| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic in higher integer powers |
| Difficulty | Standard +0.3 This question requires finding the second derivative using standard power rule differentiation, then solving a cubic equation that reduces to a quadratic via substitution (letting y = 2x + 1/x). While it involves multiple steps, the techniques are routine for C1 level—differentiation is mechanical and the substitution, though requiring some algebraic manipulation, follows a standard pattern for this type of reciprocal function. Slightly above average difficulty due to the algebraic manipulation required in part (ii). |
| Spec | 1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(24x^2\) | B1 | |
| \(kx^{-4}\) | B1 | |
| \(-3x^{-4}\) | B1 | |
| \[48x + 12x^{-5}\] | M1 A1 5 | Attempt to differentiate their (i). Fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| \[8x^3 + x^{-3} = -9\] | ||
| \[8x^6 + 1 = -9x^3\] | ||
| \[8x^6 + 9x^3 + 1 = 0\] | *M1 | Use a substitution to obtain a 3-term quadratic |
| Let \(y = x^3\) | DM1 | Correct method to solve quadratic |
| \[8y^2 + 9y + 1 = 0\] | ||
| \[(8y + 1)(y + 1) = 0\] | ||
| \[y = -\frac{1}{8}, y = -1\] | A1 | |
| \[x = -\frac{1}{2}, x = -1\] | M1 | Attempt to cube root at least one of their y-values |
| A1 5 | \(-\frac{1}{2}, -1\) | |
| SR one correct \(x\) value www | B1 | |
| SR for trial and improvement: | ||
| \(x = -1\) | B1 | |
| \(x = -\frac{1}{2}\) | B2 | |
| Justification that there are no further solutions | B2 |
## 10(i)
$24x^2$ | B1 |
$kx^{-4}$ | B1 |
$-3x^{-4}$ | B1 |
$$48x + 12x^{-5}$$ | M1 A1 5 | Attempt to differentiate their (i). Fully correct |
## 10(ii)
$$8x^3 + x^{-3} = -9$$ | | |
$$8x^6 + 1 = -9x^3$$ | | |
$$8x^6 + 9x^3 + 1 = 0$$ | *M1 | Use a substitution to obtain a 3-term quadratic |
Let $y = x^3$ | DM1 | Correct method to solve quadratic |
$$8y^2 + 9y + 1 = 0$$ | | |
$$(8y + 1)(y + 1) = 0$$ | | |
$$y = -\frac{1}{8}, y = -1$$ | A1 | |
$$x = -\frac{1}{2}, x = -1$$ | M1 | Attempt to cube root at least one of their y-values |
| A1 5 | $-\frac{1}{2}, -1$ |
**SR** one correct $x$ value **www** | B1 |
**SR for trial and improvement:** | |
$x = -1$ | B1 |
$x = -\frac{1}{2}$ | B2 |
Justification that there are no further solutions | B2 |
---10 Given that $\mathrm { f } ( x ) = 8 x ^ { 3 } + \frac { 1 } { x ^ { 3 } }$,\\
(i) find $\mathrm { f } ^ { \prime \prime } ( x )$,\\
(ii) solve the equation $\mathrm { f } ( x ) = - 9$.
\hfill \mbox{\textit{OCR C1 2008 Q10 [10]}}