Unit 11 Review Exercises

Polynomials and Inequalities

These questions mix the ideas from the whole unit. All of them are HL only AHL. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 AHL Find the coefficient of \(x^2\) in the expansion of \((2x^2 - x + 3)(x^2 + 4x - 5)\).

Pairs giving \(x^2\): \(\;2x^2 \times (-5)\), \(\;(-x) \times 4x\), \(\;3 \times x^2\).

\(-10 - 4 + 3 = -11\). See 11.1.

2 Paper 1 AHL Find the quotient and the remainder when \(2x^4 + x^3 - 3x + 5\) is divided by \(x^2 + 1\).

Write the divisor as \(x^2 + 0x + 1\) and the dividend as \(2x^4 + x^3 + 0x^2 - 3x + 5\).

\[ \begin{array}{r} 2x^2 + x - 2\phantom{)} \\[-3pt] x^2 + 0x + 1\,\enclose{longdiv}{\,2x^4 + x^3 + 0x^2 - 3x + 5} \\ \underline{2x^4 + 0x^3 + 2x^2\phantom{{} - 3x + 5}} \\ x^3 - 2x^2 - 3x\phantom{{} + 5} \\ \underline{x^3 + 0x^2 + x\phantom{{} + 5}} \\ -2x^2 - 4x + 5 \\ \underline{-2x^2 + 0x - 2} \\ -4x + 7 \end{array} \]

Quotient \(2x^2 + x - 2\), remainder \(-4x + 7\). See 11.1.

3 Paper 1 AHL Let \(f(x) = x^3 + ax^2 - 3x + b\). The factor \(x - 2\) divides \(f(x)\) exactly, and the remainder is \(12\) when \(f(x)\) is divided by \(x + 1\).

(a) Find \(a\) and \(b\).

(b) Hence solve \(f(x) = 0\), giving exact answers.

(a) \(f(2) = 8 + 4a - 6 + b = 0 \Rightarrow 4a + b = -2\). \(\;f(-1) = -1 + a + 3 + b = 12 \Rightarrow a + b = 10\).

Subtracting: \(3a = -12\), so \(a = -4\) and \(b = 14\).

(b) \(f(x) = x^3 - 4x^2 - 3x + 14\). Dividing by \(x - 2\) gives \(x^2 - 2x - 7\).

\(x^2 - 2x - 7 = 0 \Rightarrow x = \dfrac{2 \pm \sqrt{32}}{2} = 1 \pm 2\sqrt{2}\).

Roots: \(x = 2\), \(x = 1 \pm 2\sqrt{2}\). See 11.2 and 11.3.

4 Paper 1 AHL The graph of a quartic function \(y = f(x)\) touches the \(x\)-axis at \(x = -2\), crosses it at \(x = 1\) and at \(x = 3\), and meets the \(y\)-axis at \((0, -12)\).

(a) Find \(f(x)\) in factorised form.

(b) Describe the behaviour of \(y\) as \(x \to \pm\infty\).

(a) \(f(x) = a(x + 2)^2(x - 1)(x - 3)\). At \(x = 0\): \(\;a(4)(-1)(-3) = 12a = -12\), so \(a = -1\).

\[ f(x) = -(x + 2)^2(x - 1)(x - 3) \]

(b) Degree 4 (even) with leading coefficient \(-1 < 0\): \(y \to -\infty\) at both ends. See 11.3.

5 Paper 1 AHL The roots of \(x^2 - 4x + 1 = 0\) are \(\alpha\) and \(\beta\).

(a) Find the value of \(\alpha^2 + \beta^2\).

(b) Find a quadratic equation whose roots are \(\alpha^2\) and \(\beta^2\).

\(\alpha + \beta = 4\), \(\alpha\beta = 1\).

(a) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 16 - 2 = 14\).

(b) Sum \(14\), product \(\alpha^2\beta^2 = (\alpha\beta)^2 = 1\): \(\;x^2 - 14x + 1 = 0\). See 11.3.

6 Paper 1 AHL The equation \(3x^3 - 6x^2 + kx - 9 = 0\) has three roots. Two of the roots have a product of \(3\).

(a) Find the third root.

(b) Find \(k\).

(c) Show that the other two roots are not real.

(a) Product of all three roots \(= \dfrac{(-1)^3 \times (-9)}{3} = 3\). Two of them multiply to \(3\), so the third root is \(1\).

(b) \(x = 1\) is a root: \(3 - 6 + k - 9 = 0\), so \(k = 12\).

(c) \(3x^3 - 6x^2 + 12x - 9 = 3(x - 1)(x^2 - x + 3)\). The discriminant of \(x^2 - x + 3\) is \(1 - 12 = -11 < 0\), so these two roots are not real (they are complex, with product \(3\) ✓). See 11.3.

7 Paper 1 AHL Solve \(x^3 - 4x^2 + x + 6 < 0\).

\(x = -1\): \(-1 - 4 - 1 + 6 = 0\). So \(x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3)\).

Critical values \(-1, 2, 3\); a positive cubic is negative to the left of \(-1\) and between \(2\) and \(3\):

\[ x < -1 \quad\text{or}\quad 2 < x < 3 . \]

See 11.2 and 11.4.

8 Paper 1 AHL Solve \(\dfrac{2}{x - 3} \leq \dfrac{1}{x + 1}\).

\[ \begin{gathered} \frac{2}{x - 3} - \frac{1}{x + 1} \leq 0 \\[4pt] \frac{2(x + 1) - (x - 3)}{(x - 3)(x + 1)} \leq 0 \\[4pt] \frac{x + 5}{(x - 3)(x + 1)} \leq 0 . \end{gathered} \]

Critical values \(-5\) (included), \(-1\) and \(3\) (excluded). Signs from the right: \(+, -, +, -\).

\[ x \leq -5 \quad\text{or}\quad -1 < x < 3 . \]

See 11.4.

9 Paper 1 AHL

(a) Solve \(\dfrac{4}{x - 1} \geq x + 2\).

(b) Hence solve \(\dfrac{4}{\ln x - 1} \geq \ln x + 2\).

(a)

\[ \begin{gathered} \frac{4 - (x + 2)(x - 1)}{x - 1} \geq 0 \\[4pt] \frac{-(x^2 + x - 6)}{x - 1} \geq 0 \\[4pt] \frac{(x + 3)(x - 2)}{x - 1} \leq 0 . \end{gathered} \]

Critical values \(-3\), \(1\) (excluded), \(2\). Signs from the right \(+, -, +, -\):

\[ x \leq -3 \quad\text{or}\quad 1 < x \leq 2 . \]

(b) Replace \(x\) by \(\ln x\) (with \(x > 0\)): \(\;\ln x \leq -3\;\) or \(\;1 < \ln x \leq 2\), so

\[ 0 < x \leq \mathrm{e}^{-3} \quad\text{or}\quad \mathrm{e} < x \leq \mathrm{e}^{2} . \]

See 11.4.

10 Paper 1 AHL Solve \(|x + 3| > |2x|\).

Both sides are non-negative, so square:

\[ \begin{aligned} (x + 3)^2 - (2x)^2 &> 0 \\ (x + 3 - 2x)(x + 3 + 2x) &> 0 \\ (3 - x)(3x + 3) &> 0 . \end{aligned} \]

This is \(-3(x - 3)(x + 1) > 0\), that is \((x - 3)(x + 1) < 0\):

\[ -1 < x < 3 . \]

See 11.5.

11 Paper 2 AHL The height of a drone above the ground is \(h(t) = 0.5t^3 - 6t^2 + 20t\) metres, \(0 \leq t \leq 8\), where \(t\) is in seconds. Find the times when the drone is more than \(12\) m above the ground.

Graph \(y = h(t)\) and \(y = 12\) for \(0 \leq t \leq 8\). The intersections are at \(t \approx 0.764\), \(t \approx 5.24\) and \(t = 6\).

(By hand: \(0.5t^3 - 6t^2 + 20t - 12 = 0.5(t - 6)(t^2 - 6t + 4)\), with roots \(6\) and \(3 \pm \sqrt{5}\).)

The curve is above the line for

\[ 0.764 < t < 5.24 \quad\text{or}\quad 6 < t \leq 8 . \]

See 11.4 and 11.5.

12 Paper 2 AHL Solve \(\ln(x + 4) \geq |x|\).

The domain is \(x > -4\). Graph \(y = \ln(x + 4)\) and \(y = |x|\). The GDC gives intersections at \(x \approx -1.07\) and \(x \approx 1.75\).

The log curve is on or above the V between them (for example at \(x = 0\): \(\ln 4 \approx 1.39 \geq 0\)):

\[ -1.07 \leq x \leq 1.75 . \]

See 11.5.