| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent given derivative expression |
| Difficulty | Standard +0.3 This is a straightforward C1 integration question requiring substitution of x=4 into f'(x) for part (a), then integration of terms after rewriting with fractional powers in part (b), followed by using point P to find the constant. All techniques are standard with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(4) = 30 + \frac{6 - 5 \times 4^2}{\sqrt{4}}\) | M1 | Attempts to substitute \(x = 4\) into \(f'(x) = 30 + \frac{6-5x^2}{\sqrt{x}}\) or their algebraically manipulated \(f'(x)\) |
| \(f'(4) = -7\) | A1 | Gradient \(= -7\) |
| \(y - (-8) = \text{"-7"} \times (x-4)\) or \(y = \text{"-7"}x + c \Rightarrow -8 = \text{"-7"} \times 4 + c \Rightarrow c = ...\) | M1 | Attempts equation of tangent using their numeric \(f'(4)\) from substituting \(x=4\) into given \(f'(x)\), and \((4,-8)\) with 4 and \(-8\) correctly placed. If using \(y = mx+c\), must reach \(c = ...\) |
| \(y = -7x + 20\) | A1 | Allow \(y = 20 - 7x\) and allow "y =" to become "detached" but must be present at some stage. E.g. \(y = ... = -7x + 20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\Rightarrow f(x) = 30x + 6\frac{x^{\frac{1}{2}}}{0.5} - 5\frac{x^{\frac{5}{2}}}{2.5}(+c)\) | M1 A1 A1 | M1: \(30 \to 30x\) or \(\frac{6}{\sqrt{x}} \to \alpha x^{\frac{1}{2}}\) or \(-\frac{5x^2}{\sqrt{x}} \to \beta x^{\frac{5}{2}}\) (these cases only). A1: Any 2 correct terms (simplified or un-simplified), including powers — allow \(-\frac{1}{2}+1\) for \(\frac{1}{2}\) and \(\frac{3}{2}+1\) for \(\frac{5}{2}\) (with or without \(+c\)). A1: All 3 terms correct (with or without \(+c\)). Ignore any spurious integral signs |
| \(x = 4, f(x) = -8 \Rightarrow -8 = 120 + 24 - 64 + c \Rightarrow c = ...\) | M1 | Substitutes \(x=4, f(x)=-8\) into their \(f(x)\) (not \(f'(x)\)) containing \(+c\) and rearranges to obtain a value for \(c\) |
| \(\Rightarrow (f(x) =) 30x + 12x^{\frac{1}{2}} - 2x^{\frac{5}{2}} - 88\) | A1 | Allow \(\sqrt{x}\) for \(x^{\frac{1}{2}}\) and e.g. \(\sqrt{x^5}\) or \(x^2\sqrt{x}\) for \(x^{\frac{5}{2}}\). Note "f(x) =" is not needed. |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(4) = 30 + \frac{6 - 5 \times 4^2}{\sqrt{4}}$ | M1 | Attempts to substitute $x = 4$ into $f'(x) = 30 + \frac{6-5x^2}{\sqrt{x}}$ or their algebraically manipulated $f'(x)$ |
| $f'(4) = -7$ | A1 | Gradient $= -7$ |
| $y - (-8) = \text{"-7"} \times (x-4)$ or $y = \text{"-7"}x + c \Rightarrow -8 = \text{"-7"} \times 4 + c \Rightarrow c = ...$ | M1 | Attempts equation of tangent using their numeric $f'(4)$ from substituting $x=4$ into given $f'(x)$, and $(4,-8)$ with 4 and $-8$ correctly placed. If using $y = mx+c$, must reach $c = ...$ |
| $y = -7x + 20$ | A1 | Allow $y = 20 - 7x$ and allow "y =" to become "detached" but must be present at some stage. E.g. $y = ... = -7x + 20$ |
**Total: 4 marks**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Rightarrow f(x) = 30x + 6\frac{x^{\frac{1}{2}}}{0.5} - 5\frac{x^{\frac{5}{2}}}{2.5}(+c)$ | M1 A1 A1 | M1: $30 \to 30x$ or $\frac{6}{\sqrt{x}} \to \alpha x^{\frac{1}{2}}$ or $-\frac{5x^2}{\sqrt{x}} \to \beta x^{\frac{5}{2}}$ (these cases only). A1: Any 2 correct terms (simplified or un-simplified), including powers — allow $-\frac{1}{2}+1$ for $\frac{1}{2}$ and $\frac{3}{2}+1$ for $\frac{5}{2}$ (with or without $+c$). A1: All 3 terms correct (with or without $+c$). **Ignore any spurious integral signs** |
| $x = 4, f(x) = -8 \Rightarrow -8 = 120 + 24 - 64 + c \Rightarrow c = ...$ | M1 | Substitutes $x=4, f(x)=-8$ into their $f(x)$ (not $f'(x)$) containing $+c$ and rearranges to obtain a value for $c$ |
| $\Rightarrow (f(x) =) 30x + 12x^{\frac{1}{2}} - 2x^{\frac{5}{2}} - 88$ | A1 | Allow $\sqrt{x}$ for $x^{\frac{1}{2}}$ and e.g. $\sqrt{x^5}$ or $x^2\sqrt{x}$ for $x^{\frac{5}{2}}$. Note "f(x) =" is not needed. |
**Total: 5 marks (9 mark question)**
---7. The curve $C$ has equation $y = \mathrm { f } ( x ) , x > 0$, where
$$\mathrm { f } ^ { \prime } ( x ) = 30 + \frac { 6 - 5 x ^ { 2 } } { \sqrt { x } }$$
Given that the point $P ( 4 , - 8 )$ lies on $C$,
\begin{enumerate}[label=(\alph*)]
\item find the equation of the tangent to $C$ at $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
\item Find $\mathrm { f } ( x )$, giving each term in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2017 Q7 [9]}}