| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring standard techniques: differentiating a polynomial with a square root term (rewritten as x^{1/2}), finding a tangent equation by substituting a given x-value, and finding a point where the gradient equals a given value. All steps are routine applications of basic calculus with no problem-solving insight required, making it easier than average but not trivial due to the multi-part nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2x - 8x^{\frac{1}{2}} + 5\) | — | |
| \(\frac{dy}{dx} = 2 - 4x^{-\frac{1}{2}}\) \((x>0)\) | M1 A1 A1 | M1: evidence of differentiation \(x^n \to x^{n-1}\); A1: any two of three terms correct; A1: \(2-4x^{-\frac{1}{2}}\) both terms correct and simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(x=\frac{1}{4}\): \(y = 2\left(\frac{1}{4}\right) - 8\sqrt{\frac{1}{4}}+5\), so \(y=\frac{3}{2}\) | B1 | Obtaining \(y=3/2\) |
| Gradient \(= 2 - \frac{4}{\sqrt{\frac{1}{4}}} = \{-6\}\) | M1 | Attempt to substitute \(x=\frac{1}{4}\) into \(\frac{dy}{dx}\) to establish gradient |
| \(y - \frac{3}{2} = -6\left(x - \frac{1}{4}\right)\) or \(y=-6x+c\) with \(\frac{3}{2}=-6\left(\frac{1}{4}\right)+c\) | dM1 | Complete method for equation of tangent; depends on previous M |
| \(y = -6x + 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x-3y+18=0 \Rightarrow\) gradient \(= \frac{2}{3}\), so tangent gradient is \(\frac{2}{3}\) | B1 | For value \(\frac{2}{3}\); not \(\frac{2}{3}x\), not \(-\frac{3}{2}\) |
| \(2 - \frac{4}{\sqrt{x}} = \frac{2}{3}\) (sets gradient function = numerical gradient) | M1 | |
| \(\frac{4}{3} = \frac{4}{\sqrt{x}} \Rightarrow x=9\) (ignore extra answer \(x=-9\)) | A1 | |
| When \(x=9\): \(y=2(9)-8\sqrt{9}+5=-1\) | dM1 | Substitutes found \(x\) into original curve equation |
| \(y=-1\) or \((9,-1)\) | A1 |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2x - 8x^{\frac{1}{2}} + 5$ | — | |
| $\frac{dy}{dx} = 2 - 4x^{-\frac{1}{2}}$ $(x>0)$ | M1 A1 A1 | M1: evidence of differentiation $x^n \to x^{n-1}$; A1: any two of three terms correct; A1: $2-4x^{-\frac{1}{2}}$ both terms correct and simplified |
**[3 marks]**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $x=\frac{1}{4}$: $y = 2\left(\frac{1}{4}\right) - 8\sqrt{\frac{1}{4}}+5$, so $y=\frac{3}{2}$ | B1 | Obtaining $y=3/2$ |
| Gradient $= 2 - \frac{4}{\sqrt{\frac{1}{4}}} = \{-6\}$ | M1 | Attempt to substitute $x=\frac{1}{4}$ into $\frac{dy}{dx}$ to establish gradient |
| $y - \frac{3}{2} = -6\left(x - \frac{1}{4}\right)$ or $y=-6x+c$ with $\frac{3}{2}=-6\left(\frac{1}{4}\right)+c$ | dM1 | Complete method for equation of tangent; depends on previous M |
| $y = -6x + 3$ | A1 | |
**[4 marks]**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x-3y+18=0 \Rightarrow$ gradient $= \frac{2}{3}$, so tangent gradient is $\frac{2}{3}$ | B1 | For value $\frac{2}{3}$; **not** $\frac{2}{3}x$, **not** $-\frac{3}{2}$ |
| $2 - \frac{4}{\sqrt{x}} = \frac{2}{3}$ (sets gradient function = numerical gradient) | M1 | |
| $\frac{4}{3} = \frac{4}{\sqrt{x}} \Rightarrow x=9$ (ignore extra answer $x=-9$) | A1 | |
| When $x=9$: $y=2(9)-8\sqrt{9}+5=-1$ | dM1 | Substitutes found $x$ into original curve equation |
| $y=-1$ or $(9,-1)$ | A1 | |
**[5 marks] — Total: 12 marks**11. The curve $C$ has equation
$$y = 2 x - 8 \sqrt { } x + 5 , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving each term in its simplest form.
The point $P$ on $C$ has $x$-coordinate equal to $\frac { 1 } { 4 }$
\item Find the equation of the tangent to $C$ at the point $P$, giving your answer in the form $y = a x + b$, where $a$ and $b$ are constants.
The tangent to $C$ at the point $Q$ is parallel to the line with equation $2 x - 3 y + 18 = 0$
\item Find the coordinates of $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q11 [12]}}