| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining routine Factor/Remainder Theorem application with a trigonometric substitution. Part (i) is direct substitution, part (ii) is standard factorization, and part (iii) requires recognizing the substitution cos θ = x but then just solving the already-factorized equation. All techniques are standard C2 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(2) = 32 - 14 - 3 = 15\) | M1 | Attempt \(f(2)\) or equiv. M0 for using \(x = -2\). At least one of the first two terms must be of the correct sign. Must be evaluated and not just substituted. |
| Obtain 15 | A1 | Do not ISW if subsequently given as \(-15\). If using division, just seeing 15 on bottom line is fine unless subsequently contradicted by eg \(-15\) or \(\frac{15}{x-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(-\frac{1}{2}) = -\frac{1}{2} + \frac{7}{2} - 3 = 0\) AG | B1 | Confirm \(f(-\frac{1}{2}) = 0\), with at least one line of working. \(4(-\frac{1}{2})^3 - 7(-\frac{1}{2}) - 3 = 0\) is enough. B0 for just \(f(-\frac{1}{2}) = 0\). If division must show '0' on last line. If coefficient matching must show '\(R = 0\)'. |
| \(f(x) = (2x+1)(2x^2 - x - 3)\) | M1 | Attempt complete division by \((2x+1)\), or another correct factor. Must be complete method - ie all 3 terms attempted. |
| \(= (2x+1)(2x-3)(x+1)\) | A1 | Obtain \(2x^2\) and one other correct term. |
| A1 | Obtain fully correct quotient of \(2x^2 - x - 3\). | |
| M1 | Attempt to factorise their quadratic quotient from division attempt by correct factor. | |
| A1 | Obtain \((2x+1)(2x-3)(x+1)\). Final answer must be seen as a product of all three factors. Allow factorised equiv such as \(2(2x+1)(x-\frac{3}{2})(x+1)\). SR: If repeated use of factor theorem, or answer given with no working, allow possible B1 for \(f(-\frac{1}{2})=0\) with additional B5 for \((2x+1)(2x-3)(x+1)\), or B3 for a multiple such as \((2x+1)(x-\frac{3}{2})(x+1)\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2\cos\theta + 1 = 0 \quad \cos\theta + 1 = 0\) \(2\cos\theta - 3 = 0\) | M1* | Identify relationship between factors of \(f(\cos\theta)\) and factors of \(f(x)\). Replace \(x\) with \(\cos\theta\) in at least one of their factors. |
| \(\cos\theta = -\frac{1}{2} \quad \cos\theta = -1\) \(\cos\theta = \frac{3}{2}\) | M1d* | Attempt to solve \(\cos\theta = k\) at least once. Must actually attempt \(\theta\), with \(-1 \leq k \leq 1\). |
| \(\theta = \frac{2\pi}{3},\ \frac{4\pi}{3} \quad \theta = \pi\) | A1 | Obtain at least 2 correct angles. Allow angles in degrees (\(120°, 240°, 180°\)). Allow decimal equivs (2.09, 4.19, 3.14). |
| A1 | Obtain all 3 correct angles. Must be exact and in radians. A0 if additional incorrect angles in range. A0 if incorrect root used even if it doesn't affect the three solutions. |
## Question 9:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(2) = 32 - 14 - 3 = 15$ | M1 | Attempt $f(2)$ or equiv. M0 for using $x = -2$. At least one of the first two terms must be of the correct sign. Must be evaluated and not just substituted. |
| Obtain 15 | A1 | Do not ISW if subsequently given as $-15$. If using division, just seeing 15 on bottom line is fine unless subsequently contradicted by eg $-15$ or $\frac{15}{x-2}$ |
---
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(-\frac{1}{2}) = -\frac{1}{2} + \frac{7}{2} - 3 = 0$ **AG** | B1 | Confirm $f(-\frac{1}{2}) = 0$, with at least one line of working. $4(-\frac{1}{2})^3 - 7(-\frac{1}{2}) - 3 = 0$ is enough. B0 for just $f(-\frac{1}{2}) = 0$. If division must show '0' on last line. If coefficient matching must show '$R = 0$'. |
| $f(x) = (2x+1)(2x^2 - x - 3)$ | M1 | Attempt complete division by $(2x+1)$, or another correct factor. Must be complete method - ie all 3 terms attempted. |
| $= (2x+1)(2x-3)(x+1)$ | A1 | Obtain $2x^2$ and one other correct term. |
| | A1 | Obtain fully correct quotient of $2x^2 - x - 3$. |
| | M1 | Attempt to factorise their quadratic quotient from division attempt by correct factor. |
| | A1 | Obtain $(2x+1)(2x-3)(x+1)$. Final answer must be seen as a product of all three factors. Allow factorised equiv such as $2(2x+1)(x-\frac{3}{2})(x+1)$. **SR**: If repeated use of factor theorem, or answer given with no working, allow possible B1 for $f(-\frac{1}{2})=0$ with additional **B5** for $(2x+1)(2x-3)(x+1)$, or **B3** for a multiple such as $(2x+1)(x-\frac{3}{2})(x+1)$. |
---
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2\cos\theta + 1 = 0 \quad \cos\theta + 1 = 0$ $2\cos\theta - 3 = 0$ | M1* | Identify relationship between factors of $f(\cos\theta)$ and factors of $f(x)$. Replace $x$ with $\cos\theta$ in at least one of their factors. |
| $\cos\theta = -\frac{1}{2} \quad \cos\theta = -1$ $\cos\theta = \frac{3}{2}$ | M1d* | Attempt to solve $\cos\theta = k$ at least once. Must actually attempt $\theta$, with $-1 \leq k \leq 1$. |
| $\theta = \frac{2\pi}{3},\ \frac{4\pi}{3} \quad \theta = \pi$ | A1 | Obtain at least 2 correct angles. Allow angles in degrees ($120°, 240°, 180°$). Allow decimal equivs (2.09, 4.19, 3.14). |
| | A1 | Obtain all 3 correct angles. Must be exact and in radians. A0 if additional incorrect angles in range. A0 if incorrect root used even if it doesn't affect the three solutions. |9 The cubic polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 4 x ^ { 3 } - 7 x - 3$.\\
(i) Find the remainder when $\mathrm { f } ( x )$ is divided by ( $x - 2$ ).\\
(ii) Show that ( $2 x + 1$ ) is a factor of $\mathrm { f } ( x )$ and hence factorise $\mathrm { f } ( x )$ completely.\\
(iii) Solve the equation
$$4 \cos ^ { 3 } \theta - 7 \cos \theta - 3 = 0$$
for $0 \leqslant \theta \leqslant 2 \pi$. Give each solution for $\theta$ in an exact form.
\hfill \mbox{\textit{OCR C2 2013 Q9 [12]}}