| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Integration after differentiation |
| Difficulty | Moderate -0.8 Part (a) is a standard textbook exercise in differentiation from first principles with a simple polynomial requiring routine algebraic manipulation. Part (b) is basic integration with a constant of integration found from given coordinates. Both parts are mechanical applications of core techniques with no problem-solving or insight required, making this easier than average. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x+h) - f(x) = ((x+h)^2 + 2(x+h)) - (x^2 + 2x)\) | M1 | Attempt expression for \(f(x+h) - f(x)\). Allow sign error from no bracket around final term, ie \((x+h)^2 + 2(x+h) - x^2 + 2x\) is M1, but no other errors allowed. If considering \(x^2\) and \(2x\) separately then expressions for both must be seen |
| \(= x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x = 2xh + h^2 + 2h\) | M1 | Expand and simplify \(f(x+h) - f(x)\). Expand and gather like terms (either separately, or single expression). Condone sign errors only, so M0 if collecting like terms after an incorrect attempt to divide by \(h\). Allow BOD if \(2x... + 2x\) becomes 0 rather than \(4x\) |
| \(\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2+2h}{h} = 2x+h+2\) | M1 | Attempt \(\frac{f(x+h)-f(x)}{h}\). Divide all terms by \(h\). Allow BOD if previous error results in a term with a denominator of \(h\) |
| \(f'(x) = \lim_{h \to 0}(2x+h+2) = 2x+2\) | A1 | Complete proof by considering limit as \(h \to 0\). www, including correct signs throughout. Must divide by \(h\) before \(h \to 0\). Must see 'lim', '\(h \to 0\)', and \(f'(x)\) at some point in their solution. Allow BOD for \(f'(x) = \frac{f(x+h)-f(x)}{h}\) followed by \(=..., =...\) on subsequent lines. A0 if 'lim' still in final answer. Condone \(\frac{dy}{dx}\) in place of \(f'(x)\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x^2 + 2x + c\) | B1 | State or imply correct equation, including \(+ c\). '\(y=\)' could be implied by use of 5; \(c\) may be implied by later work |
| \(5 = 1 - 2 + c\), \(c = 6\) | M1 | Attempt \(c\) using \((-1, 5)\). Allow M1 if equation incorrect, as long as from attempt at integrating \(2x+2\) ie of form \(y = kx^2 + 2x + c\). Must use \(x\) and \(y\) the correct way around. As far as attempting a value for \(c\) |
| \(y = x^2 + 2x + 6\) | A1 | Obtain correct equation, including \(y = ...\). Equation must be stated, and not just implied by \(c = 6\) seen. Allow \(f(x) = ...\). A0 for 'equation' \(= x^2 + 2x + 6\). Just stating \(y = x^2 + 2x + 6\) or \(y = (x+1)^2 + 5\) gets full marks (may come from observing that \((-1, 5)\) is the minimum point) |
| [3] |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x+h) - f(x) = ((x+h)^2 + 2(x+h)) - (x^2 + 2x)$ | **M1** | Attempt expression for $f(x+h) - f(x)$. Allow sign error from no bracket around final term, ie $(x+h)^2 + 2(x+h) - x^2 + 2x$ is M1, but no other errors allowed. If considering $x^2$ and $2x$ separately then expressions for both must be seen |
| $= x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x = 2xh + h^2 + 2h$ | **M1** | Expand and simplify $f(x+h) - f(x)$. Expand and gather like terms (either separately, or single expression). Condone sign errors only, so M0 if collecting like terms after an incorrect attempt to divide by $h$. Allow BOD if $2x... + 2x$ becomes 0 rather than $4x$ |
| $\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2+2h}{h} = 2x+h+2$ | **M1** | Attempt $\frac{f(x+h)-f(x)}{h}$. Divide **all** terms by $h$. Allow BOD if previous error results in a term with a denominator of $h$ |
| $f'(x) = \lim_{h \to 0}(2x+h+2) = 2x+2$ | **A1** | Complete proof by considering limit as $h \to 0$. www, including correct signs throughout. Must divide by $h$ before $h \to 0$. Must see 'lim', '$h \to 0$', and $f'(x)$ at some point in their solution. Allow BOD for $f'(x) = \frac{f(x+h)-f(x)}{h}$ followed by $=..., =...$ on subsequent lines. **A0** if 'lim' still in final answer. Condone $\frac{dy}{dx}$ in place of $f'(x)$ |
| **[4]** | | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^2 + 2x + c$ | **B1** | State or imply correct equation, including $+ c$. '$y=$' could be implied by use of 5; $c$ may be implied by later work |
| $5 = 1 - 2 + c$, $c = 6$ | **M1** | Attempt $c$ using $(-1, 5)$. Allow M1 if equation incorrect, as long as from attempt at integrating $2x+2$ ie of form $y = kx^2 + 2x + c$. Must use $x$ and $y$ the correct way around. As far as attempting a value for $c$ |
| $y = x^2 + 2x + 6$ | **A1** | Obtain correct equation, including $y = ...$. Equation must be stated, and not just implied by $c = 6$ seen. Allow $f(x) = ...$. **A0** for 'equation' $= x^2 + 2x + 6$. Just stating $y = x^2 + 2x + 6$ or $y = (x+1)^2 + 5$ gets full marks (may come from observing that $(-1, 5)$ is the minimum point) |
| **[3]** | | |
---3
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { f } ( x ) = x ^ { 2 } + 2 x$, use differentiation from first principles to show that $\mathrm { f } ^ { \prime } ( x ) = 2 x + 2$.
\item The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x + 2$ and the curve passes through the point $( - 1,5 )$.
Find the equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q3 [7]}}