| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Difficulty | Challenging +1.8 This is a Further Maths partial fractions question requiring factorization of a cubic denominator (likely by grouping to get (x-1)(x²+3)), decomposition into partial fractions with a quadratic factor, integration involving both logarithmic and arctangent forms, and numerical evaluation. The 10-mark allocation and multiple technical steps (factorization, partial fractions setup, integration of rational functions, numerical computation) make this substantially harder than average A-level questions, though it follows a standard Further Maths technique without requiring novel insight. |
| Spec | 1.08j Integration using partial fractions4.05c Partial fractions: extended to quadratic denominators |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{6x^2 + 2x + 16}{x^3 - x^2 + 3x - 3} = \frac{6x^2 + 2x + 16}{(x-1)(x^2+3)}\) | M1 A1 | Linear \(\times\) Quadratic |
$\frac{6x^2 + 2x + 16}{x^3 - x^2 + 3x - 3} = \frac{6x^2 + 2x + 16}{(x-1)(x^2+3)}$ | M1 A1 | Linear $\times$ Quadratic
$\frac{6x^2Evaluate the integral
$$\int_2^4 \frac{6x^2 + 2x + 16}{x^3 - x^2 + 3x - 3} dx,$$
giving your answer correct to three decimal places. [10]
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q14 [10]}}