| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Moderate -0.8 This is a straightforward multi-part differentiation question testing standard techniques: finding dy/dx, substituting to verify a stationary point, using the second derivative test, and finding where a horizontal tangent meets the y-axis. All parts are routine applications of C1 content with no problem-solving insight required, making it easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| i) \(\frac{dy}{dx} = 9x^2 - 7 - 2x^{-2}\); When \(x = 1, \frac{dy}{dx} = 9 - 7 - 2 = 0\); Therefore a stationary point | M1*, A1, A1, M1dep, A1 [5] | Attempt to differentiate, any term correct. Two correct terms. Fully correct. Substitute \(x = 1\) into their derivative. Correctly obtain zero www and state conclusion AG |
| ii) \(\frac{d^2y}{dx^2} = 18x + 4x^{-3}\); When \(x = 1, \frac{d^2y}{dx^2} > 0\) so minimum | M1, A1 [2] | Correct method to find nature of stationary point e.g. substituting \(x = 1\) into second derivative (at least one term correct from their first derivative in (i)). No incorrect working seen in this part i.e. if second derivative is evaluated, it must be 22. Alternative valid methods include: 1) Evaluating gradient at either side of 1(\(x > 0\)); 2) Evaluating \(y\) at 1 and either side of 1 (\(x > 0\)). If using alternatives, working must be fully correct to obtain the A mark |
| iii) When \(x = 1, y = -2\); \((0, -2)\) | B1, B1 [2] | Finding \(y = -2\) at \(x = 1\). Correct coordinate www |
**i)** $\frac{dy}{dx} = 9x^2 - 7 - 2x^{-2}$; When $x = 1, \frac{dy}{dx} = 9 - 7 - 2 = 0$; Therefore a stationary point | M1*, A1, A1, M1dep, A1 [5] | Attempt to differentiate, any term correct. Two correct terms. Fully correct. Substitute $x = 1$ into their derivative. Correctly obtain zero www and state conclusion AG
**ii)** $\frac{d^2y}{dx^2} = 18x + 4x^{-3}$; When $x = 1, \frac{d^2y}{dx^2} > 0$ so minimum | M1, A1 [2] | Correct method to find nature of stationary point e.g. substituting $x = 1$ into second derivative (at least one term correct from their first derivative in (i)). No incorrect working seen in this part i.e. if second derivative is evaluated, it must be 22. Alternative valid methods include: 1) Evaluating gradient at either side of 1($x > 0$); 2) Evaluating $y$ at 1 and either side of 1 ($x > 0$). If using alternatives, working must be fully correct to obtain the A mark
**iii)** When $x = 1, y = -2$; $(0, -2)$ | B1, B1 [2] | Finding $y = -2$ at $x = 1$. Correct coordinate wwwA curve has equation $y = 3x^3 - 7x + \frac{2}{x}$.
\begin{enumerate}[label=(\roman*)]
\item Verify that the curve has a stationary point when $x = 1$. [5]
\item Determine the nature of this stationary point. [2]
\item The tangent to the curve at this stationary point meets the $y$-axis at the point $Q$. Find the coordinates of $Q$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2014 Q8 [9]}}