OCR FP2 2015 June — Question 7 10 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypeRational function curve sketching
DifficultyChallenging +1.2 This is a Further Maths question requiring partial fractions (standard technique), asymptote identification, and curve sketching of y²=f(x). While part (iv) requires careful analysis of where f(x)≥0 and consideration of ±√f(x), the techniques are all within FP2 syllabus and follow a structured progression. The improper fraction adds mild complexity but is routine for this level.
Spec1.02n Sketch curves: simple equations including polynomials1.02y Partial fractions: decompose rational functions

7 It is given that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } - 25 } { ( x - 1 ) ( x + 2 ) }\).
  1. Express \(\mathrm { f } ( x )\) in partial fractions.
  2. Write down the equations of the asymptotes of the curve \(y = \mathrm { f } ( x )\).
  3. Find the value of \(x\) where the graph of \(y = \mathrm { f } ( x )\) cuts the horizontal asymptote.
  4. Sketch the graph of \(y ^ { 2 } = \mathrm { f } ( x )\).

Question 7:
Part (i):
AnswerMarks Guidance
\(\frac{x^2 - 25}{(x-1)(x+2)} = A + \frac{B}{(x-1)} + \frac{C}{(x+2)}\)M1 Splitting in correct way to give partial fractions (may be seen anywhere)
\(x^2 - 25 = A(x-1)(x+2) + B(x+2) + C(x-1)\)
3 processes of equating coefficients or substituting:
\(x = 1 \Rightarrow -24 = 3B \Rightarrow B = -8\)
\(x = -2 \Rightarrow -21 = -3C \Rightarrow C = 7\)
coeff of \(x^2\): \(A = 1\)
AnswerMarks Guidance
\(\frac{x^2-25}{(x-1)(x+2)} = 1 - \frac{8}{(x-1)} + \frac{7}{(x+2)}\)B1 For \(A\)
A1For \(B\)
A1For \(C\)

Total: 4

Part (ii):
AnswerMarks
\(x = 1,\ x = -2\)B1
\(y = 1\)B1

Total: 2

Part (iii):
AnswerMarks
\(y = 1 \Rightarrow (x-1)(x+2) = x^2 - 25\)M1
\(x^2 + x - 2 = x^2 - 25 \Rightarrow x = -23\)A1

Total: 2

Part (iv):
AnswerMarks Guidance
4 bits as shown, roughly symmetric about axes, approaching asymptotesB1
Lh side crosses asymptotes and upper section approaches from above and lower section approaches from belowB1 Ignore any graph of \(y = f(x)\)

Total: 2

## Question 7:

### Part (i):
$\frac{x^2 - 25}{(x-1)(x+2)} = A + \frac{B}{(x-1)} + \frac{C}{(x+2)}$ | **M1** | Splitting in correct way to give partial fractions (may be seen anywhere)

$x^2 - 25 = A(x-1)(x+2) + B(x+2) + C(x-1)$

3 processes of equating coefficients or substituting:

$x = 1 \Rightarrow -24 = 3B \Rightarrow B = -8$

$x = -2 \Rightarrow -21 = -3C \Rightarrow C = 7$

coeff of $x^2$: $A = 1$

$\frac{x^2-25}{(x-1)(x+2)} = 1 - \frac{8}{(x-1)} + \frac{7}{(x+2)}$ | **B1** | For $A$
| **A1** | For $B$
| **A1** | For $C$

Total: **4**

### Part (ii):
$x = 1,\ x = -2$ | **B1** |

$y = 1$ | **B1** |

Total: **2**

### Part (iii):
$y = 1 \Rightarrow (x-1)(x+2) = x^2 - 25$ | **M1** |

$x^2 + x - 2 = x^2 - 25 \Rightarrow x = -23$ | **A1** |

Total: **2**

### Part (iv):
4 bits as shown, roughly symmetric about axes, approaching asymptotes | **B1** |

Lh side crosses asymptotes and upper section approaches from above and lower section approaches from below | **B1** | Ignore any graph of $y = f(x)$

Total: **2**

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7 It is given that $\mathrm { f } ( x ) = \frac { x ^ { 2 } - 25 } { ( x - 1 ) ( x + 2 ) }$.\\
(i) Express $\mathrm { f } ( x )$ in partial fractions.\\
(ii) Write down the equations of the asymptotes of the curve $y = \mathrm { f } ( x )$.\\
(iii) Find the value of $x$ where the graph of $y = \mathrm { f } ( x )$ cuts the horizontal asymptote.\\
(iv) Sketch the graph of $y ^ { 2 } = \mathrm { f } ( x )$.

\hfill \mbox{\textit{OCR FP2 2015 Q7 [10]}}