OCR C4 — Question 7 9 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeFactor polynomial then partial fractions
DifficultyStandard +0.3 This is a straightforward C4 partial fractions question with clear guidance. Part (i) involves routine factor verification and algebraic division. Part (ii) requires standard partial fractions decomposition and integration of logarithmic forms. The 'show that' structure removes problem-solving demands, making this slightly easier than average for C4.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08j Integration using partial fractions

7. (i) Show that ( \(2 x + 3\) ) is a factor of ( \(\left. 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 \right)\) and hence, simplify $$\frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$ (ii) Show that $$\int _ { 2 } ^ { 5 } \frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } \mathrm {~d} x = \ln k$$ where \(k\) is an integer.

AnswerMarks Guidance
(i) \(f(x) = 2x^3 - x^2 + 4x + 15\); \(f(-\frac{3}{2}) = -\frac{27}{4} - \frac{9}{4} - 6 + 15 = 0\)B1 \((2x + 3)\) is a factor
Polynomial division: \(2x^3 - x^2 + 4x + 15 = (2x + 3)(x^2 - 2x + 5)\)M1 A1
(ii) \(\frac{2x^2 + x - 3}{2x^3 - x^2 + 4x + 15} = \frac{(2x + 3)(x - 1)}{(2x + 3)(x^2 - 2x + 5)} = \frac{x - 1}{x^2 - 2x + 5}\)M1 A1
\(\int_2^5 \frac{x - 1}{x^2 - 2x + 5} dx = \left[\frac{1}{2}\lnx^2 - 2x + 5 \right]_2^5\)
\(= \frac{1}{2}(\ln 20 - \ln 5) = \frac{1}{2}\ln 4 = \ln 2\)M1 A1 (9)
**(i)** $f(x) = 2x^3 - x^2 + 4x + 15$; $f(-\frac{3}{2}) = -\frac{27}{4} - \frac{9}{4} - 6 + 15 = 0$ | B1 | $(2x + 3)$ is a factor

Polynomial division: $2x^3 - x^2 + 4x + 15 = (2x + 3)(x^2 - 2x + 5)$ | M1 A1 |

**(ii)** $\frac{2x^2 + x - 3}{2x^3 - x^2 + 4x + 15} = \frac{(2x + 3)(x - 1)}{(2x + 3)(x^2 - 2x + 5)} = \frac{x - 1}{x^2 - 2x + 5}$ | M1 A1 |

$\int_2^5 \frac{x - 1}{x^2 - 2x + 5} dx = \left[\frac{1}{2}\ln|x^2 - 2x + 5|\right]_2^5$ | M1 A1 |

$= \frac{1}{2}(\ln 20 - \ln 5) = \frac{1}{2}\ln 4 = \ln 2$ | M1 A1 | **(9)**

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7. (i) Show that ( $2 x + 3$ ) is a factor of ( $\left. 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 \right)$ and hence, simplify

$$\frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$

(ii) Show that

$$\int _ { 2 } ^ { 5 } \frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } \mathrm {~d} x = \ln k$$

where $k$ is an integer.\\

\hfill \mbox{\textit{OCR C4  Q7 [9]}}