| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| 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 straightforward C1 differentiation application requiring standard techniques: differentiate y = k√x, use the perpendicular gradient relationship (normal ⊥ tangent), solve for k, find the normal equation, and calculate a triangle area. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of normal \(= -\dfrac{2}{3}\) | B1 | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}kx^{-\frac{1}{2}}\) | M1*, A1 | Attempt to differentiate; \(\dfrac{1}{2}kx^{-\frac{1}{2}}\) |
| When \(x = 4\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{k}{4}\) | M1dep* | Attempt to substitute \(x = 4\) into \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) |
| \(\therefore \dfrac{k}{4} = \dfrac{3}{2}\) | M1dep* | Equate gradient expression to negative reciprocal of gradient of normal |
| \(k = 6\) | A1 [6] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\) is point \((4, 12)\) | B1ft | |
| \(Q\) is point \((22, 0)\) | M1, A1 | Correct method to find coordinates of \(Q\); Correct \(x\) coordinate |
| Area of triangle \(= \dfrac{1}{2} \times 12 \times 22\) | M1 | Must use \(y\) coordinate of \(P\) and \(x\) coordinate of \(Q\) |
| \(= 132\) sq. units | A1 [5] |
## Question 11:
### Part (i):
Gradient of normal $= -\dfrac{2}{3}$ | B1 |
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}kx^{-\frac{1}{2}}$ | M1*, A1 | Attempt to differentiate; $\dfrac{1}{2}kx^{-\frac{1}{2}}$
When $x = 4$, $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{k}{4}$ | M1dep* | Attempt to substitute $x = 4$ into $\dfrac{\mathrm{d}y}{\mathrm{d}x}$
$\therefore \dfrac{k}{4} = \dfrac{3}{2}$ | M1dep* | Equate gradient expression to negative reciprocal of gradient of normal
$k = 6$ | A1 [6] | cao
### Part (ii):
$P$ is point $(4, 12)$ | B1ft |
$Q$ is point $(22, 0)$ | M1, A1 | Correct method to find coordinates of $Q$; Correct $x$ coordinate
Area of triangle $= \dfrac{1}{2} \times 12 \times 22$ | M1 | Must use $y$ coordinate of $P$ and $x$ coordinate of $Q$
$= 132$ sq. units | A1 [5] |11 The point $P$ on the curve $y = k \sqrt { x }$ has $x$-coordinate 4 . The normal to the curve at $P$ is parallel to the line $2 x + 3 y = 0$.\\
(i) Find the value of $k$.\\
(ii) This normal meets the $x$-axis at the point $Q$. Calculate the area of the triangle $O P Q$, where $O$ is the point $( 0,0 )$.
RECOGNISING ACHIEVEMENT
\hfill \mbox{\textit{OCR C1 2009 Q11 [11]}}