| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (extended problem with normals, stationary points, or further geometry) |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring splitting a fraction, integrating powers of x, and using a boundary condition to find the constant. Part (b) is routine tangent-finding. Slightly easier than average due to mechanical nature with no conceptual challenges. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{5x^2+4}{2\sqrt{x}} = \frac{5}{2}x^{\frac{3}{2}} + 2x^{-\frac{1}{2}}\) | B1 | May have unsimplified coefficients. Allow decimal indices. May be implied by later work |
| \(f(x) = \frac{\frac{5}{2}x^{\frac{5}{2}}}{\frac{5}{2}} + \frac{2x^{\frac{1}{2}}}{\frac{1}{2}} - 5x \quad (+c)\) | M1 A1 A1 | M1: attempt to integrate, one power increased by one. A1: two of three terms correct unsimplified. A1: all three terms correct unsimplified, no need for \(+c\) |
| Uses \(f(4)=14\) to find \(c\) | dM1 | Uses \(x=4\), \(f(x)=14\) to find numerical \(c\); must have attempted to integrate |
| \(c=-6\) and so \(f(x) = x^{\frac{5}{2}} + 4x^{\frac{1}{2}} - 5x - 6\) | A1 | All four terms correct simplified with \(-6\) included |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient at \((4,14)\): \(f'(4) = \frac{84}{4} - 5 = 16\) | M1 A1 | M1: substitute \(x=4\) into \(f'(x)=\frac{5x^2+4}{2\sqrt{x}}-5\). A1: \(f'(4)=16\) |
| \((y-14) = 16(x-4)\) and \(y = 16x - 50\) | dM1 A1 | dM1: linear equation through \((4,14)\) with their gradient. A1: cao \(y=16x-50\) |
# Question 11:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{5x^2+4}{2\sqrt{x}} = \frac{5}{2}x^{\frac{3}{2}} + 2x^{-\frac{1}{2}}$ | B1 | May have unsimplified coefficients. Allow decimal indices. May be implied by later work |
| $f(x) = \frac{\frac{5}{2}x^{\frac{5}{2}}}{\frac{5}{2}} + \frac{2x^{\frac{1}{2}}}{\frac{1}{2}} - 5x \quad (+c)$ | M1 A1 A1 | M1: attempt to integrate, one power increased by one. A1: two of three terms correct unsimplified. A1: all three terms correct unsimplified, no need for $+c$ |
| Uses $f(4)=14$ to find $c$ | dM1 | Uses $x=4$, $f(x)=14$ to find numerical $c$; must have attempted to integrate |
| $c=-6$ and so $f(x) = x^{\frac{5}{2}} + 4x^{\frac{1}{2}} - 5x - 6$ | A1 | All four terms correct simplified with $-6$ included |
**[6 marks]**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient at $(4,14)$: $f'(4) = \frac{84}{4} - 5 = 16$ | M1 A1 | M1: substitute $x=4$ into $f'(x)=\frac{5x^2+4}{2\sqrt{x}}-5$. A1: $f'(4)=16$ |
| $(y-14) = 16(x-4)$ and $y = 16x - 50$ | dM1 A1 | dM1: linear equation through $(4,14)$ with their gradient. A1: cao $y=16x-50$ |
**[4 marks]**
---11. The curve $C$ has equation $y = \mathrm { f } ( x ) , x > 0$, where
$$f ^ { \prime } ( x ) = \frac { 5 x ^ { 2 } + 4 } { 2 \sqrt { x } } - 5$$
It is given that the point $P ( 4,14 )$ lies on $C$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { f } ( x )$, writing each term in a simplified form.
\item Find the equation of the tangent to $C$ at the point $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q11 [10]}}