OCR MEI Paper 1 2022 June — Question 2 3 marks

Exam BoardOCR MEI
ModulePaper 1 (Paper 1)
Year2022
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeBasic partial fractions decomposition
DifficultyEasy -1.2 This is a straightforward partial fractions decomposition with two distinct linear factors in the denominator. It requires only routine application of the standard method (cover-up rule or equating coefficients) with no complications like repeated factors, improper fractions, or quadratic terms. This is a textbook exercise testing basic recall and technique.
Spec1.02y Partial fractions: decompose rational functions
2 Express \(\frac { 13 - x } { ( x - 3 ) ( x + 2 ) }\) in partial fractions.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\dfrac{13-x}{(x-3)(x+2)} = \dfrac{A}{(x-3)} + \dfrac{B}{(x+2)}\)
\(13-x = A(x+2) + B(x-3)\)M1 Clearing the denominators oe
\(A = 2,\ B = -3\)
So \(\dfrac{2}{(x-3)} - \dfrac{3}{(x+2)}\)A1 For one correct coefficient
A1Correct partial fractions seen. Accept just values for \(A\) and \(B\) if defined
[3] 
# Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{13-x}{(x-3)(x+2)} = \dfrac{A}{(x-3)} + \dfrac{B}{(x+2)}$ | | |
| $13-x = A(x+2) + B(x-3)$ | M1 | Clearing the denominators oe |
| $A = 2,\ B = -3$ | | |
| So $\dfrac{2}{(x-3)} - \dfrac{3}{(x+2)}$ | A1 | For one correct coefficient |
| | A1 | Correct partial fractions seen. Accept just values for $A$ and $B$ if defined |
| **[3]** | | |

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2 Express $\frac { 13 - x } { ( x - 3 ) ( x + 2 ) }$ in partial fractions.

\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q2 [3]}}
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Q1 4 Q2 3 Q3 8 Q4 4 Q5 5 Q6 9 Q7 6 Q8 10 Q9 8 Q10 8 Q11 6 Q12 4 Q13 12 Q14 13