| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring integration with a boundary condition, then finding a normal line. Part (a) involves routine integration of powers of x and applying a point to find the constant. Part (b) requires evaluating the derivative at x=9, finding the negative reciprocal for the normal's gradient, and writing in the specified form. All techniques are standard C1/C2 material with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 3\sqrt{x} - \frac{9}{\sqrt{x}} + 2 = 3x^{\frac{1}{2}} - 9x^{-\frac{1}{2}} + 2\) | B1 | Expression written as \(3x^{\frac{1}{2}} - 9x^{-\frac{1}{2}} + 2\) (may be implied by correct integration) |
| \(f(x) = \int(3x^{\frac{1}{2}} - 9x^{-\frac{1}{2}} + 2)\,dx = \frac{3x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{9x^{\frac{1}{2}}}{\frac{1}{2}} + 2x \, (+c)\) | M1, A1, A1 | M1: evidence of integration, \(x^n \to x^{n+1}\) at least once (accept on the 2). A1: two terms of \(\frac{3x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{9x^{\frac{1}{2}}}{\frac{1}{2}} + 2x\) correct. A1: all three terms correct (no need to simplify or have \(+c\)) |
| When \(x=9\), \(y=14\), so \(c=\) ... \(\Rightarrow f(x) = 2x^{\frac{3}{2}} - 18x^{\frac{1}{2}} + 2x - 4\) | M1, A1 | M1: substitutes \(x=9\), \(y=14\) into changed expression (with \(+c\)) to obtain \(c\). A1: all terms simplified and \(c\) correct. Allow \(y = 2x\sqrt{x} - 18\sqrt{x} + 2x - 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of curve at \(P\) is \(f'(9) = 8\) | B1 | |
| Gradient of normal is \(k = -\frac{1}{f'(9)}\) | M1 | Uses or states gradient of normal is \(-\frac{1}{f'(9)}\) |
| Equation of normal: \((y-14) = k(x-9)\) or \(y = kx + c'\) with \((9,14)\) to find \(c'\), where \(k = -\frac{1}{8}\) | dM1, A1 | dM1: dependent on previous M; finds equation of line using \((9,14)\) and their \(-\frac{1}{f'(9)}\) as gradient. A1: correct unsimplified answer e.g. \((y-14) = -\frac{1}{8}(x-9)\) or \(y = -\frac{1}{8}x + \frac{121}{8}\) |
| \(x + 8y - 121 = 0\) | A1 | Any multiple of \(x + 8y - 121 = 0\) |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 3\sqrt{x} - \frac{9}{\sqrt{x}} + 2 = 3x^{\frac{1}{2}} - 9x^{-\frac{1}{2}} + 2$ | B1 | Expression written as $3x^{\frac{1}{2}} - 9x^{-\frac{1}{2}} + 2$ (may be implied by correct integration) |
| $f(x) = \int(3x^{\frac{1}{2}} - 9x^{-\frac{1}{2}} + 2)\,dx = \frac{3x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{9x^{\frac{1}{2}}}{\frac{1}{2}} + 2x \, (+c)$ | M1, A1, A1 | M1: evidence of integration, $x^n \to x^{n+1}$ at least once (accept on the 2). A1: two terms of $\frac{3x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{9x^{\frac{1}{2}}}{\frac{1}{2}} + 2x$ correct. A1: all three terms correct (no need to simplify or have $+c$) |
| When $x=9$, $y=14$, so $c=$ ... $\Rightarrow f(x) = 2x^{\frac{3}{2}} - 18x^{\frac{1}{2}} + 2x - 4$ | M1, A1 | M1: substitutes $x=9$, $y=14$ into changed expression (with $+c$) to obtain $c$. A1: all terms simplified and $c$ correct. Allow $y = 2x\sqrt{x} - 18\sqrt{x} + 2x - 4$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of curve at $P$ is $f'(9) = 8$ | B1 | |
| Gradient of normal is $k = -\frac{1}{f'(9)}$ | M1 | Uses or states gradient of normal is $-\frac{1}{f'(9)}$ |
| Equation of normal: $(y-14) = k(x-9)$ or $y = kx + c'$ with $(9,14)$ to find $c'$, where $k = -\frac{1}{8}$ | dM1, A1 | dM1: dependent on previous M; finds equation of line using $(9,14)$ and their $-\frac{1}{f'(9)}$ as gradient. A1: correct unsimplified answer e.g. $(y-14) = -\frac{1}{8}(x-9)$ or $y = -\frac{1}{8}x + \frac{121}{8}$ |
| $x + 8y - 121 = 0$ | A1 | Any multiple of $x + 8y - 121 = 0$ |\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x ) , x > 0$, where
\end{enumerate}
$$f ^ { \prime } ( x ) = 3 \sqrt { x } - \frac { 9 } { \sqrt { x } } + 2$$
Given that the point $P ( 9,14 )$ lies on $C$,\\
(a) find $\mathrm { f } ( x )$, simplifying your answer,\\
(b) 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.
\hfill \mbox{\textit{Edexcel C12 2015 Q11 [11]}}