| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining polynomial factorization with trigonometric substitution. Part (a) involves routine verification that f(1)=0 and standard factor theorem application to solve a cubic. Part (b) requires recognizing the substitution sin θ = x and finding angles, which is a standard technique. The question is slightly above average difficulty due to the multi-step nature and the trigonometric extension, but requires no novel insight—it's a textbook exercise testing standard A-level techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Show \(f(1) = 0\) | B1 [1] | Correct working and result seen |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-1)\) is a factor | B1 | stated or implied, eg by attempt \(\div\) by \((x-1)\); or \((2x-1)\) or \((x+3)\) is factor |
| Attempt find other (quadratic) factor by any method | M1 | or show that \(f(-3)=0\) or \(f(\frac{1}{2})=0\); Inspection: Must have \(2x^2\) and \(\pm 3\) oe |
| \((x-1)(2x^2+5x-3)\) | A1f* | or \(x = \frac{-5 \pm \sqrt{25-4\times2\times(-3)}}{4}\); ft their quad factor |
| \((x-1)(2x-1)(x+3)\) | A1f | |
| \(x=1\) or \(\frac{1}{2}\) or \(-3\) | A1 [4] | Dep A1*; eg correct factors no working: B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\theta = 1\) or \(\sin\theta = \frac{1}{2}\) | M1 | Use of a root from (a). May be implied |
| \(\theta = 90°, 30°, 150°\) | A3f [3.1a, 1.1, 1.1] | Three correct none incorrect: A3; Three correct and \(\geq 1\) incorrect: A2; Two correct: A2; One correct: A1; ft their (a) for all A-marks; Ignore any "correct" solutions outside range |
| \(\sin\theta = -3\) gives no solution, or it doesn't exist oe or outside range or impossible or not acceptable oe | B1 [5] | Statement needed. Ignore all else; Not incorrect reason eg "no solution because minus": B0; Just "X" or "Error" not enough because DR |
## Question 3:
### Part (a)(i):
Show $f(1) = 0$ | **B1** [1] | Correct working and result seen
### Part (a)(ii):
$(x-1)$ is a factor | **B1** | stated or implied, eg by attempt $\div$ by $(x-1)$; or $(2x-1)$ or $(x+3)$ is factor
Attempt find other (quadratic) factor by any method | **M1** | or show that $f(-3)=0$ or $f(\frac{1}{2})=0$; Inspection: Must have $2x^2$ and $\pm 3$ oe
$(x-1)(2x^2+5x-3)$ | **A1f*** | or $x = \frac{-5 \pm \sqrt{25-4\times2\times(-3)}}{4}$; ft their quad factor
$(x-1)(2x-1)(x+3)$ | **A1f** |
$x=1$ or $\frac{1}{2}$ or $-3$ | **A1** [4] | Dep A1*; eg correct factors no working: B1 only
### Part (b):
$\sin\theta = 1$ or $\sin\theta = \frac{1}{2}$ | **M1** | Use of a root from (a). May be implied
$\theta = 90°, 30°, 150°$ | **A3f** [3.1a, 1.1, 1.1] | Three correct none incorrect: A3; Three correct and $\geq 1$ incorrect: A2; Two correct: A2; One correct: A1; ft their (a) for all A-marks; **Ignore** any "correct" solutions outside range
$\sin\theta = -3$ gives no solution, or it doesn't exist oe or outside range or impossible or not acceptable oe | **B1** [5] | Statement needed. Ignore all else; Not incorrect reason eg "no solution because minus": B0; Just "X" or "Error" not enough because **DR**
---3 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 8 x + 3$.
\begin{enumerate}[label=(\roman*)]
\item Show that $f ( 1 ) = 0$.
\item Solve the equation $\mathrm { f } ( x ) = 0$.
\end{enumerate}\item Hence solve the equation $2 \sin ^ { 3 } \theta + 3 \sin ^ { 2 } \theta - 8 \sin \theta + 3 = 0$ for $0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q3 [10]}}