| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Prove root count with given polynomial |
| Difficulty | Standard +0.8 This is a multi-part question requiring factor theorem application, substitution reasoning (y² = x), and trigonometric analysis across multiple periods. Part (a) is routine, but parts (b) and (c) require careful reasoning about solution counts through substitution and periodic behavior, going beyond standard textbook exercises. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2)=-24+32-18+10 \Rightarrow f(2)=0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \((x-2)\) or \((2-x)\) is a factor and attempts quadratic factor by long division or factorisation | M1 | Using long division to obtain \(\pm3x^2 \pm kx+...\), \(k\neq0\), or factorising to obtain \((\pm3x^2 \pm kx \pm c)\) |
| \((x-2)(-3x^2+2x-5)\) or \((2-x)(3x^2-2x+5)\) or \(-(x-2)(3x^2-2x+5)\) stated as a product | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-3y^6+8y^4-9y^2+10=0 \Rightarrow (y^2-2)(-3y^4+2y^2-5)=0\); explains \(-3y^4+2y^2-5=0\) has no real solutions e.g. \(b^2-4ac=(2)^2-4(-3)(-5)=-56<0\), or \(y^2=2\) has 2 real solutions \(y=\pm\sqrt{2}\) | M1 | 2.4; correct follow-through for discriminant of their \(-3y^4+2y^2-5\); \(<0\) must be stated for A1 |
| Complete proof that equation has exactly two real solutions | A1 | Proof must be correct with no errors; do not allow for incorrect working e.g. \((2)^2-4(-3)(-5)=-54\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\tan^3\theta-8\tan^2\theta+9\tan\theta-10=0\); \(7\pi \leq \theta < 10\pi\); deduces there are 3 solutions | B1 | Give B0 for stating \(\theta=\) awrt 23.1, awrt 26.2, awrt 29.4 without reference to 3 solutions; do not recover work from (b) in (c) |
## Question 6:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2)=-24+32-18+10 \Rightarrow f(2)=0$ | B1 | |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $(x-2)$ or $(2-x)$ is a factor and attempts quadratic factor by long division or factorisation | M1 | Using long division to obtain $\pm3x^2 \pm kx+...$, $k\neq0$, or factorising to obtain $(\pm3x^2 \pm kx \pm c)$ |
| $(x-2)(-3x^2+2x-5)$ or $(2-x)(3x^2-2x+5)$ or $-(x-2)(3x^2-2x+5)$ stated as a product | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-3y^6+8y^4-9y^2+10=0 \Rightarrow (y^2-2)(-3y^4+2y^2-5)=0$; explains $-3y^4+2y^2-5=0$ has no real solutions e.g. $b^2-4ac=(2)^2-4(-3)(-5)=-56<0$, or $y^2=2$ has 2 real solutions $y=\pm\sqrt{2}$ | M1 | 2.4; correct follow-through for discriminant of their $-3y^4+2y^2-5$; $<0$ must be stated for A1 |
| Complete proof that equation has exactly two real solutions | A1 | Proof must be correct **with no errors**; do not allow for incorrect working e.g. $(2)^2-4(-3)(-5)=-54$ |
**Notes:** Using formula on $-3y^4+2y^2-5=0$ gives $y^2=\frac{-2\pm\sqrt{-56}}{-6}$; completing square on $-3x^2+2x-5=0$ gives $x=\frac{1}{3}\pm\sqrt{\frac{-14}{9}}$
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\tan^3\theta-8\tan^2\theta+9\tan\theta-10=0$; $7\pi \leq \theta < 10\pi$; deduces there are 3 solutions | B1 | Give B0 for stating $\theta=$ awrt 23.1, awrt 26.2, awrt 29.4 **without** reference to 3 solutions; do not recover work from (b) in (c) |
---6.
$$f ( x ) = - 3 x ^ { 3 } + 8 x ^ { 2 } - 9 x + 10 , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Calculate f(2)
\item Write $\mathrm { f } ( x )$ as a product of two algebraic factors.
Using the answer to (a)(ii),
\end{enumerate}\item prove that there are exactly two real solutions to the equation
$$- 3 y ^ { 6 } + 8 y ^ { 4 } - 9 y ^ { 2 } + 10 = 0$$
\item deduce the number of real solutions, for $7 \pi \leqslant \theta < 10 \pi$, to the equation
$$3 \tan ^ { 3 } \theta - 8 \tan ^ { 2 } \theta + 9 \tan \theta - 10 = 0$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q6 [6]}}