Unit 10 Review Exercises
Further Functions
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. AHL marks questions for HL students only.
1 Paper 1 Let \(f(x) = \dfrac{3x - 2}{x + 4}\). Find the equations of the asymptotes and the axis intercepts of the graph of \(y = f(x)\).
Vertical asymptote \(x = -4\); horizontal asymptote \(y = \dfrac{3}{1} = 3\).
\(x\)-intercept: \(3x - 2 = 0 \Rightarrow x = \dfrac{2}{3}\). \(\;y\)-intercept: \(f(0) = -\dfrac{2}{4} = -\dfrac{1}{2}\). See 10.1.
2 Paper 1 Let \(f(x) = \dfrac{4x + 3}{2x - 4}\), \(x \neq 2\). Find \(f^{-1}(x)\), and hence show that \(f\) is self-inverse.
\(y(2x - 4) = 4x + 3 \Rightarrow 2xy - 4x = 4y + 3 \Rightarrow x(2y - 4) = 4y + 3 \Rightarrow x = \dfrac{4y + 3}{2y - 4}\).
So \(f^{-1}(x) = \dfrac{4x + 3}{2x - 4} = f(x)\), and \(f\) is self-inverse. See 10.1.
3 Paper 1 The graph of \(y = \dfrac{ax + b}{x + c}\) has a vertical asymptote \(x = 3\), a horizontal asymptote \(y = -2\), and an \(x\)-intercept at \((1, 0)\). Find \(a\), \(b\) and \(c\).
\(x = -c = 3\), so \(c = -3\). \(\;y = \dfrac{a}{1} = -2\), so \(a = -2\).
\(x\)-intercept \(1\): \(-2(1) + b = 0\), so \(b = 2\). The function is \(y = \dfrac{2 - 2x}{x - 3}\). See 10.1.
4 Paper 1 AHL Let \(f(x) = \dfrac{x - 1}{x^2 + x - 6}\).
(a) Find the equations of all asymptotes and the axis intercepts.
(b) Draw a sign diagram for \(f(x)\).
(a) \(x^2 + x - 6 = (x + 3)(x - 2)\): vertical asymptotes \(x = -3\) and \(x = 2\). Horizontal asymptote \(y = 0\). \(x\)-intercept \(1\); \(y\)-intercept \(f(0) = \dfrac{-1}{-6} = \dfrac{1}{6}\).
(b)
| \(x\) | \(x < -3\) | \(-3\) | \(-3 < x < 1\) | \(1\) | \(1 < x < 2\) | \(2\) | \(x > 2\) |
|---|---|---|---|---|---|---|---|
| sign | \(-\) | undefined | \(+\) | \(0\) | \(-\) | undefined | \(+\) |
See 10.2.
5 Paper 1 AHL Let \(g(x) = \dfrac{x^2 - 3x + 4}{x - 1}\).
(a) Write \(g(x)\) in the form \(px + q + \dfrac{r}{x - 1}\).
(b) Write down the equations of the asymptotes.
(c) Show that the graph does not meet the \(x\)-axis, and find the \(y\)-intercept.
(a) \(x^2 - 3x + 4 = (x - 1)(x - 2) + 2\), so \(g(x) = x - 2 + \dfrac{2}{x - 1}\).
(b) \(x = 1\) and \(y = x - 2\).
(c) For \(x^2 - 3x + 4 = 0\), \(\Delta = 9 - 16 = -7 < 0\): no real roots, so no \(x\)-intercepts. \(y\)-intercept: \(g(0) = \dfrac{4}{-1} = -4\). See 10.2.
6 Paper 1 AHL The graph of \(f(x) = \dfrac{x + 3}{x^2 - 4x + k}\) has exactly one vertical asymptote. Find the two possible values of \(k\), and the vertical asymptote in each case.
Case 1: repeated root. \(\Delta = 16 - 4k = 0 \Rightarrow k = 4\). Then \(x^2 - 4x + 4 = (x - 2)^2\): asymptote \(x = 2\).
Case 2: one root cancels with the numerator. \(x = -3\) is a root when \(9 + 12 + k = 0\), so \(k = -21\). Then \(x^2 - 4x - 21 = (x + 3)(x - 7)\) and \(f(x) = \dfrac{1}{x - 7}\) for \(x \neq -3\): one asymptote \(x = 7\) (and a hole at \(x = -3\)).
So \(k = 4\) or \(k = -21\). See 10.2.
7 Paper 1 AHL Solve \(|3x - 2| = |x + 6|\).
\(3x - 2 = x + 6 \Rightarrow x = 4\), or \(3x - 2 = -(x + 6) \Rightarrow 4x = -4 \Rightarrow x = -1\).
Check: \(|10| = |10|\) ✓, \(|-5| = |5|\) ✓. See 10.3.
8 Paper 1 AHL Solve \(|x - 3| > 2x\).
If \(x < 0\), the right side is negative and \(|x - 3| \geq 0\), so the inequality is always true.
If \(x \geq 0\), both sides are non-negative, so square: \((x - 3)^2 > 4x^2 \Rightarrow 3x^2 + 6x - 9 < 0 \Rightarrow x^2 + 2x - 3 < 0 \Rightarrow (x + 3)(x - 1) < 0 \Rightarrow -3 < x < 1\). With \(x \geq 0\): \(0 \leq x < 1\).
Together: \(x < 1\). See 11.5.
9 Paper 1 AHL Solve \(\left|\dfrac{x - 1}{x + 3}\right| \geq 1\).
\(x \neq -3\). Multiply by \(|x + 3| > 0\): \(|x - 1| \geq |x + 3|\). Square:
\[ x^2 - 2x + 1 \geq x^2 + 6x + 9 \;\Longrightarrow\; -8x \geq 8 \;\Longrightarrow\; x \leq -1 . \]
So \(x \leq -1\), \(x \neq -3\). See 11.5.
10 Paper 2 AHL Solve \(|x^2 - 4| < 2^x\).
Draw \(y = |x^2 - 4|\) and \(y = 2^x\). The GDC gives three intersections: \(x = -2.06\), \(x = -1.93\) and \(x = 1.26\) (3 s.f.).
The exponential curve is above the modulus graph only in a narrow interval around \(x = -2\) (where \(|x^2 - 4| = 0\)) and for all \(x\) to the right of \(1.26\). So
\[ -2.06 < x < -1.93 \quad\text{or}\quad x > 1.26 . \]
See 11.5.