IBDP Maths AHL 2.12 Polynomial functions and their graphs AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) Showing (x+1) is a factor:
Solution:
Using Factor Theorem:
\( P(-1) = 3(-1)^3 + 5(-1)^2 + (-1) – 1 = -3 + 5 – 1 – 1 = 0 \)
Thus, \( (x+1) \) is a factor of \( P(x) \).
[2 marks]
(b) Factoring P(x):
Solution:
Using polynomial division or synthetic division:
\( P(x) = (x+1)(3x^2 + 2x – 1) \)
Further factoring the quadratic:
\( = (x+1)(x+1)(3x-1) = (x+1)^2(3x-1) \)
[3 marks]
(c) Partial fraction decomposition:
Solution:
Set up:
\( \frac{1}{(x+1)(2x+1)} = \frac{A}{x+1} + \frac{B}{2x+1} \)
Solving gives \( A = -1 \) and \( B = 2 \)
Thus:
\( \frac{1}{Q(x)} = -\frac{1}{x+1} + \frac{2}{2x+1} \)
[3 marks]
(d) Further decomposition:
Solution:
Using result from (c):
\( \frac{1}{(x+1)Q(x)} = \frac{1}{(x+1)^2(2x+1)} = \frac{4}{2x+1} – \frac{2}{x+1} – \frac{1}{(x+1)^2} \)
[3 marks]
(e) Integration:
Solution:
Using the decomposition from (d):
\( \int \frac{1}{(x+1)^2(2x+1)} dx = 2\ln|2x+1| – 2\ln|x+1| + \frac{1}{x+1} + C \)
[4 marks]
(f) Limits:
(i) Solution:
Method 1: Factor cancellation
\( \lim_{x \to -1} \frac{(x+1)^2(3x-1)}{(x+1)^2(2x+1)} = \frac{3(-1)-1}{2(-1)+1} = \frac{-4}{-1} = 4 \)
(ii) Solution:
Method 1: Divide by highest power
\( \lim_{x \to \infty} \frac{3x^3}{2x^3} = \frac{3}{2} \)
[4 marks]
▶️ Answer/Explanation
Method 1: Using Sum and Product of Roots
Solution:
1. The other two roots must be complex conjugates: \( a – bi \) and \( b – ai \)
2. Sum of roots: \( (a+bi) + (a-bi) + (b+ai) + (b-ai) = 2a + 2b = -4 \)
⇒ \( a + b = -2 \)
3. Product of roots: \( (a^2 + b^2)^2 = 400 \) ⇒ \( a^2 + b^2 = 20 \)
4. Solve the system:
\( b = -2 – a \)
\( a^2 + (-2-a)^2 = 20 \) ⇒ \( 2a^2 + 4a – 16 = 0 \) ⇒ \( a^2 + 2a – 8 = 0 \)
Solutions: \( a = 2 \) or \( a = -4 \)
[5 marks]
Method 2: Polynomial Factorization
Solution:
1. Factor the polynomial using the given roots:
\( (z – (a+bi))(z – (a-bi))(z – (b+ai))(z – (b-ai)) = 0 \)
2. This becomes:
\( [(z-a)^2 + b^2][(z-b)^2 + a^2] = 0 \)
3. Expand and compare coefficients:
Coefficient of \( z^3 \): \( -2a -2b = -4 \) ⇒ \( a + b = 2 \)
Constant term: \( (a^2 + b^2)^2 = 400 \) ⇒ \( a^2 + b^2 = 20 \)
4. Solve the system to get \( a = 2 \) or \( a = -4 \)
[5 marks]
▶️ Answer/Explanation
(a) \(4{(x – 0.5)^2} + 4\) A1A1
Note: A1 for two correct parameters, A2 for all three correct.
[2 marks]
(b)
- Translation \(\left( \begin{array}{c} 0.5 \\ 0 \end{array} \right)\) (allow “0.5 to the right”) A1
- Stretch parallel to y-axis, scale factor 4 (allow vertical stretch or similar) A1
- Translation \(\left( \begin{array}{c} 0 \\ 4 \end{array} \right)\) (allow “4 up”) A1
Note: All transformations must state magnitude and direction.
Note: First two transformations can be in either order.
[3 marks]
(c)
- General shape (including asymptote and single maximum in first quadrant) A1
- Intercept \(\left( 0,\frac{1}{5} \right)\) or maximum \(\left( \frac{1}{2},\frac{1}{4} \right)\) shown A1
[2 marks]
(d) \(0 < f(x) \leqslant \frac{1}{4}\) A1A1
Note: A1 for \( \leqslant \frac{1}{4}\), A1 for \(0 < \).
[2 marks]
(e)
Let \(u = x – \frac{1}{2}\) A1
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\) (or \(\mathrm{d}u = \mathrm{d}x\)) A1
\(\int \frac{1}{4{x^2} – 4x + 5}\,\mathrm{d}x = \int \frac{1}{4{\left( x – \frac{1}{2} \right)}^2 + 4}\,\mathrm{d}x\) A1
\(\int \frac{1}{4{u^2} + 4}\,\mathrm{d}u = \frac{1}{4}\int \frac{1}{{u^2} + 1}\,\mathrm{d}u\) AG
Note: If following through an incorrect answer to part (a), do not award final A1 mark.
[3 marks]
(f)
\(\int_1^{3.5} \frac{1}{4{x^2} – 4x + 5}\,\mathrm{d}x = \frac{1}{4}\int_{0.5}^3 \frac{1}{{u^2} + 1}\,\mathrm{d}u\) A1
Note: A1 for correct change of limits.
\(\frac{1}{4}\left[ \arctan(u) \right]_{0.5}^3\) (M1)
\(\frac{1}{4}\left( \arctan(3) – \arctan\left( \frac{1}{2} \right) \right)\) A1
Let the integral = I
\(\tan 4I = \tan \left( \arctan(3) – \arctan\left( \frac{1}{2} \right) \right)\) M1
\(\frac{3 – 0.5}{1 + 3 \times 0.5} = \frac{2.5}{2.5} = 1\) (M1)A1
\(4I = \frac{\pi}{4} \Rightarrow I = \frac{\pi}{16}\) A1AG
[7 marks]
▶️ Answer/Explanation
Solution:
(a) Finding the value of k:
Since \((x – 4)\) is a factor, by the Factor Theorem we know that: \[ q(4) = 0 \] Substituting \(x = 4\) into the polynomial: \[ 3(4)^3 – 11(4)^2 + k(4) + 8 = 0 \] \[ 192 – 176 + 4k + 8 = 0 \] \[ 24 + 4k = 0 \] \[ k = -6 \]
(b) Factorizing the polynomial:
With \(k = -6\), the polynomial becomes: \[ q(x) = 3x^3 – 11x^2 – 6x + 8 \]
Using the known factor \((x – 4)\), we perform polynomial division or inspection: \[ q(x) = (x – 4)(3x^2 + x – 2) \] The quadratic can be further factored: \[ 3x^2 + x – 2 = (3x – 2)(x + 1) \] Therefore, the complete factorization is: \[ q(x) = (x – 4)(3x – 2)(x + 1) \]
Markscheme:
(a)
\(q(4) = 0\) (M1)
\(192 – 176 + 4k + 8 = 0{\text{ }}(24 + 4k = 0)\) A1
\(k = -6\) A1
[3 marks]
(b)
\(3x^3 – 11x^2 – 6x + 8 = (x – 4)(3x^2 + px – 2)\)
equate coefficients of \(x^2\): (M1)
\(-12 + p = -11\)
\(p = 1\)
\((x – 4)(3x^2 + x – 2)\) (A1)
\((x – 4)(3x – 2)(x + 1)\) A1
Note: Allow part (b) marks if any of this work is seen in part (a).
Note: Allow equivalent methods (eg, synthetic division) for the M marks in each part.
[3 marks]
Total: [6 marks]