| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify, factorise, solve with substitution |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on the Factor Theorem requiring routine verification that (2x+1) is a factor, then factorisation (likely quadratic factor remaining), followed by a trigonometric equation that converts to the polynomial. The proof of no real solutions likely involves showing the polynomial has no real roots via discriminant or completing the square. Standard techniques throughout 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=1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p\left(-\frac{1}{2}\right) = 30\times\left(-\frac{1}{2}\right)^3 - 7\left(-\frac{1}{2}\right)^2 - 7\left(-\frac{1}{2}\right) + 2 = 0\) | M1 (AO1.1a) | Begins a proof using a valid method e.g. Factor theorem, algebraic division, multiplication of correct factors |
| \(\therefore 2x+1\) is a factor of \(p(x)\) | R1 (AO2.1) | Constructs rigorous mathematical proof. Must clearly substitute and state that \(p(-1/2)=0\) and clearly state that this implies \(2x+1\) is a factor. Method must be completely correct with a concluding statement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p(x) = (2x+1)(15x^2-11x+2)\) | M1 (AO1.1a) | Obtains quadratic factor PI |
| \(= (2x+1)(5x-2)(3x-1)\) | A1 (AO1.1b) | Obtains second linear factor |
| A1 (AO1.1b) | Writes \(p(x)\) as the product of the correct three linear factors. NMS correct answer 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{30\sec^2 x + 2\cos x}{7} = \sec x + 1\) | M1 (AO3.1a) | Rearranges to achieve a cubic equation in \(\sec x\) (or \(\cos x\)) |
| \(\Rightarrow 30\sec^2 x + 2\cos x = 7\sec x + 7\) | M1 (AO1.1a) | Equates to zero and uses result from (b) or factorises |
| \(\Rightarrow 30\sec^3 x + 2 = 7\sec^2 x + 7\sec x\) | ||
| \(30\sec^3 x - 7\sec^2 x - 7\sec x + 2 = 0\) | ||
| \(\Rightarrow (2\sec x+1)(5\sec x-2)(3\sec x-1)=0\) | ||
| \(\Rightarrow \sec x = -\frac{1}{2}, \frac{1}{3}, \frac{2}{5}\) | A1 (AO2.2a) | Deduces that if solutions exist they must be of the form \(\sec x = -\frac{1}{2}\), \(\sec x = \frac{1}{3}\) or \(\sec x = \frac{2}{5}\) OE |
| These values do not fall within the range of \(\sec x\) as they are between \(-1\) and \(1\) | E1 (AO2.4) | Explains that the range of \(\sec x\) is \((-\infty,-1]\cup[1,\infty)\) OE; OE for \(\cos x\) |
| \(\therefore \frac{30\sec^2 x + 2\cos x}{7} = \sec x + 1\) has no real solutions | R1 (AO2.1) | Completes argument explaining that there cannot be any real solutions as values are outside of the function's range |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p\left(-\frac{1}{2}\right) = 30\times\left(-\frac{1}{2}\right)^3 - 7\left(-\frac{1}{2}\right)^2 - 7\left(-\frac{1}{2}\right) + 2 = 0$ | M1 (AO1.1a) | Begins a proof using a valid method e.g. Factor theorem, algebraic division, multiplication of correct factors |
| $\therefore 2x+1$ is a factor of $p(x)$ | R1 (AO2.1) | Constructs rigorous mathematical proof. Must clearly substitute and state that $p(-1/2)=0$ **and** clearly state that this implies $2x+1$ is a factor. Method must be completely correct with a concluding statement |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p(x) = (2x+1)(15x^2-11x+2)$ | M1 (AO1.1a) | Obtains quadratic factor PI |
| $= (2x+1)(5x-2)(3x-1)$ | A1 (AO1.1b) | Obtains second linear factor |
| | A1 (AO1.1b) | Writes $p(x)$ as the product of the correct three linear factors. NMS correct answer 3/3 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{30\sec^2 x + 2\cos x}{7} = \sec x + 1$ | M1 (AO3.1a) | Rearranges to achieve a cubic equation in $\sec x$ (or $\cos x$) |
| $\Rightarrow 30\sec^2 x + 2\cos x = 7\sec x + 7$ | M1 (AO1.1a) | Equates to zero and uses result from (b) or factorises |
| $\Rightarrow 30\sec^3 x + 2 = 7\sec^2 x + 7\sec x$ | | |
| $30\sec^3 x - 7\sec^2 x - 7\sec x + 2 = 0$ | | |
| $\Rightarrow (2\sec x+1)(5\sec x-2)(3\sec x-1)=0$ | | |
| $\Rightarrow \sec x = -\frac{1}{2}, \frac{1}{3}, \frac{2}{5}$ | A1 (AO2.2a) | Deduces that if solutions exist they must be of the form $\sec x = -\frac{1}{2}$, $\sec x = \frac{1}{3}$ or $\sec x = \frac{2}{5}$ OE |
| These values do not fall within the range of $\sec x$ as they are between $-1$ and $1$ | E1 (AO2.4) | Explains that the range of $\sec x$ is $(-\infty,-1]\cup[1,\infty)$ OE; OE for $\cos x$ |
| $\therefore \frac{30\sec^2 x + 2\cos x}{7} = \sec x + 1$ has no real solutions | R1 (AO2.1) | Completes argument explaining that there cannot be any real solutions as values are outside of the function's range |
---12
\begin{enumerate}[label=(\alph*)]
\item Prove that ( $2 x + 1$ ) is a factor of $\mathrm { p } ( x )$\\
12
\item Factorise $\mathrm { p } ( x )$ completely.\\
12
\item Prove that there are no real solutions to the equation
$$\frac { 30 \sec ^ { 2 } x + 2 \cos x } { 7 } = \sec x + 1$$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q12 [10]}}