Unit 11 Exam Practice
Polynomials and Inequalities
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
This whole unit is HL only AHL 2.12 AHL 2.15 AHL 2.16, so there is no SL set.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
HL questions
These questions use the remainder and factor theorems, the sum and product of roots, and polynomial, rational and modulus inequalities, and combine them with arithmetic sequences from Unit 2 and exponentials and logarithms from Unit 1.
HL 1 Paper 1
Let \(f(x) = x^{3} + ax^{2} + bx + c\), where \(a\), \(b\) and \(c\) are constants. It is given that \(x - 2\) is a factor of \(f(x)\), and that the remainder is \(9 - 6x\) when \(f(x)\) is divided by \(x^{2} - 1\).
(a) Find the values of \(a\), \(b\) and \(c\). [5]
(b) Hence solve the equation \(f(x) = 0\), giving your answers in exact form. [3]
Total 8 marks
(a)
\(f(x) \equiv (x^{2} - 1)\,Q(x) + 9 - 6x\), so \(f(1) = 3\) and \(f(-1) = 15\) M1
\(1 + a + b + c = 3\) and \(-1 + a - b + c = 15\) A1
\(f(2) = 0\): \(\;8 + 4a + 2b + c = 0\) A1
subtracting the first two equations: \(2 + 2b = -12\), so \(b = -7\) A1
\(a + c = 9\) and \(4a + c = 6\), so \(a = -1\) and \(c = 10\) A1
[5 marks]
(b)
\(x^{3} - x^{2} - 7x + 10 = (x - 2)(x^{2} + x - 5)\) M1 A1
\(x = 2\) or \(x = \dfrac{-1 \pm \sqrt{21}}{2}\) A1
Award M1 for any valid method of division by \(x - 2\) (long division, synthetic division or comparing coefficients).
[3 marks]
HL 2 Paper 1
The roots of the equation \(3x^{2} - 2x + k = 0\) are \(\alpha\) and \(\beta\), where \(k\) is a constant. It is given that \(\alpha^{2} + \beta^{2} = \dfrac{28}{9}\).
(a) Find the value of \(k\). [3]
(b) Find a quadratic equation, with integer coefficients, whose roots are \(\alpha + \dfrac{1}{\beta}\) and \(\beta + \dfrac{1}{\alpha}\). [5]
Total 8 marks
(a)
\(\alpha + \beta = \dfrac{2}{3}\) and \(\alpha\beta = \dfrac{k}{3}\) A1
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = \dfrac{4}{9} - \dfrac{2k}{3}\) M1
\(\dfrac{4}{9} - \dfrac{2k}{3} = \dfrac{28}{9}\), so \(k = -4\) A1
[3 marks]
(b)
\(\alpha\beta = -\dfrac{4}{3}\); sum of new roots \(= (\alpha + \beta) + \dfrac{\alpha + \beta}{\alpha\beta}\) M1
\(= \dfrac{2}{3} - \dfrac{1}{2} = \dfrac{1}{6}\) A1
product of new roots \(= \alpha\beta + 1 + 1 + \dfrac{1}{\alpha\beta}\) M1
\(= -\dfrac{4}{3} + 2 - \dfrac{3}{4} = -\dfrac{1}{12}\) A1
\(x^{2} - \dfrac{1}{6}x - \dfrac{1}{12} = 0\), that is, \(12x^{2} - 2x - 1 = 0\) A1
Accept any integer multiple of \(12x^{2} - 2x - 1 = 0\). The final A1 needs “\(= 0\)”.
[5 marks]
See 11.3.
HL 3 Paper 1
(a) Solve the inequality \(\dfrac{2x - 7}{x - 2} \geq \dfrac{x - 1}{x + 2}\). [5]
(b) Hence, by using the substitution \(u = \dfrac{1}{x}\), solve the inequality \(\dfrac{2 - 7x}{1 - 2x} \geq \dfrac{1 - x}{1 + 2x}\). [3]
Total 8 marks
(a)
\(\dfrac{2x - 7}{x - 2} - \dfrac{x - 1}{x + 2} \geq 0 \;\Rightarrow\; \dfrac{(2x - 7)(x + 2) - (x - 1)(x - 2)}{(x - 2)(x + 2)} \geq 0\) M1
numerator \(= x^{2} - 16 = (x - 4)(x + 4)\) A1
critical values \(-4\), \(-2\), \(2\), \(4\), where \(x = \pm 2\) are excluded A1
\(x \leq -4\) or \(-2 < x < 2\) or \(x \geq 4\) A2
Do not award the M1 for cross-multiplying without considering the signs of the denominators. Award A1 (not A2) for two correct intervals, or for three intervals with an error in the end points.
[5 marks]
(b)
for \(x \neq 0\), multiply top and bottom by \(u = \dfrac{1}{x}\): \(\dfrac{2u - 7}{u - 2} \geq \dfrac{u - 1}{u + 2}\), which is (a) in \(u\) M1
\(u \leq -4\), \(-2 < u < 2\), \(u \geq 4\) give \(-\dfrac{1}{4} \leq x < 0\), \(|x| > \dfrac{1}{2}\), \(0 < x \leq \dfrac{1}{4}\) A1
\(x = 0\) also works (\(2 \geq 1\)), so \(x < -\dfrac{1}{2}\) or \(-\dfrac{1}{4} \leq x \leq \dfrac{1}{4}\) or \(x > \dfrac{1}{2}\) A1
Award the final A1 only if \(x = 0\) is included. \(|x| > \dfrac{1}{2}\) comes from \(0 < u < 2\) and \(-2 < u < 0\) (\(u\) is never \(0\)).
[3 marks]
HL 4 Paper 2
(a) Solve the inequality \(|x - 3| > 2|x + 1|\). [4]
(b) Solve the inequality \(|x - 3| > 2x + 2\). [3]
(c) Hence solve the inequality \(|\mathrm{e}^{x} - 3| > 2\mathrm{e}^{x} + 2\). [2]
(d) Solve the inequality \(|x - 3| \geq 2\ln(x + 2)\). [4]
Total 13 marks
(a)
both sides are non-negative, so square: \((x - 3)^{2} > 4(x + 1)^{2}\) M1
\((x - 3)^{2} - (2x + 2)^{2} > 0 \;\Rightarrow\; (-x - 5)(3x - 1) > 0\) A1
\((x + 5)(3x - 1) < 0\), critical values \(-5\) and \(\dfrac{1}{3}\) A1
\(-5 < x < \dfrac{1}{3}\) A1
Accept the expanded form \(3x^{2} + 14x - 5 < 0\) for the second A1.
[4 marks]
(b)
\(2x + 2\) can be negative, so consider the two cases of the modulus M1
\(x \geq 3\): \(\;x - 3 > 2x + 2\) gives \(x < -5\), so no solutions A1
\(x < 3\): \(\;3 - x > 2x + 2\) gives \(x < \dfrac{1}{3}\); answer \(x < \dfrac{1}{3}\) A1
Do not award the M1 for squaring both sides.
[3 marks]
(c)
this is (b) with \(x\) replaced by \(\mathrm{e}^{x}\), so \(\mathrm{e}^{x} < \dfrac{1}{3}\) M1
\(x < \ln \dfrac{1}{3}\), that is, \(x < -\ln 3\) A1
[2 marks]
(d)
intersection of \(y = |x - 3|\) and \(y = 2\ln(x + 2)\) (for example, using a GDC) M1
\(x = 0.883\) and \(x = 7.50\) \((0.882608\ldots,\ 7.50327\ldots)\) A1
\(\ln(x + 2)\) is defined only for \(x > -2\) R1
\(-2 < x \leq 0.883\) or \(x \geq 7.50\) A1
Do not award the final A1 for \(x \leq 0.883\) without the lower bound \(-2\).
[4 marks]
HL 5 Paper 1
Let \(P(x) = 2x^{3} - 9x^{2} + kx - 6\), where \(k\) is a constant. The three roots of the equation \(P(x) = 0\) are real, and they form an arithmetic sequence.
(a) Show that \(\dfrac{3}{2}\) is a root of \(P(x) = 0\). [2]
(b) Hence find the value of \(k\). [2]
(c) Find the other two roots. [3]
(d) Show that \(P(x) \equiv (2x - 3)(x^{2} - 3x + 1) + 2x - 3\). [2]
(e) Hence solve the inequality \(P(x) > 2x - 3\). [4]
(f) Hence solve the inequality \(P(|x|) > 2|x| - 3\). [3]
Total 16 marks
(a)
let the roots be \(m - d\), \(m\), \(m + d\); their sum is \(3m\) M1
sum of roots \(= \dfrac{-(-9)}{2} = \dfrac{9}{2}\) A1
\(3m = \dfrac{9}{2}\), so \(m = \dfrac{3}{2}\) is a root AG
[2 marks]
(b)
\(P\left(\dfrac{3}{2}\right) = \dfrac{27}{4} - \dfrac{81}{4} + \dfrac{3k}{2} - 6 = 0\) M1
\(\dfrac{3k}{2} = \dfrac{39}{2}\), so \(k = 13\) A1
[2 marks]
(c)
product of roots: \(\left(\dfrac{3}{2} - d\right) \times \dfrac{3}{2} \times \left(\dfrac{3}{2} + d\right) = \dfrac{(-1)^{3}(-6)}{2}\) M1
\(\dfrac{3}{2}\left(\dfrac{9}{4} - d^{2}\right) = 3\), so \(d^{2} = \dfrac{1}{4}\) A1
the other two roots are \(1\) and \(2\) A1
Accept division of \(P(x)\) by \(2x - 3\), giving \(x^{2} - 3x + 2 = (x - 1)(x - 2)\).
[3 marks]
(d)
\(P(x) = (2x - 3)(x - 1)(x - 2) = (2x - 3)(x^{2} - 3x + 2)\) M1
\(= (2x - 3)(x^{2} - 3x + 1) + (2x - 3)\) A1 AG
Accept expanding the right-hand side to obtain \(2x^{3} - 9x^{2} + 13x - 6\).
[2 marks]
(e)
\(P(x) - (2x - 3) > 0 \;\Rightarrow\; (2x - 3)(x^{2} - 3x + 1) > 0\) M1
critical values \(x = \dfrac{3}{2}\) and \(x = \dfrac{3 \pm \sqrt{5}}{2}\) A1
sign diagram, or sketch of a cubic with positive leading coefficient M1
\(\dfrac{3 - \sqrt{5}}{2} < x < \dfrac{3}{2}\) or \(x > \dfrac{3 + \sqrt{5}}{2}\) A1
Do not award the first M1 for dividing both sides by \(2x - 3\).
[4 marks]
(f)
it is (e) with \(x\) replaced by \(|x|\) M1
\(\dfrac{3 - \sqrt{5}}{2} < |x| < \dfrac{3}{2}\): \(\;-\dfrac{3}{2} < x < -\dfrac{3 - \sqrt{5}}{2}\) or \(\dfrac{3 - \sqrt{5}}{2} < x < \dfrac{3}{2}\) A1
\(|x| > \dfrac{3 + \sqrt{5}}{2}\): \(\;x < -\dfrac{3 + \sqrt{5}}{2}\) or \(x > \dfrac{3 + \sqrt{5}}{2}\) A1
[3 marks]