| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| 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 C2 question requiring integration to find the curve equation, using the tangent gradient to find a point, and analyzing x-intercepts. All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(1 + \frac{2}{\sqrt{x}} = 2\) | M1 | |
| \(\sqrt{x} = 2\) | M1 | |
| \(x = 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x = 4 \therefore y = 2(4) - 1 = 7\) | B1 | |
| \(y = \int (1 + \frac{2}{\sqrt{x}}) \, dx\) | ||
| \(y = x + 4x^{\frac{1}{2}} + c\) | M1 A2 | |
| \((4, 7) \therefore 7 = 4 + 8 + c\) | ||
| \(c = -5\) | M1 | |
| \(y = x + 4x^{\frac{1}{2}} - 5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x + 4x^{\frac{1}{2}} - 5 = 0\) | ||
| \((x^{\frac{1}{2}} + 5)(x^{\frac{1}{2}} - 1) = 0\) | M1 | |
| \(x^{\frac{1}{2}} = -5\) (no real solutions), \(1\) | A1 | |
| \(x = 1 \therefore (1, 0)\) and no other point | A1 | (13) |
# Question 9:
## Part (i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $2$ | B1 | |
## Part (ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $1 + \frac{2}{\sqrt{x}} = 2$ | M1 | |
| $\sqrt{x} = 2$ | M1 | |
| $x = 4$ | A1 | |
## Part (iii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x = 4 \therefore y = 2(4) - 1 = 7$ | B1 | |
| $y = \int (1 + \frac{2}{\sqrt{x}}) \, dx$ | | |
| $y = x + 4x^{\frac{1}{2}} + c$ | M1 A2 | |
| $(4, 7) \therefore 7 = 4 + 8 + c$ | | |
| $c = -5$ | M1 | |
| $y = x + 4x^{\frac{1}{2}} - 5$ | A1 | |
## Part (iv):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x + 4x^{\frac{1}{2}} - 5 = 0$ | | |
| $(x^{\frac{1}{2}} + 5)(x^{\frac{1}{2}} - 1) = 0$ | M1 | |
| $x^{\frac{1}{2}} = -5$ (no real solutions), $1$ | A1 | |
| $x = 1 \therefore (1, 0)$ 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 $P$.\\
(i) State the gradient of $C$ at $P$.\\
(ii) Find the $x$-coordinate of $P$.\\
(iii) Find an equation for $C$.\\
(iv) Show that $C$ crosses the $x$-axis at the point $( 1,0 )$ and at no other point.
\hfill \mbox{\textit{OCR C2 Q9 [13]}}