| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Standard integral of 1/√(x²-a²) |
| Difficulty | Challenging +1.2 Part (a) is routine completing the square. Part (b) requires recognizing the standard integral form ∫dx/√(a²x² + bx + c) leads to inverse hyperbolic or logarithmic functions—a Further Maths topic but fairly standard once the substitution is identified. The multi-step nature and Further Maths content elevate it above average, but it's a textbook application rather than requiring novel insight. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| \(4x^2 + 10x - 24 = 4\left[x^2 + \frac{5}{2}x - 6\right]\) | M1 | |
| \(= 4\left[\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}\right]\) | m1 | oe |
| \(= 4\left(x + \frac{5}{4}\right)^2 - \frac{121}{4}\) | A1 | |
| Therefore, \(a = 4\), \(b = \frac{5}{4}\), \(c = -\frac{121}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_3^5 \frac{6}{\sqrt{4x^2 + 10x - 24}} \, dx\) | M1 | M0 no working. FT (a) for equivalent difficulty |
| \(= \int_3^5 \frac{6}{\sqrt{4\left(x + \frac{5}{4}\right)^2 - \frac{121}{4}}} \, dx\) | m1 | Extracting a factor of \(\sqrt{4}\) from denominator |
| \(= \int_3^5 \frac{6}{2\sqrt{\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}}} \, dx\) | A1 | oe |
| \(= \left[3\cosh^{-1}\left(\frac{x + \frac{5}{4}}{\frac{\sqrt{121}}{\sqrt{16}}}\right)\right]_3^5\) | A1 | |
| \(= \left[3\cosh^{-1}\left(\frac{4x + 5}{11}\right)\right]_3^5\) | m1 | |
| \(= 3\cosh^{-1}\left(\frac{25}{11}\right) - 3\cosh^{-1}\left(\frac{17}{11}\right)\) | A1 | cao Must be 3d.p. |
| \(= 1.379\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_3^5 \frac{6}{\sqrt{4x^2 + 10x - 24}} \, dx\) | (M1) | M0 no working. FT (a) for equivalent difficulty |
| \(= \int_3^5 \frac{6}{\sqrt{4\left(x + \frac{5}{4}\right)^2 - \frac{121}{4}}} \, dx\) | (m1) | Extracting a factor of \(\sqrt{4}\) from denominator |
| \(= \int_3^5 \frac{6}{2\sqrt{\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}}} \, dx\) | (A1) | |
| \(= \left[3\ln\left | x + \frac{5}{4} + \sqrt{\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}}\right | \right]_3^5\) |
| \(= 3\ln\left | \frac{25}{4} + \sqrt{\frac{504}{16}}\right | - 3\ln\left |
| \(= 3\ln\left | \frac{25 + \sqrt{504}}{17 + \sqrt{168}}\right | = 3\ln\left |
| \(= 1.379\) | (A1) |
**Part a)**
$4x^2 + 10x - 24 = 4\left[x^2 + \frac{5}{2}x - 6\right]$ | M1 |
$= 4\left[\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}\right]$ | m1 | oe
$= 4\left(x + \frac{5}{4}\right)^2 - \frac{121}{4}$ | A1 |
Therefore, $a = 4$, $b = \frac{5}{4}$, $c = -\frac{121}{4}$ | |
**Part b)**
METHOD 1:
$\int_3^5 \frac{6}{\sqrt{4x^2 + 10x - 24}} \, dx$ | M1 | M0 no working. FT (a) for equivalent difficulty
$= \int_3^5 \frac{6}{\sqrt{4\left(x + \frac{5}{4}\right)^2 - \frac{121}{4}}} \, dx$ | m1 | Extracting a factor of $\sqrt{4}$ from denominator
$= \int_3^5 \frac{6}{2\sqrt{\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}}} \, dx$ | A1 | oe
$= \left[3\cosh^{-1}\left(\frac{x + \frac{5}{4}}{\frac{\sqrt{121}}{\sqrt{16}}}\right)\right]_3^5$ | A1 |
$= \left[3\cosh^{-1}\left(\frac{4x + 5}{11}\right)\right]_3^5$ | m1 |
$= 3\cosh^{-1}\left(\frac{25}{11}\right) - 3\cosh^{-1}\left(\frac{17}{11}\right)$ | A1 | cao Must be 3d.p.
$= 1.379$ | A1 |
METHOD 2:
$\int_3^5 \frac{6}{\sqrt{4x^2 + 10x - 24}} \, dx$ | (M1) | M0 no working. FT (a) for equivalent difficulty
$= \int_3^5 \frac{6}{\sqrt{4\left(x + \frac{5}{4}\right)^2 - \frac{121}{4}}} \, dx$ | (m1) | Extracting a factor of $\sqrt{4}$ from denominator
$= \int_3^5 \frac{6}{2\sqrt{\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}}} \, dx$ | (A1) |
$= \left[3\ln\left|x + \frac{5}{4} + \sqrt{\left(x + \frac{5}{4}\right)^2 - \frac{121}{16}}\right|\right]_3^5$ | (A1) |
$= 3\ln\left|\frac{25}{4} + \sqrt{\frac{504}{16}}\right| - 3\ln\left|\frac{17}{4} + \sqrt{\frac{168}{16}}\right|$ | (m1) |
$= 3\ln\left|\frac{25 + \sqrt{504}}{17 + \sqrt{168}}\right| = 3\ln\left|\frac{25 + 6\sqrt{14}}{17 + 2\sqrt{42}}\right|$ | (A1) | cao Must be 3d.p.
$= 1.379$ | (A1) |
---\begin{enumerate}[label=(\alph*)]
\item Express $4x^2 + 10x - 24$ in the form $a(x + b)^2 + c$, where $a$, $b$, $c$ are constants whose values are to be found. [3]
\item Hence evaluate the integral
$$\int_3^5 \frac{6}{\sqrt{4x^2 + 10x - 24}} dx.$$
Give your answer correct to 3 decimal places. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q7 [8]}}