| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Moderate -0.8 This is a straightforward application of standard differentiation techniques. Part (a) requires finding dy/dx, setting it to zero, and showing the discriminant is negative. Part (b) requires finding d²y/dx², setting it to zero, and verifying the sign change. Both parts are routine calculus exercises with no problem-solving insight required, making this easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx}=3x^2-6x+4=0\) | M1 | Differentiate & equate to 0. May be implied by calc of D. or \(x=\frac{6\pm\sqrt{36-48}}{6}\) or \(x=\frac{6\pm\sqrt{12}}{6}\) oe |
| \(b^2-4ac=-12\) or \(D=-12\) or \(3(x-1)^2+1=0\) oe; No (real) roots or no value of \(x\), or can't \(\sqrt{\text{negative}}\), or gradient always \(+\)ve | A1 | Must see justification as line above, no errors, & statement. Other correct forms of quadratic equation and justification may be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2}=6x-6=0\) | M1 | Differentiate their \(\frac{dy}{dx}\) and \(=0\). Can be implied by \(x=1\) |
| \(x=1\) gives a point of inflection | A1 | Statement "\(x=1\) gives a point of inflection" is enough. or: This equation has one root (so curve has one inflection). Not just "\(x=1\)". Ignore \(y\)-coordinate |
| or \(x=1\) & show that, either side of this point, gradient does not change sign, or second derivative does change sign |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx}=3x^2-6x+4=0$ | M1 | Differentiate & equate to 0. May be implied by calc of D. or $x=\frac{6\pm\sqrt{36-48}}{6}$ or $x=\frac{6\pm\sqrt{12}}{6}$ oe |
| $b^2-4ac=-12$ or $D=-12$ or $3(x-1)^2+1=0$ oe; No (real) roots or no value of $x$, or can't $\sqrt{\text{negative}}$, or gradient always $+$ve | A1 | Must see justification as line above, no errors, & statement. Other correct forms of quadratic equation and justification may be seen |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2y}{dx^2}=6x-6=0$ | M1 | Differentiate their $\frac{dy}{dx}$ and $=0$. Can be implied by $x=1$ |
| $x=1$ gives a point of inflection | A1 | Statement "$x=1$ gives a point of inflection" is enough. or: This equation has one root (so curve has one inflection). Not just "$x=1$". Ignore $y$-coordinate |
| or $x=1$ & show that, either side of this point, gradient does not change sign, or second derivative does change sign | | |
---5 In this question you must show detailed reasoning.
A curve has equation $y = x ^ { 3 } - 3 x ^ { 2 } + 4 x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the curve has no stationary points.
\item Show that the curve has exactly one point of inflection.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2022 Q5 [4]}}