Past Exam Questions

(i) To express \(8x - x^2\) in the form \(a - (x + b)^2\), we complete the square:

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

Thus, \(a = 16\) and \(b = -4\).

(ii) The stationary point occurs where \(\frac{dy}{dx} = 0\). Differentiating, \(\frac{dy}{dx} = 8 - 2x\).

Setting \(8 - 2x = 0\) gives \(x = 4\).

Substituting \(x = 4\) into \(y = 8x - x^2\), we get \(y = 16\).

Thus, the stationary point is \((4, 16)\).

(iii) Solve \(8x - x^2 \geq -20\):

\(x^2 - 8x - 20 \leq 0\).

Factorizing gives \((x - 10)(x + 2) \leq 0\).

The solution is \(-2 \leq x \leq 10\).

(iv) From the function \(g(x) = 8x - x^2\) for \(x \geq 4\), the maximum value is at \(x = 4\), giving \(y = 16\).

Thus, the domain of \(g^{-1}\) is \(x \leq 16\) and the range is \(x \geq 4\).

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

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

Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 4y}}{2}\).

Since \(x \geq 4\), we take the positive root: \(x = 4 + \sqrt{16 - y}\).

Thus, \(g^{-1}(x) = 4 + \sqrt{16 - x}\).

(a) To find the inverse, start with \(y = (x + a)^2 - a\).

Rearrange to solve for \(x\):

\(y + a = (x + a)^2\)

\(x + a = \pm \sqrt{y + a}\)

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

\(x = -\sqrt{y + a} - a\)

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

(b)(i) The domain of \(f^{-1}\) is the range of \(f\), which is \(x > -a\).

(b)(ii) The range of \(f^{-1}\) is the domain of \(f\), which is \(y \leq -a\).

(a) The function \(f(x) = 2 - \frac{5}{x+2}\) has a horizontal asymptote at \(y = 2\). As \(x \to \infty\), \(f(x) \to 2\). Therefore, the range of \(f\) is \(y < 2\).

(b) To find the inverse, start with \(y = 2 - \frac{5}{x+2}\).

Rearrange to solve for \(x\):

\(y(x+2) = 2(x+2) - 5\)

\(yx + 2y = 2x + 4 - 5\)

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

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

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

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

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

The domain of \(f^{-1}\) is determined by the range of \(f\), which is \(x < 2\).

First, complete the square for \(g(x) = x^2 - 6x + 7\).

Rewrite \(x^2 - 6x + 7\) as \((x-3)^2 - 9 + 7\).

This simplifies to \((x-3)^2 - 2\).

Set \(y = (x-3)^2 - 2\) and solve for \(x\).

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

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

Since \(x > 4\), choose the positive root: \(x = 3 + \sqrt{y + 2}\).

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

The domain of \(g^{-1}\) is \(x > -1\) because \(x + 2 > 0\).

(i) To express \(9x^2 - 6x + 6\) in the form \((ax + b)^2 + c\), complete the square:

\(9x^2 - 6x + 6 = 9(x^2 - \frac{2}{3}x) + 6\)

\(= 9((x - \frac{1}{3})^2 - \frac{1}{9}) + 6\)

\(= 9(x - \frac{1}{3})^2 - 1 + 6\)

\(= 9(x - \frac{1}{3})^2 + 5\)

Thus, \(a = 3\), \(b = -1\), \(c = 5\).

(ii) The function \(f(x) = 9x^2 - 6x + 6\) is a quadratic function that opens upwards. For \(f\) to be one-one, it must be restricted to the domain where it is either increasing or decreasing. The vertex of the parabola is at \(x = \frac{1}{3}\). Therefore, the smallest value of \(p\) is \(\frac{1}{3}\).

(iii) For \(p = \frac{1}{3}\), the function is one-one for \(x \geq \frac{1}{3}\). To find \(f^{-1}(x)\), start with \(y = (3x-1)^2 + 5\):

\(y - 5 = (3x-1)^2\)

\(\sqrt{y-5} = \pm (3x-1)\)

Since \(x \geq \frac{1}{3}\), we take the positive root:

\(3x - 1 = \sqrt{y-5}\)

\(3x = \sqrt{y-5} + 1\)

\(x = \frac{1}{3}\sqrt{y-5} + \frac{1}{3}\)

Thus, \(f^{-1}(x) = \frac{1}{3}\sqrt{x-5} + \frac{1}{3}\) with domain \(x \geq 5\).

(iv) The equation \(f(x) = q\) has no solution when \(q < 5\) because the minimum value of \(f(x)\) is 5, which occurs at the vertex \(x = \frac{1}{3}\).

(i) To express \(-x^2 + 6x - 5\) in the form \(a(x + b)^2 + c\), we complete the square:

\(-x^2 + 6x - 5 = -1(x^2 - 6x) - 5\)

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

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

Thus, \(-1(x^2 - 6x) = -1((x - 3)^2 - 9) = -1(x - 3)^2 + 9\)

Therefore, \(-x^2 + 6x - 5 = -1(x - 3)^2 + 9 - 5 = -1(x - 3)^2 + 4\)

(ii) The function \(f(x) = -x^2 + 6x - 5\) is a downward-opening parabola. It is one-one when restricted to the domain where \(x\) is greater than or equal to the vertex. The vertex is at \(x = 3\), so the smallest \(m\) is 3.

(iii) For \(m = 5\), we find the inverse:

Start with \(y = -x^2 + 6x - 5\)

Rearrange to solve for \(x\):

\(y = -(x - 3)^2 + 4\)

\((x - 3)^2 = 4 - y\)

\(x - 3 = \pm \sqrt{4 - y}\)

Since \(x \geq 3\), we take the positive root:

\(x = 3 + \sqrt{4 - y}\)

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

The domain of \(f^{-1}\) is \(x \leq 0\) because the range of \(f\) for \(x \geq 5\) is \(-\infty \leq y \leq 0\).

Let \(y = \frac{15}{2x+3}\).

Rearrange to solve for \(x\):

\(2x + 3 = \frac{15}{y}\)

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

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

Thus, \(f^{-1}(x) = \frac{15 - 3x}{2x}\).

For the range of \(f^{-1}\), since \(0 \leq x \leq 6\) for \(f(x)\), the range of \(f^{-1}\) is \(0 \leq f^{-1}(x) \leq 6\).

For the domain of \(f^{-1}\), solve \(0 \leq \frac{15}{2x+3} \leq 6\) for \(x\):

\(1 \leq x \leq 5\).

Let \(y = \frac{5}{1 - 3x}\).

Rearrange to make \(x\) the subject:

\(3x = 1 - \frac{5}{y}\)

\(3x = \frac{y - 5}{y}\)

\(x = \frac{y - 5}{3y}\)

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

The range of \(f\) is \(x \geq 1\), so the domain of \(f^{-1}\) is \(-2.5 \leq x < 0\).

The domain of \(f\) is \(x \geq 1\), so the range of \(f^{-1}\) is \(x \geq 1\).

(i) To express \(2x^2 - 12x + 13\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 12x + 13 = 2(x^2 - 6x) + 13\)

\(= 2((x-3)^2 - 9) + 13\)

\(= 2(x-3)^2 - 18 + 13\)

\(= 2(x-3)^2 - 5\)

Thus, \(a = 2, b = -3, c = -5\).

(ii) For \(f(x)\) to be one-one, the derivative \(f'(x) = 4x - 12\) must be non-negative for \(x \geq k\). Solving \(4x - 12 \geq 0\) gives \(x \geq 3\). Therefore, the smallest possible value of \(k\) is 3.

(iii) With \(k = 7\), the minimum value of \(f(x)\) is \(f(7)\):

\(f(7) = 2(7)^2 - 12(7) + 13 = 98 - 84 + 13 = 27\)

Thus, the range of \(f\) is \(y \geq 27\).

(iv) To find \(f^{-1}(x)\), start with \(y = 2(x-3)^2 - 5\) and solve for \(x\):

\(2(x-3)^2 = y + 5\)

\((x-3)^2 = \frac{1}{2}(y + 5)\)

\(x-3 = \pm \sqrt{\frac{1}{2}(y + 5)}\)

Since \(x \geq 7\), choose the positive root:

\(x = 3 + \sqrt{\frac{1}{2}(y + 5)}\)

Thus, \(f^{-1}(x) = 3 + \sqrt{\frac{1}{2}(x + 5)}\) for \(x \geq 27\).

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

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

Substitute into \(g(x) = 2x - 2\):

\(gf(x) = 2\left(1 + \frac{3}{x-2}\right) - 2\).

Simplify the expression:

\(gf(x) = 2 + \frac{6}{x-2} - 2\).

Combine like terms:

\(gf(x) = \frac{6}{x-2}\).

\((a) To solve the equation f(x) = 0, we have:\)

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

\(Let y = x^{\frac{1}{2}}, then x = y^2. Substitute to get:\)

\(y^2 - 4y + 1 = 0\)

\(Using the quadratic formula, y = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}\)

\(Thus, x = (2 \pm \sqrt{3})^2 = 7 \pm 4\sqrt{3}\)

(b) Given f(x) \equiv gh(x), we have:

\(gh(x) = m\left(x^{\frac{1}{2}} - 2\right)^2 + n\)

Expanding, we get:

\(gh(x) = m(x - 4x^{\frac{1}{2}} + 4) + n\)

\(Since f(x) = x - 4x^{\frac{1}{2}} + 1, equate the expressions:\)

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

\(Comparing coefficients, m = 1 and n = -3.\)

(b) Start with \(y = \frac{-x}{\sqrt{4-x^2}}\). Square both sides to get \(x^2 = y^2(4-x^2)\).

Rearrange to \(x^2(1+y^2) = 4y^2\).

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

Select the correct square root: \(f^{-1}(x) = \frac{-2x}{\sqrt{1+x^2}}\).

(c) The maximum possible value of \(a\) is 1, as \(-1 < x < 1\) is required for \(fg\) to be formed.

(d) Substitute \(g(x) = 2x\) into \(f(x)\): \(fg(x) = f(2x) = \frac{-2x}{\sqrt{4-(2x)^2}}\).

Simplify to \(fg(x) = \frac{-x}{\sqrt{1-x^2}}\).

(a) To find \(ff(5)\), we first calculate \(f(5)\):

\(f(5) = \frac{5+3}{5-1} = \frac{8}{4} = 2\).

Now, calculate \(f(f(5)) = f(2)\):

\(f(2) = \frac{2+3}{2-1} = \frac{5}{1} = 5\).

Thus, \(ff(5) = 5\).

(b) To find \(f^{-1}(x)\), set \(y = f(x) = \frac{x+3}{x-1}\).

Swap \(x\) and \(y\):

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

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

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

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

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

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

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

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

(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:

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

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

\(9x = 3a\)

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