| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Curve properties and tangent/normal |
| Difficulty | Moderate -0.3 This is a straightforward C1 integration question requiring standard techniques: integrating polynomial and power terms to find f(x) using the given point to find the constant, then finding the tangent equation using the derivative at x=2. While it involves multiple steps, each step uses routine A-level methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(3x^2 \to cx^3\) or \(-6 \to cx\) or \(-8x^{-2} \to cx^{-1}\) | M1 | |
| \(f(x) = \frac{3x^3}{3} - 6x - \frac{8x^{-1}}{-1}\) | \((+C)\) | \(\left(x^3 - 6x + \frac{8}{x}\right)\) |
| Substitute \(x = 2\) and \(y = 1\) into a 'changed function' to form an equation in C. | M1 | |
| \(1 = 8 - 12 + 4 + C\), \(C = 1\) | A1cso | (5 marks) |
| (b) \(3x \times 2^2 - 6 - \frac{8}{2^2}\) | M1 | |
| \(= 4\) | A1 | |
| Eqn. of tangent: \(y - 1 = 4(x - 2)\) | M1 | |
| \(y = 4x - 7\) | (Must be in this form) | A1 |
**(a)** $3x^2 \to cx^3$ or $-6 \to cx$ or $-8x^{-2} \to cx^{-1}$ | M1 |
$f(x) = \frac{3x^3}{3} - 6x - \frac{8x^{-1}}{-1}$ | $(+C)$ | $\left(x^3 - 6x + \frac{8}{x}\right)$ | A1, A1 |
Substitute $x = 2$ and $y = 1$ into a 'changed function' to form an equation in C. | M1 |
$1 = 8 - 12 + 4 + C$, $C = 1$ | A1cso | (5 marks)
**(b)** $3x \times 2^2 - 6 - \frac{8}{2^2}$ | | M1 |
$= 4$ | | A1 |
Eqn. of tangent: $y - 1 = 4(x - 2)$ | | M1 |
$y = 4x - 7$ | (Must be in this form) | A1 | (4 marks)
**Total: 9 marks**
**Guidance:**
- (a) First 2 A marks: + C is not required, and coefficients need not be simplified, but powers must be simplified.
- All 3 terms correct: A1 A1
- Two terms correct: A1 A0
- Only one term correct: A0 A0
- Allow the M1 A1 for finding C to be scored either in part (a) or in part (b).
- Correct solution only (cso): any wrong working seen loses the A mark.
- (b) 1st M: Substituting $x = 2$ into $3x^2 - 6 - \frac{8}{x^2}$ (must be this function).
- 2nd M: Awarded generously for attempting the equation of a straight line through (2, 1) with any value of m, however found.
- 2nd M: Alternative is to use (2, 1) or (1, 2) in $y = mx + c$ to find a value for c.
- If calculation for the gradient value is seen in part (a), it must be used in part (b) to score the first M1 A1 in (b).
- Using (1, 2) instead of (2, 1): Loses the 2nd method mark in (a). Gains the 2nd method mark in (b).
---7. The curve $C$ has equation $y = \mathrm { f } ( x ) , x \neq 0$, and the point $P ( 2,1 )$ lies on $C$. Given that
$$f ^ { \prime } ( x ) = 3 x ^ { 2 } - 6 - \frac { 8 } { x ^ { 2 } } ,$$
\begin{enumerate}[label=(\alph*)]
\item find $\mathrm { f } ( x )$.
\item Find an equation for the tangent to $C$ at the point $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2007 Q7 [9]}}