| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.3 This is a straightforward C1 integration question requiring reverse differentiation of a polynomial plus a power of x, finding a constant using a given point, then applying standard tangent line formula. All techniques are routine for this level with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)=\frac{2x^2}{2}+\frac{3x^{-1}}{-1}(+c)\), \(-\frac{3}{x}\) is OK | M1A1 | For some attempt to integrate \(x^n\to x^{n+1}\); for both \(x\) terms as printed or better. Ignore \((+c)\) |
| \((3,7\tfrac{1}{2})\) gives \(\frac{15}{2}=9-\frac{3}{3}+c\); \(3^2\) or \(3^{-1}\) are OK instead of 9 or \(\frac{1}{3}\) | M1A1f.t. | For use of \((3,7\tfrac{1}{2})\) or \((-2,5)\) to form equation for \(c\). No \(+c\) is M0. Some changes in \(x\) terms needed |
| \(c=-\frac{1}{2}\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(-2)=4+\frac{3}{2}-\frac{1}{2}\,(*)\) | B1c.s.o. | (1). If \((-2,5)\) is used to find \(c\) in (a), B0 here unless they verify \(f(3)=7.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m=-4+\frac{3}{4}=-3.25\) | M1, A1 | For attempting \(m=f'(\pm2)\); for \(-\frac{13}{4}\) or \(-3.25\) |
| Equation of tangent: \(y-5=-3.25(x+2)\) | M1 | For attempting equation of tangent at \((-2,5)\), f.t. their \(m\), based on \(\frac{dy}{dx}\) |
| \(4y+13x+6=0\) | A1 | (4). o.e. must have \(a\), \(b\) and \(c\) integers and \(=0\) |
## Question 10:
### Part (a):
| $f(x)=\frac{2x^2}{2}+\frac{3x^{-1}}{-1}(+c)$, $-\frac{3}{x}$ is OK | M1A1 | For some attempt to integrate $x^n\to x^{n+1}$; for both $x$ terms as printed or better. Ignore $(+c)$ |
| $(3,7\tfrac{1}{2})$ gives $\frac{15}{2}=9-\frac{3}{3}+c$; $3^2$ or $3^{-1}$ are OK instead of 9 or $\frac{1}{3}$ | M1A1f.t. | For use of $(3,7\tfrac{1}{2})$ or $(-2,5)$ to form equation for $c$. No $+c$ is M0. Some changes in $x$ terms needed |
| $c=-\frac{1}{2}$ | A1 | (5) |
### Part (b):
| $f(-2)=4+\frac{3}{2}-\frac{1}{2}\,(*)$ | B1c.s.o. | (1). If $(-2,5)$ is used to find $c$ in (a), B0 here unless they verify $f(3)=7.5$ |
### Part (c):
| $m=-4+\frac{3}{4}=-3.25$ | M1, A1 | For attempting $m=f'(\pm2)$; for $-\frac{13}{4}$ or $-3.25$ |
| Equation of tangent: $y-5=-3.25(x+2)$ | M1 | For attempting equation of tangent at $(-2,5)$, f.t. their $m$, based on $\frac{dy}{dx}$ |
| $4y+13x+6=0$ | A1 | (4). o.e. must have $a$, $b$ and $c$ integers and $=0$ |10. The curve $C$ with equation $y = \mathrm { f } ( x ) , x \neq 0$, passes through the point ( $3,7 \frac { 1 } { 2 }$ ).
Given that $\mathrm { f } ^ { \prime } ( x ) = 2 x + \frac { 3 } { x ^ { 2 } }$,
\begin{enumerate}[label=(\alph*)]
\item find $\mathrm { f } ( x )$.
\item Verify that $f ( - 2 ) = 5$.
\item Find an equation for the tangent to $C$ at the point ( $- 2,5$ ), giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2006 Q10 [10]}}