| Exam Board | OCR |
|---|---|
| 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 straightforward C1 differentiation question requiring chain rule application (shown for students), gradient calculation, and normal line equation. Part (iii) adds mild algebraic verification but follows standard technique. Slightly above average due to the three-part structure and normal (not tangent) requirement, but all steps are routine for C1 level. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = x - 6\sqrt{x} + 9\); \(\frac{dy}{dx} = 1 - 3x^{-\frac{1}{2}} = 1 - \frac{3}{\sqrt{x}}\) | M1, A1; M1, A1 | |
| (ii) \(x = 4\) \(\therefore y = 1\); grad of tangent \(= 1 - \frac{3}{2} = -\frac{1}{2}\); grad of normal \(= \frac{-1}{-\frac{1}{2}} = 2\); \(\therefore y - 1 = 2(x-4)\); \(y = 2x - 7\) | B1; M1; A1; M1; A1 | |
| (iii) at intersect: \(x - 6\sqrt{x} + 9 = 2x - 7\); \(x + 6\sqrt{x} - 16 = 0\); \((\sqrt{x} + 8)(\sqrt{x} - 2) = 0\); \(\sqrt{x} = -8, 2\); \(\sqrt{x} = 2 \Rightarrow x = 4\) (at P); \(\sqrt{x} = -8 \Rightarrow\) no real solutions \(\therefore\) normal does not intersect again | M1; M1; A1; A1 | (13) |
**(i)** $y = x - 6\sqrt{x} + 9$; $\frac{dy}{dx} = 1 - 3x^{-\frac{1}{2}} = 1 - \frac{3}{\sqrt{x}}$ | M1, A1; M1, A1 |
**(ii)** $x = 4$ $\therefore y = 1$; grad of tangent $= 1 - \frac{3}{2} = -\frac{1}{2}$; grad of normal $= \frac{-1}{-\frac{1}{2}} = 2$; $\therefore y - 1 = 2(x-4)$; $y = 2x - 7$ | B1; M1; A1; M1; A1 |
**(iii)** at intersect: $x - 6\sqrt{x} + 9 = 2x - 7$; $x + 6\sqrt{x} - 16 = 0$; $(\sqrt{x} + 8)(\sqrt{x} - 2) = 0$; $\sqrt{x} = -8, 2$; $\sqrt{x} = 2 \Rightarrow x = 4$ (at P); $\sqrt{x} = -8 \Rightarrow$ no real solutions $\therefore$ normal does not intersect again | M1; M1; A1; A1 | (13)
**Total: (72)**10. A curve has the equation $y = ( \sqrt { x } - 3 ) ^ { 2 } , x \geq 0$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - \frac { 3 } { \sqrt { x } }$.
The point $P$ on the curve has $x$-coordinate 4 .\\
(ii) Find an equation for the normal to the curve at $P$ in the form $y = m x + c$.\\
(iii) Show that the normal to the curve at $P$ does not intersect the curve again.
\hfill \mbox{\textit{OCR C1 Q10 [13]}}