OCR C4 2007 June — Question 1 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2007
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypePartial fractions then differentiate
DifficultyModerate -0.3 This is a straightforward two-part question requiring standard partial fractions decomposition followed by term-by-term differentiation. Part (ii) requires showing all terms in f'(x) are negative, which is routine algebraic verification. Slightly easier than average as it follows a predictable template with no conceptual surprises.
Spec1.02y Partial fractions: decompose rational functions1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx

1 The equation of a curve is \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 3 x + 1 } { ( x + 2 ) ( x - 3 ) }\).
  1. Express \(\mathrm { f } ( x )\) in partial fractions.
  2. Hence find \(\mathrm { f } ^ { \prime } ( x )\) and deduce that the gradient of the curve is negative at all points on the curve.

AnswerMarks Guidance
\(\frac{A}{x+2} + \frac{B}{x-3}\)M1 s.o.i. in answer
\(A=1\) and \(B=2\)A1 2 for both
\(-A(x+2)^2 - B(x-3)^2\)M1 f.t.
Convincing statement that each denom \(> 0\) State whole exp \(< 0\)AG
Accept \(\geq 0\). Do not accept \(x^2 > 0\). Dep on previous 4 marks.
$\frac{A}{x+2} + \frac{B}{x-3}$ | M1 | s.o.i. in answer
$A=1$ and $B=2$ | A1 | 2 for both
$-A(x+2)^2 - B(x-3)^2$ | M1 | f.t.
Convincing statement that each denom $> 0$ State whole exp $< 0$ | AG | 
| | | Accept $\geq 0$. Do not accept $x^2 > 0$. Dep on previous 4 marks.

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1 The equation of a curve is $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 3 x + 1 } { ( x + 2 ) ( x - 3 ) }$.\\
(i) Express $\mathrm { f } ( x )$ in partial fractions.\\
(ii) Hence find $\mathrm { f } ^ { \prime } ( x )$ and deduce that the gradient of the curve is negative at all points on the curve.

\hfill \mbox{\textit{OCR C4 2007 Q1 [5]}}