| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a C1 question requiring differentiation using chain rule (or expansion), finding a normal equation, and showing non-intersection. Part (a) is routine differentiation with 'show that' guidance. Part (b) is standard normal-finding. Part (c) requires solving a quadratic and interpreting the discriminant, which is slightly beyond pure routine but still well within standard C1 problem-solving. The multi-step nature and part (c)'s requirement to prove non-intersection elevates it slightly above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y = x - 6\sqrt{x} + 9\) | M1 A1 | |
| \(\frac{dy}{dx} = 1 - 3x^{-\frac{1}{2}} = 1 - \frac{3}{\sqrt{x}}\) | M1 A1 | |
| (b) \(x = 4 \therefore y = 1\) | ||
| grad of tangent \(= 1 - \frac{3}{2} = -\frac{1}{2}\) | M1 | |
| grad of normal \(= \frac{-1}{-\frac{1}{2}} = 2\) | M1 A1 | |
| \(\therefore y - 1 = 2(x - 4)\) | M1 | |
| \(y = 2x - 7\) | A1 | |
| (c) at intersect: \(x - 6\sqrt{x} + 9 = 2x - 7\) | ||
| \(x + 6\sqrt{x} - 16 = 0\) | M1 | |
| \((\sqrt{x} + 8)(\sqrt{x} - 2) = 0\) | M1 | |
| \(\sqrt{x} = -8, 2\) | A1 | |
| \(\sqrt{x} = 2 \Rightarrow x = 4\) (at P) | ||
| \(\sqrt{x} = -8 \Rightarrow\) no real solutions \(\therefore\) normal does not intersect again | A1 | (13) |
**(a)** $y = x - 6\sqrt{x} + 9$ | M1 A1 |
$\frac{dy}{dx} = 1 - 3x^{-\frac{1}{2}} = 1 - \frac{3}{\sqrt{x}}$ | M1 A1 |
**(b)** $x = 4 \therefore y = 1$ | |
grad of tangent $= 1 - \frac{3}{2} = -\frac{1}{2}$ | M1 |
grad of normal $= \frac{-1}{-\frac{1}{2}} = 2$ | M1 A1 |
$\therefore y - 1 = 2(x - 4)$ | M1 |
$y = 2x - 7$ | A1 |
**(c)** at intersect: $x - 6\sqrt{x} + 9 = 2x - 7$ | |
$x + 6\sqrt{x} - 16 = 0$ | M1 |
$(\sqrt{x} + 8)(\sqrt{x} - 2) = 0$ | M1 |
$\sqrt{x} = -8, 2$ | A1 |
$\sqrt{x} = 2 \Rightarrow x = 4$ (at P) | |
$\sqrt{x} = -8 \Rightarrow$ no real solutions $\therefore$ normal does not intersect again | A1 | (13)A curve has the equation $y = (\sqrt{x} - 3)^2$, $x \geq 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dy}{dx} = 1 - \frac{3}{\sqrt{x}}$. [4]
\end{enumerate}
The point $P$ on the curve has $x$-coordinate 4.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an equation for the normal to the curve at $P$ in the form $y = mx + c$. [5]
\item Show that the normal to the curve at $P$ does not intersect the curve again. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q9 [13]}}