| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: polynomial with multiple terms |
| Difficulty | Standard +0.3 Part (a) is a standard first-principles differentiation of a simple polynomial requiring algebraic expansion and cancellation. Part (b) is routine sketching of a quadratic. Part (c) requires finding where f''(x)=0 and verifying it's not stationary, which is straightforward but adds a small conceptual step beyond pure routine work. |
| Spec | 1.07c Sketch gradient function: for given curve1.07e Second derivative: as rate of change of gradient1.07f Convexity/concavity: points of inflection1.07g Differentiation from first principles: for small positive integer powers of x1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^3-(x+h)-(x^3-x)}{h}\) | M1 (2.1) | Substituting into \(\frac{f(x+h)-f(x)}{h}\) and attempt to expand \((x+h)^3\) |
| \(= \frac{x^3+3hx^2+3h^2x+h^3-x-h-(x^3-x)}{h}\) | A1 (2.1) | Correct expansion of \((x+h)^3\). Allow correct 6 terms not simplified |
| \(= \frac{3x^2h+3xh^2+h^3-h}{h} = 3x^2+3xh+h^2-1\) | M1 (2.1) | Simplifying the fraction to eliminate a denominator |
| \(f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}(3x^2-1+3xh+h^2) = 3x^2-1\) | E1 (2.1) [4] | Must include the idea of limit as \(h\) tends to zero AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct shape with vertex on negative \(y\)-axis | B1 (1.1a) | |
| \((0,-1)\) labelled and indication graph crosses \(x\)-axis at \(\left(\pm\frac{1}{\sqrt{3}}, 0\right)\) | B1 (dep) (1.1) [2] | Allow without \(\pm\frac{1}{\sqrt{3}}\) if clear that the points are between \((-1,0)\) and \((1,0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Point of inflection when \(f''(x) = 0\) | M1 (2.1) | Equating their second derivative to zero |
| \(f''(x) = 6x = 0\) has only one root \(x = 0\) | A1 (2.1) | Must explain that this is the only point of inflection |
| When \(x=0\), \(f'(x) = -1 \neq 0\) so the point of inflection is not a stationary point | E1 (2.1) [3] | Must prove that the point is not stationary from correct value for \(f'(0)\). Also allow if shown that the stationary points are at \(\left(\pm\frac{1}{\sqrt{3}}, 0\right)\) |
# Question 12:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^3-(x+h)-(x^3-x)}{h}$ | M1 (2.1) | Substituting into $\frac{f(x+h)-f(x)}{h}$ and attempt to expand $(x+h)^3$ |
| $= \frac{x^3+3hx^2+3h^2x+h^3-x-h-(x^3-x)}{h}$ | A1 (2.1) | Correct expansion of $(x+h)^3$. Allow correct 6 terms not simplified |
| $= \frac{3x^2h+3xh^2+h^3-h}{h} = 3x^2+3xh+h^2-1$ | M1 (2.1) | Simplifying the fraction to eliminate a denominator |
| $f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}(3x^2-1+3xh+h^2) = 3x^2-1$ | E1 (2.1) [4] | Must include the idea of limit as $h$ tends to zero AG |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape with vertex on negative $y$-axis | B1 (1.1a) | |
| $(0,-1)$ labelled and indication graph crosses $x$-axis at $\left(\pm\frac{1}{\sqrt{3}}, 0\right)$ | B1 (dep) (1.1) [2] | Allow without $\pm\frac{1}{\sqrt{3}}$ if clear that the points are between $(-1,0)$ and $(1,0)$ |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Point of inflection when $f''(x) = 0$ | M1 (2.1) | Equating their second derivative to zero |
| $f''(x) = 6x = 0$ has only one root $x = 0$ | A1 (2.1) | Must explain that this is the only point of inflection |
| When $x=0$, $f'(x) = -1 \neq 0$ so the point of inflection is not a stationary point | E1 (2.1) [3] | Must prove that the point is not stationary from correct value for $f'(0)$. Also allow if shown that the stationary points are at $\left(\pm\frac{1}{\sqrt{3}}, 0\right)$ |
---12 A function is defined by $\mathrm { f } ( x ) = x ^ { 3 } - x$.
\begin{enumerate}[label=(\alph*)]
\item By considering $\frac { f ( x + h ) - f ( x ) } { h }$, show from first principles that $f ^ { \prime } ( x ) = 3 x ^ { 2 } - 1$.
\item Sketch the gradient function $\mathrm { f } ^ { \prime } ( x )$.
\item Show that the curve $y = f ( x )$ has a single point of inflection which is not a stationary point.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q12 [9]}}