| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete square then find vertex/turning point |
| Difficulty | Moderate -0.8 This is a straightforward completing the square question with standard follow-up parts. Part (a) is routine algebraic manipulation, part (b) is direct reading from the completed square form, and part (c) requires recognizing the substitution x = tan θ and using basic trigonometry. While multi-part, each step follows a well-practiced procedure with no novel problem-solving required, making it easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(p = 2\) | B1 | Could be implied by \(2(x+q)^2 + r\) |
| \(q = 1.5\) | B1 | Could be implied by \(p(x+1.5)^2 + r\) |
| \(r = 2.5\) giving \(2(x+1.5)^2 + 2.5\) | B1FT | FT on their \(p\) and \(q\) ie \(7 - pq^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((-1.5, 2.5)\) — Correct \(x\)-coordinate | B1FT | FT on their (a). Could come from differentiation |
| Correct \(y\)-coordinate | B1FT | FT on their (a). No FT on incorrect \(x\)-value from differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| minimum value of the function \(= 2.5\) | B1FT | FT on their minimum value. Allow BOD if different answers in (a) and (b). 2.5 must be stated as, or clearly intended to be, the minimum value. Just \((\ldots, 2.5)\) is insufficient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan\theta = -1.5\), \(\theta = -56.3°\) | M1 | Attempt to solve \(\tan\theta =\) their \((-1.5)\); to obtain numerical value for \(\theta\); allow radians (expect \(-0.983\) rad); allow BOD if different answers in (a) and (b) |
| \(\theta = 124°\) | A1 [3] | Obtain \(124°\) or better; A0 if additional solutions; condone approaches other than 'hence' |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $p = 2$ | **B1** | Could be implied by $2(x+q)^2 + r$ |
| $q = 1.5$ | **B1** | Could be implied by $p(x+1.5)^2 + r$ |
| $r = 2.5$ giving $2(x+1.5)^2 + 2.5$ | **B1FT** | FT on their $p$ and $q$ ie $7 - pq^2$ |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(-1.5, 2.5)$ — Correct $x$-coordinate | **B1FT** | FT on their (a). Could come from differentiation |
| Correct $y$-coordinate | **B1FT** | FT on their (a). No FT on incorrect $x$-value from differentiation |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| minimum value of the function $= 2.5$ | **B1FT** | FT on their minimum value. Allow BOD if different answers in (a) and (b). 2.5 must be stated as, or clearly intended to be, the minimum value. Just $(\ldots, 2.5)$ is insufficient |
# Question (Previous - tan θ):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan\theta = -1.5$, $\theta = -56.3°$ | M1 | Attempt to solve $\tan\theta =$ their $(-1.5)$; to obtain numerical value for $\theta$; allow radians (expect $-0.983$ rad); allow BOD if different answers in (a) and (b) |
| $\theta = 124°$ | A1 [3] | Obtain $124°$ or better; **A0** if additional solutions; condone approaches other than 'hence' |
---4
\begin{enumerate}[label=(\alph*)]
\item Write $2 x ^ { 2 } + 6 x + 7$ in the form $p ( x + q ) ^ { 2 } + r$, where $p , q$ and $r$ are constants.
\item State the coordinates of the minimum point on the graph of $y = 2 x ^ { 2 } + 6 x + 7$.
\item Hence deduce
\begin{itemize}
\item the minimum value of $2 \tan ^ { 2 } \theta + 6 \tan \theta + 7$,
\item the smallest positive value of $\theta$, in degrees, for which the minimum value occurs.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2022 Q4 [8]}}