| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| 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 and tangent question with clear signposting through parts (a)-(d). Part (a) is immediate recognition that gradient equals 2, part (b) requires solving a simple equation with surds, part (c) is routine integration with one constant to find, and part (d) involves basic root-finding. While multi-part, each step follows naturally from the previous with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2\) | B1 | |
| (b) \(1 + \frac{2}{\sqrt{x}} = 2\) | M1 | |
| \(\sqrt{x} = 2\) | M1 | |
| \(x = 4\) | A1 | |
| (c) At \(x = 4\): \(y = 2(4) - 1 = 7\) | B1 | |
| \(y = \int(1 + \frac{2}{\sqrt{x}}) dx\) | ||
| \(y = x + 4x^{\frac{1}{2}} + c\) | M1 A2 | |
| \((4, 7)\): \(7 = 4 + 8 + c\), so \(c = -5\) | M1 | |
| \(y = x + 4\sqrt{x} - 5\) | A1 | |
| (d) \(x + 4x^{\frac{1}{2}} - 5 = 0\) | ||
| \((x^{\frac{1}{2}} + 5)(x^{\frac{1}{2}} - 1) = 0\) | M1 | |
| \(x^{\frac{1}{2}} = -5\) (no real solutions), \(x = 1\) | A1 | |
| \((1, 0)\) and no other point | A1 | (13) |
**(a)** $2$ | B1 |
**(b)** $1 + \frac{2}{\sqrt{x}} = 2$ | M1 |
$\sqrt{x} = 2$ | M1 |
$x = 4$ | A1 |
**(c)** At $x = 4$: $y = 2(4) - 1 = 7$ | B1 |
$y = \int(1 + \frac{2}{\sqrt{x}}) dx$ | |
$y = x + 4x^{\frac{1}{2}} + c$ | M1 A2 |
$(4, 7)$: $7 = 4 + 8 + c$, so $c = -5$ | M1 |
$y = x + 4\sqrt{x} - 5$ | A1 |
**(d)** $x + 4x^{\frac{1}{2}} - 5 = 0$ | |
$(x^{\frac{1}{2}} + 5)(x^{\frac{1}{2}} - 1) = 0$ | M1 |
$x^{\frac{1}{2}} = -5$ (no real solutions), $x = 1$ | A1 |
$(1, 0)$ and no other point | A1 | (13)9. The curve $C$ has the equation $y = \mathrm { f } ( x )$ where
$$f ^ { \prime } ( x ) = 1 + \frac { 2 } { \sqrt { x } } , \quad x > 0$$
The straight line $l$ has the equation $y = 2 x - 1$ and is a tangent to $C$ at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item State the gradient of $C$ at $P$.
\item Find the $x$-coordinate of $P$.
\item Find an equation for $C$.
\item Show that $C$ crosses the $x$-axis at the point $( 1,0 )$ and at no other point.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q9 [13]}}