| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Exponential substitution equations |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring standard techniques: factor theorem to find k (routine substitution), polynomial factorization (straightforward once k is known), exponential substitution (recognizing e^{-t} as the variable), and logarithm manipulation. While it spans multiple topics and requires careful algebra, each step follows predictable A-level methods with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06e Logarithm as inverse: ln(x) inverse of e^x1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f\!\left(\frac{1}{2}\right) = 6\!\left(\frac{1}{2}\right)^3 + k\!\left(\frac{1}{2}\right)^2 + 57\!\left(\frac{1}{2}\right) - 20\) | M1 | Substitutes \(x = \pm 0.5\) into \(f(x)\); allow one slip |
| \(\frac{3}{4} + \frac{k}{4} + \frac{57}{2} - 20 = 0 \Rightarrow k = -37\) | A1 | AG – \(f\!\left(\frac{1}{2}\right) = 0\) and at least one line of intermediate working before given answer. Long division in this part is no marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = 6x^3 - 37x^2 + 57x - 20\) \(\Rightarrow f(x) = (2x-1)(3x^2 + kx + 20)\) | M1 | Quadratic factor found with correct coefficient of \(x^2\) and constant term (or second correct root found from factor theorem). Or from long division: \(3x^2\) and an \(x\) term at least for M1 |
| \(f(x) = (2x-1)(3x^2 - 17x + 20)\) | A1 | Or second factor stated correctly |
| \(= (2x-1)(3x-5)(x-4)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{-t} = \frac{1}{2}, \frac{5}{3}, 4\) | M1* | Correctly relates \(e^{-t}\) to at least one of the roots of \(f(x) = 0\) |
| \(t = -\ln\!\left(\frac{1}{2}\right), -\ln\!\left(\frac{5}{3}\right), -\ln 4\) | A1 | Correctly takes logs and obtains correct values of \(t\). Any equivalent form, allow 3 s.f. or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum t = -\!\left(\ln\!\left(\frac{1}{2}\right) + \ln\!\left(\frac{5}{3}\right) + \ln 4\right) = -\ln\!\left(\frac{5 \times 4}{2 \times 3}\right)\) | M1dep* | Correctly uses log laws to add their three values of \(t\) together. Dependent on M mark in part (i) |
| \(= \ln\!\left(\frac{3}{10}\right)\) or \(-\ln\!\left(\frac{10}{3}\right)\) | A1 | cao |
# Question 4:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f\!\left(\frac{1}{2}\right) = 6\!\left(\frac{1}{2}\right)^3 + k\!\left(\frac{1}{2}\right)^2 + 57\!\left(\frac{1}{2}\right) - 20$ | M1 | Substitutes $x = \pm 0.5$ into $f(x)$; allow one slip |
| $\frac{3}{4} + \frac{k}{4} + \frac{57}{2} - 20 = 0 \Rightarrow k = -37$ | A1 | AG – $f\!\left(\frac{1}{2}\right) = 0$ and at least one line of intermediate working before given answer. Long division in this part is no marks |
**[2 marks]**
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = 6x^3 - 37x^2 + 57x - 20$ $\Rightarrow f(x) = (2x-1)(3x^2 + kx + 20)$ | M1 | Quadratic factor found with correct coefficient of $x^2$ and constant term (or second correct root found from factor theorem). Or from long division: $3x^2$ and an $x$ term at least for M1 |
| $f(x) = (2x-1)(3x^2 - 17x + 20)$ | A1 | Or second factor stated correctly |
| $= (2x-1)(3x-5)(x-4)$ | A1 | |
**[3 marks]**
## Part (c)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{-t} = \frac{1}{2}, \frac{5}{3}, 4$ | M1* | Correctly relates $e^{-t}$ to at least one of the roots of $f(x) = 0$ |
| $t = -\ln\!\left(\frac{1}{2}\right), -\ln\!\left(\frac{5}{3}\right), -\ln 4$ | A1 | Correctly takes logs and obtains correct values of $t$. Any equivalent form, allow 3 s.f. or better |
**[2 marks]**
## Part (c)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum t = -\!\left(\ln\!\left(\frac{1}{2}\right) + \ln\!\left(\frac{5}{3}\right) + \ln 4\right) = -\ln\!\left(\frac{5 \times 4}{2 \times 3}\right)$ | M1dep* | Correctly uses log laws to add their three values of $t$ together. Dependent on M mark in part (i) |
| $= \ln\!\left(\frac{3}{10}\right)$ or $-\ln\!\left(\frac{10}{3}\right)$ | A1 | cao |
**[2 marks]**
---4 In this question you must show detailed reasoning.
The cubic polynomial $6 x ^ { 3 } + k x ^ { 2 } + 57 x - 20$ is denoted by $\mathrm { f } ( x )$. It is given that $( 2 x - 1 )$ is a factor of $\mathrm { f } ( x )$.
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $k = - 37$.
\item Using this value of $k$, factorise $\mathrm { f } ( x )$ completely.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find the three values of $t$ satisfying the equation $6 \mathrm { e } ^ { - 3 t } - 37 \mathrm { e } ^ { - 2 t } + 57 \mathrm { e } ^ { - t } - 20 = 0$.
\item Express the sum of the three values found in part (c)(i) as a single logarithm.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q4 [9]}}