| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| 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 standard C1 differentiation question with three routine parts: finding a gradient, finding a normal equation, and finding tangent conditions. Part (iii) requires solving a discriminant condition (b²-4ac=0) which elevates it slightly above pure recall, but all techniques are standard textbook exercises with no novel insight required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 2x+1\) | M1 | Attempt to differentiate \(y\) |
| \(= 5\) | A1 (2) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of normal \(= -\frac{1}{5}\) | B1 ft | ft from a non-zero numerical value in (i) |
| When \(x=2\), \(y=6\) | B1 | May be embedded in equation of line |
| \(y-6=-\frac{1}{5}(x-2)\) | M1 | Equation of line, any non-zero gradient, their \(y\) coordinate |
| \(x+5y-32=0\) | A1 (4) | Correct equation in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x^2+x=kx-4\) | *M1 | Equating \(y_1=y_2\) |
| \(x^2+(1-k)x+4=0\) | ||
| One solution \(\Rightarrow b^2-4ac=0\) | DM1 | Statement that discriminant \(=0\) |
| \((1-k)^2-4\times1\times4=0\) | DM1 | Attempt (involving \(k\)) to use a, b, c from their equation |
| \((1-k)^2=16\) | A1 | Correct equation (may be unsimplified) |
| \(1-k=\pm4\) | DM1 | Correct method to find \(k\), dep on 1st 3Ms |
| \(k=-3\) or \(5\) | A1 (6) | Both values correct |
## Question 10:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 2x+1$ | M1 | Attempt to differentiate $y$ |
| $= 5$ | A1 (2) | cao |
**Part (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of normal $= -\frac{1}{5}$ | B1 ft | ft from a non-zero numerical value in (i) |
| When $x=2$, $y=6$ | B1 | May be embedded in equation of line |
| $y-6=-\frac{1}{5}(x-2)$ | M1 | Equation of line, any non-zero gradient, their $y$ coordinate |
| $x+5y-32=0$ | A1 (4) | Correct equation in correct form |
**Part (iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x^2+x=kx-4$ | *M1 | Equating $y_1=y_2$ |
| $x^2+(1-k)x+4=0$ | | |
| One solution $\Rightarrow b^2-4ac=0$ | DM1 | Statement that discriminant $=0$ |
| $(1-k)^2-4\times1\times4=0$ | DM1 | Attempt (involving $k$) to use a, b, c from their equation |
| $(1-k)^2=16$ | A1 | Correct equation (may be unsimplified) |
| $1-k=\pm4$ | DM1 | Correct method to find $k$, dep on 1st 3Ms |
| $k=-3$ or $5$ | A1 (6) | Both values correct |10 A curve has equation $y = x ^ { 2 } + x$.\\
(i) Find the gradient of the curve at the point for which $x = 2$.\\
(ii) Find the equation of the normal to the curve at the point for which $x = 2$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
(iii) Find the values of $k$ for which the line $y = k x - 4$ is a tangent to the curve.
\hfill \mbox{\textit{OCR C1 2009 Q10 [12]}}