| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring standard application of power rule (including rewriting 1/x as x^{-1}), point verification by substitution, finding a gradient, and determining the normal line equation. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the fractional and negative powers. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{3}{2}x^2 - \frac{27}{2}x^{\frac{1}{2}} - 8x^{-2}\) | M1A1A1A1 | M1: attempt to differentiate \(x^n \to x^{n-1}\); 1st A1: one correct term in \(x\); 2nd A1: 2 terms correct; 3rd A1: all correct \(x\) terms, no 30 term and no \(+c\) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=4 \Rightarrow y = \frac{1}{2}\times 64 - 9\times 2^3 + \frac{8}{4} + 30\) | M1 | Substituting \(x=4\) into \(y\) and attempting \(4^{\frac{3}{2}}\) |
| \(= 32 - 72 + 2 + 30 = \underline{-8}\) * | A1cso | Note this is a printed answer |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=4 \Rightarrow y' = \frac{3}{2}\times 4^2 - \frac{27}{2}\times 2 - \frac{8}{16}\) | M1 | Substitute \(x=4\) into \(y'\) (allow slips) |
| \(= 24 - 27 - \frac{1}{2} = -\frac{7}{2}\) | A1 | Obtains \(-3.5\) or equivalent |
| Gradient of normal \(= -1 \div \left(-\frac{7}{2}\right)\) | M1 | Correct use of perpendicular gradient rule using their gradient. Gradient must come from \(y'\) |
| Equation of normal: \(y - {-8} = \frac{2}{7}(x-4)\) | M1A1ft | 3rd M1: attempt at equation of tangent or normal at \(P\); A1ft: correct use of changed gradient to find normal at \(P\). Depends on 1st, 2nd and 3rd Ms |
| \(\underline{7y - 2x + 64 = 0}\) | A1 | Any equivalent form with integer coefficients |
| (6) |
## Question 11:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{3}{2}x^2 - \frac{27}{2}x^{\frac{1}{2}} - 8x^{-2}$ | M1A1A1A1 | M1: attempt to differentiate $x^n \to x^{n-1}$; 1st A1: one correct term in $x$; 2nd A1: 2 terms correct; 3rd A1: all correct $x$ terms, no 30 term and no $+c$ |
| | **(4)** | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=4 \Rightarrow y = \frac{1}{2}\times 64 - 9\times 2^3 + \frac{8}{4} + 30$ | M1 | Substituting $x=4$ into $y$ and attempting $4^{\frac{3}{2}}$ |
| $= 32 - 72 + 2 + 30 = \underline{-8}$ * | A1cso | Note this is a printed answer |
| | **(2)** | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=4 \Rightarrow y' = \frac{3}{2}\times 4^2 - \frac{27}{2}\times 2 - \frac{8}{16}$ | M1 | Substitute $x=4$ into $y'$ (allow slips) |
| $= 24 - 27 - \frac{1}{2} = -\frac{7}{2}$ | A1 | Obtains $-3.5$ or equivalent |
| Gradient of normal $= -1 \div \left(-\frac{7}{2}\right)$ | M1 | Correct use of perpendicular gradient rule using their gradient. Gradient must come from $y'$ |
| Equation of normal: $y - {-8} = \frac{2}{7}(x-4)$ | M1A1ft | 3rd M1: attempt at equation of tangent or normal at $P$; A1ft: correct use of changed gradient to find **normal** at $P$. Depends on 1st, 2nd and 3rd Ms |
| $\underline{7y - 2x + 64 = 0}$ | A1 | Any equivalent form with integer coefficients |
| | **(6)** | |11. The curve $C$ has equation
$$y = \frac { 1 } { 2 } x ^ { 3 } - 9 x ^ { \frac { 3 } { 2 } } + \frac { 8 } { x } + 30 , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that the point $P ( 4 , - 8 )$ lies on $C$.
\item Find an equation of the normal to $C$ at the point $P$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
\includegraphics[max width=\textwidth, alt={}, center]{95e11fd7-765c-477d-800b-7574bc1af81f-15_113_129_2405_1816}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2011 Q11 [12]}}