| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Exponential substitution equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard factor theorem application, polynomial factorization, and exponential substitution. Part (a) is routine verification, part (b) requires algebraic division and factoring a quadratic, and part (c) involves recognizing the substitution x = 2^y. While the exponential substitution adds a small twist, the overall question requires only standard techniques with clear signposting, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR | ||
| \(f(0.5) = 0.25 - 0.75 - 5.5 + 6 = 0\) | B1 | Attempt \(f(0.5)\) and show equal to 0; must be using factor theorem so B0 for alternative methods; condone \(2(0.5)^3 - 3(0.5)^2 - 11(0.5) + 6 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR | ||
| \(f(x) = (2x-1)(x^2 - x - 6)\) | M1 | Attempt complete division by \((2x-1)\); allow equivalent complete methods eg coefficient matching/inspection/grid method; condone slip(s) in otherwise correct method |
| A1 | Obtain correct quadratic factor; seen in division/correct coeffs eg \(A = 1\) etc/at top of grid | |
| \(f(x) = (2x-1)(x-3)(x+2)\) | A1 | Obtain correct fully factorised \(f(x)\); must be seen as a product of all 3 factors; SC B1 for correct factorisation with no DR |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR | ||
| \(x = 2^y\) | B1 | State or imply that \(x = 2^y\); could be implied by equating \(2^y\) to at least one of their roots |
| \(2^y = 0.5,\ y = -1\) and \(2^y = 3,\ y = 1.58\) | M1 | Attempt to find at least one value of \(y\); exact or decimal |
| \(2^y = -2\), no solutions as \(2^y > 0\) for all \(y\); hence \(y = -1,\ y = \log_2 3\) | A1 | Obtain both correct values and no others; 1.58 or better, or \(\frac{\log_n 3}{\log_n 2}\) for \(\log_2 3\); must give reason for \(2^y = -2\) having no solution: eg cannot log a negative number / \(2^y\) always greater than 0 |
# Question 4:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** | | |
| $f(0.5) = 0.25 - 0.75 - 5.5 + 6 = 0$ | B1 | Attempt $f(0.5)$ and show equal to 0; must be using factor theorem so B0 for alternative methods; condone $2(0.5)^3 - 3(0.5)^2 - 11(0.5) + 6 = 0$ |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** | | |
| $f(x) = (2x-1)(x^2 - x - 6)$ | M1 | Attempt complete division by $(2x-1)$; allow equivalent complete methods eg coefficient matching/inspection/grid method; condone slip(s) in otherwise correct method |
| | A1 | Obtain correct quadratic factor; seen in division/correct coeffs eg $A = 1$ etc/at top of grid |
| $f(x) = (2x-1)(x-3)(x+2)$ | A1 | Obtain correct fully factorised $f(x)$; must be seen as a product of all 3 factors; **SC B1** for correct factorisation with no DR |
---
# Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** | | |
| $x = 2^y$ | B1 | State or imply that $x = 2^y$; could be implied by equating $2^y$ to at least one of their roots |
| $2^y = 0.5,\ y = -1$ and $2^y = 3,\ y = 1.58$ | M1 | Attempt to find at least one value of $y$; exact or decimal |
| $2^y = -2$, no solutions as $2^y > 0$ for all $y$; hence $y = -1,\ y = \log_2 3$ | A1 | Obtain both correct values and no others; 1.58 or better, or $\frac{\log_n 3}{\log_n 2}$ for $\log_2 3$; must give **reason** for $2^y = -2$ having no solution: eg cannot log a negative number / $2^y$ always greater than 0 |
---4 In this question you must show detailed reasoning.\\
The cubic polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6$.
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $( 2 x - 1 )$ is a factor of $\mathrm { f } ( x )$.
\item Express $\mathrm { f } ( x )$ in fully factorised form.
\item Hence solve the equation $2 \times 8 ^ { y } - 3 \times 4 ^ { y } - 11 \times 2 ^ { y } + 6 = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2021 Q4 [7]}}