| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 10 |
| Topic | Partial Fractions |
| Type | Factor polynomial then partial fractions |
| Difficulty | Standard +0.8 This is a well-structured multi-part question requiring substitution, factorization, partial fractions decomposition, and integration with careful handling of limits. While each technique is standard A-level material, the question requires sustained accuracy across multiple steps and the substitution in part (i) is non-routine. The integration involves logarithms and requires exact values, adding moderate challenge. This is harder than typical textbook exercises but not exceptionally difficult for strong students. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(y = x - 1\), expand and simplify ⇒ \(x^3 - 3x^2\) (\(= x^2(x - 3)\)) | B1 | 1 mark |
| (ii) Write \(\frac{2y + 5}{y^3 - 3y - 2} = \frac{2x + 3}{x^2(x - 3)} = \frac{A}{x - 3} + \frac{B}{x} + \frac{C}{x^2}\) | B1 | Appropriate method to find coefficients, based on \(2x + 3 = Ax^2 + Bx(x - 3) + C(x - 3)\) |
| (iii) Attempt to integrate all of their three (or two) partial fractions, condoning lack of modulus signs | M1 | Obtain (with or without modulus signs) \(\left[\ln |
(i) Substitute $y = x - 1$, expand and simplify ⇒ $x^3 - 3x^2$ ($= x^2(x - 3)$) | B1 | 1 mark
(ii) Write $\frac{2y + 5}{y^3 - 3y - 2} = \frac{2x + 3}{x^2(x - 3)} = \frac{A}{x - 3} + \frac{B}{x} + \frac{C}{x^2}$ | B1 | Appropriate method to find coefficients, based on $2x + 3 = Ax^2 + Bx(x - 3) + C(x - 3)$ | M1 | Obtain $A = 1$ and $B = -1$ and $C = -1$ | A1A1A1 | Result: $\left(\frac{2y + 5}{y^3 - 3y - 2} = \frac{1}{y - 2} - \frac{1}{y + 1} - \frac{1}{(y + 1)^2}\right)$ | 5 marks
(iii) Attempt to integrate all of their three (or two) partial fractions, condoning lack of modulus signs | M1 | Obtain (with or without modulus signs) $\left[\ln|y - 2| - \ln|y + 1| + \frac{1}{y + 1}\right]_0^1$ | A1 | Correct substitution of both limits in their integrated expressions: $(\ln 1 - \ln 2 + 1) - (\ln|-2| - \ln|1|)$ | M1(dep) | $-\frac{1}{4} - 2\ln 2$ | A1 | 4 marks\begin{enumerate}[label=(\roman*)]
\item Express $y^3 - 3y - 2$ in terms of $x$, where $x = y + 1$. [1]
\item Hence express
$$\frac{2y + 5}{y^3 - 3y - 2}$$
in partial fractions. [5]
\item Find the exact value of
$$\int_0^1 \frac{2y + 5}{y^3 - 3y - 2} dy.$$ [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q6 [10]}}