OCR H240/02 2018 June — Question 7 7 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
TopicFunction Transformations
TypeInverse transformation or reflection
DifficultyStandard +0.3 This question tests understanding of function transformations through reflections, requiring students to identify which curve corresponds to which reflection. While it involves multiple transformations (y=x, y=-x, y=0), these are standard A-level topics with visual support provided. The question requires conceptual understanding rather than just calculation, but the transformations themselves are routine and well-practiced at this level.
Spec1.02w Graph transformations: simple transformations of f(x)

7 The diagram shows a part \(A B C\) of the curve \(y = 3 - 2 x ^ { 2 }\), together with its reflections in the lines \(y = x\), \(y = - x\) and \(y = 0\). \includegraphics[max width=\textwidth, alt={}, center]{65d9d34c-8c78-45fe-b9f0-dab071ae56bb-05_691_673_1957_678}

Question 7:
AnswerMarks Guidance
Attempt find \(x\) at intersection of curvesM1 AO 3.1a
\(x=1\)A1 AO 1.1
Correct integral, any limitsM1 AO 3.1a
Correct numerical resultA1 AO 1.1
Attempt area of part or all of \(2\times 2\) squareM1 AO 1.1
Wholly correct methodM1 AO 2.1
\(\frac{44}{3}\)A1 [7] AO 1.1
Method 1:
AnswerMarks Guidance
\(3-2x^2=x \Rightarrow 2x^2+x-3=0 \Rightarrow x=1\)M1, A1
\(\int_0^1(3-2x^2)dx\) or \(\int_{-1}^1(3-2x^2)dx\)M1
\(=\left[3x-\frac{2x^3}{3}\right]_0^1\) or \(\left[3x-\frac{2x^3}{3}\right]_{-1}^1 = \frac{7}{3}\) or \(\frac{14}{3}\)A1
\(``\frac{7}{3}``-1\left(=\frac{4}{3}\right)\) or \(``\frac{14}{3}``-2\left(=\frac{8}{3}\right)\)M1
\(8\times``\frac{4}{3}``+4\) or \(4\times``\frac{8}{3}``+4\)M1
\(=\frac{44}{3}\)A1
## Question 7:

Attempt find $x$ at intersection of curves | **M1** | AO 3.1a | Can be implied
$x=1$ | **A1** | AO 1.1 |
Correct integral, any limits | **M1** | AO 3.1a |
Correct numerical result | **A1** | AO 1.1 | from correct limits
Attempt area of part or all of $2\times 2$ square | **M1** | AO 1.1 |
Wholly correct method | **M1** | AO 2.1 |
$\frac{44}{3}$ | **A1** [7] | AO 1.1 |

**Method 1:**
$3-2x^2=x \Rightarrow 2x^2+x-3=0 \Rightarrow x=1$ | **M1, A1** | | Ignore other root
$\int_0^1(3-2x^2)dx$ or $\int_{-1}^1(3-2x^2)dx$ | **M1** | | Correct integrand with any limits
$=\left[3x-\frac{2x^3}{3}\right]_0^1$ or $\left[3x-\frac{2x^3}{3}\right]_{-1}^1 = \frac{7}{3}$ or $\frac{14}{3}$ | **A1** |
$``\frac{7}{3}``-1\left(=\frac{4}{3}\right)$ or $``\frac{14}{3}``-2\left(=\frac{8}{3}\right)$ | **M1** | | Attempt area above $y=1$ or above $y=x$
$8\times``\frac{4}{3}``+4$ or $4\times``\frac{8}{3}``+4$ | **M1** | | Complete correct method
$=\frac{44}{3}$ | **A1** |

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7 The diagram shows a part $A B C$ of the curve $y = 3 - 2 x ^ { 2 }$, together with its reflections in the lines $y = x$, $y = - x$ and $y = 0$.\\
\includegraphics[max width=\textwidth, alt={}, center]{65d9d34c-8c78-45fe-b9f0-dab071ae56bb-05_691_673_1957_678}

\hfill \mbox{\textit{OCR H240/02 2018 Q7 [7]}}