1 = ax\)

\(x = \frac{y - 1}{a}\)

For \(a = 5\), \(g^{-1}(x) = \frac{x - 1}{5}\).

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

\(g^{-1}f(x) = \frac{x + \frac{1}{x} - 1}{5}\).

(e) The composite function \(fg\) cannot be formed because the domain of \(f\) (\(x > 0\)) does not include the whole range of \(g\), which is all real numbers.

To find \(fg(x)\), we substitute \(g(x) = 2x + 4\) into \(f(x)\).

\(fg(x) = f(2x + 4) = 2((2x + 4)^2) - 16(2x + 4) + 23\).

First, calculate \((2x + 4)^2 = 4x^2 + 16x + 16\).

Then, \(2(4x^2 + 16x + 16) = 8x^2 + 32x + 32\).

Next, calculate \(-16(2x + 4) = -32x - 64\).

Combine these results: \(8x^2 + 32x + 32 - 32x - 64 + 23\).

Simplify: \(8x^2 + 32 - 64 + 23 = 8x^2 - 9\).

(a) The domain of \(f^{-1}\) is determined by the range of \(f(x)\). Since \(f(x)\) is undefined at \(x = \frac{1}{2}\), the domain of \(f^{-1}\) is \(x \neq 1\) or \(x < 1\), \(x > 1\) or \((-\infty, 1) \cup (1, \infty)\).

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

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

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

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

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

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

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

(c) Substitute \(x = 3\) into \(f^{-1}(x)\):

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

Then, \(g(f^{-1}(3)) = g(2) = 2^2 + 4 = 8\).

(d) \(g(x) = x^2 + 4\) is not one-to-one because it is a quadratic function, which means \(g^{-1}(x)\) cannot be found.

(e) Show that \(1 + \frac{2}{2x-1} = \frac{2x+1}{2x-1}\):

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

Find the equation of the tangent at \(x = 1\):

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

The derivative \(f'(x) = \frac{-4}{(2x-1)^2}\) gives \(f'(1) = -4\)

The equation of the tangent is \(y - 3 = -4(x - 1)\) or \(y = -4x + 7\)

The tangent crosses the x-axis at \(x = \frac{7}{4}\) and the y-axis at \(y = 7\).

The area of the triangle is \(\frac{1}{2} \times \frac{7}{4} \times 7 = \frac{49}{8}\).

Answer: \(k=\frac{7}{2}\).

For the line \(y=g(x)\) to be tangent to the curve \(y=f(x)\), the equation \(f(x)=g(x)\) must have exactly one repeated root.

So

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

Rearranging,

\(2x^2+6x+1+k=0.\)

For a repeated root, the discriminant is zero:

\(6^2-4(2)(1+k)=0.\)

Hence

\(36-8-8k=0,\)

so

\(28=8k\)

and therefore

\(k=\frac{28}{8}=\frac{7}{2}.\)

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

\(fg(x) = 4\left(\frac{9}{2-x}\right) - 2k = \frac{36}{2-x} - 2k\).

Set \(fg(x) = x\):

\(\frac{36}{2-x} - 2k = x\).

Multiply through by \(2-x\) to clear the fraction:

\(36 - 2kx = x(2-x)\).

Rearrange to form a quadratic equation:

\(x^2 + 2kx - 2x + 36 - 4k = 0\).

For the equation to have two equal roots, the discriminant must be zero:

\((2k-2)^2 = 4(36-4k)\).

Solve for \(k\):

\(4k^2 - 8k + 4 = 144 - 16k\).

\(4k^2 + 8k - 140 = 0\).

\(k^2 + 2k - 35 = 0\).

Factorize:

\((k-5)(k+7) = 0\).

Thus, \(k = 5\) or \(k = -7\).

(ii) Substitute \(k = 5\) and \(k = -7\) into the quadratic equation:

For \(k = 5\):

\(x^2 + 8x + 16 = 0\).

Roots are \(x = -4\) (double root).

For \(k = -7\):

\(x^2 - 16x + 64 = 0\).

Roots are \(x = 8\) (double root).

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

\(k - x = \frac{9}{x+2}\)

Multiply through by \(x+2\):

\((k-x)(x+2) = 9\)

\(kx + 2k - x^2 - 2x = 9\)

Rearrange to form a quadratic:

\(x^2 + (2-k)x + 9 - 2k = 0\)

For two equal roots, the discriminant must be zero:

\(b^2 - 4ac = 0\)

\((2-k)^2 - 4 \times 1 \times (9-2k) = 0\)

Solve for \(k\):

\(k = 4\) or \(k = -8\)

For \(k = 4\), the root is \(x = 1\).

For \(k = -8\), the root is \(x = -5\).

(ii) Solve \(fg(x) = 5\) when \(k = 6\):

\((6-x) \cdot \frac{9}{x+2} = 5\)

\(54 - 9x = 5x + 10\)

\(44 = 14x\)

\(x = 7\)

(iii) Express \(g^{-1}(x)\) in terms of \(x\):

Let \(y = \frac{9}{x+2}\)

\(x = \frac{9}{y} - 2\)

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

(i) Given \(f(x) = 2x - 3\), we need to find \(ff(x)\).

First, calculate \(f(f(x)) = f(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\).

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

Solve for \(x\):

\(4x - 9 = 11\)

\(4x = 20\)

\(x = 5\)

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

\(2x - a = x^2 - 6x\)

Rearrange to form a quadratic equation:

\(x^2 - 8x + a = 0\)

For exactly one real solution, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(b = -8\), \(a = 1\), \(c = a\).

\((-8)^2 - 4 \times 1 \times a = 0\)

\(64 - 4a = 0\)

\(4a = 64\)

\(a = 16\)

First, find the expression for \(gf(x)\):

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

Simplify the expression:

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

To find if there are real solutions, solve \(2x^2 - 4x + 3 = 0\).

Calculate the discriminant \(b^2 - 4ac\) for the quadratic equation \(ax^2 + bx + c = 0\):

Here, \(a = 2\), \(b = -4\), \(c = 3\).

\(b^2 - 4ac = (-4)^2 - 4 \times 2 \times 3 = 16 - 24 = -8\).

Since the discriminant is negative, the quadratic equation has no real solutions.

Therefore, the equation \(gf(x) = 0\) has no real solutions.

(ii) To find \(f^{-1}(x)\), start with \(y = 2x - 5\). Solve for \(x\):

\(y + 5 = 2x\)

\(x = \frac{1}{2}(y + 5)\)

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

To find \(g^{-1}(x)\), start with \(y = \frac{4}{2-x}\). Solve for \(x\):

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

\(2y - xy = 4\)

\(xy = 2y - 4\)

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

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

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

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

\(\frac{1}{2}(x + 5) = 2 - \frac{4}{x}\)

Multiply through by 2x to clear fractions:

\(x(x + 5) = 4x - 8\)

\(x^2 + 5x = 4x - 8\)

\(x^2 + x + 8 = 0\)

Use the discriminant \(b^2 - 4ac\):

\(1^2 - 4 \times 1 \times 8 = 1 - 32 = -31\)

Since the discriminant is negative, there are no real roots.

(i) To find \(f^{-1}(x)\), start with \(y = \frac{2}{3 - 2x}\). Rearrange to get \(y(3 - 2x) = 2\), which simplifies to \(3y - 2xy = 2\). Solving for \(x\), we get \(2xy = 3y - 2\), so \(x = \frac{3y - 2}{2y}\). Thus, \(f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\).

(ii) For \(gf(-1) = 3\), calculate \(f(-1) = \frac{2}{3 - 2(-1)} = \frac{2}{5}\). Then \(g(f(-1)) = 4 \times \frac{2}{5} + a = \frac{8}{5} + a\). Set \(\frac{8}{5} + a = 3\) to find \(a = \frac{7}{5}\).

(iii) Set \(f^{-1}(x) = g^{-1}(x)\). From (i), \(f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\). For \(g^{-1}(x) = \frac{x - a}{4}\), equate \(\frac{3}{2} - \frac{1}{x} = \frac{x - a}{4}\). Rearrange to form a quadratic equation: \(x^2 - x(a + 6) + 4 = 0\). For two equal roots, the discriminant must be zero: \((a + 6)^2 = 16\). Solving gives \(a + 6 = 4\) or \(a + 6 = -4\), so \(a = -2\) or \(a = -10\).