| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (find unknown constant in derivative first) |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring only standard integral rules (power rule for x^(-2)) and finding a constant using given conditions. It involves routine techniques with no problem-solving insight needed, making it easier than average but not trivial since it requires two steps and algebraic manipulation. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f(x) = \int (5 + \frac{4}{x^3}) dx\) | M1 A2 | |
| \(f(x) = 5x - 4x^{-1} + c\) | M1 | |
| (ii) \(f(1) = 5 - 4 + c = 1 + c\) | M1 | |
| \(f(2) = 10 - 2 + c = 8 + c\) | M1 | |
| \(f(2) = 2f(1) \therefore 8 + c = 2(1 + c)\) | A1 | |
| \(c = 6\) | ||
| \(f(x) = 5x - 4x^{-1} + 6\) | ||
| \(f(4) = 20 - 1 + 6 = 25\) | M1 A1 | (8) |
**(i)** $f(x) = \int (5 + \frac{4}{x^3}) dx$ | M1 A2 |
$f(x) = 5x - 4x^{-1} + c$ | M1 |
**(ii)** $f(1) = 5 - 4 + c = 1 + c$ | M1 |
$f(2) = 10 - 2 + c = 8 + c$ | M1 |
$f(2) = 2f(1) \therefore 8 + c = 2(1 + c)$ | A1 |
$c = 6$ | |
$f(x) = 5x - 4x^{-1} + 6$ | |
$f(4) = 20 - 1 + 6 = 25$ | M1 A1 | (8)
---6. Given that
$$\mathrm { f } ^ { \prime } ( x ) = 5 + \frac { 4 } { x ^ { 2 } } , \quad x \neq 0$$
(i) find an expression for $\mathrm { f } ( x )$.
Given also that
$$\mathrm { f } ( 2 ) = 2 \mathrm { f } ( 1 ) ,$$
(ii) find $\mathrm { f } ( 4 )$.\\
\hfill \mbox{\textit{OCR C2 Q6 [8]}}