| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Moderate -0.8 This is a straightforward product rule application with standard functions (polynomial and exponential), followed by a routine sign analysis. Part (a) requires direct application of the product rule with no algebraic complications, and part (b) is a standard 'hence show' that simply requires demonstrating the derivative is always positive—a typical textbook exercise with minimal problem-solving demand. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}(e^{2x}) = 2e^{2x}\) | B1 | Seen anywhere in solution |
| \(6xe^{2x} + (2 + 3x^2)(2e^{2x})\) | M1 | Attempt product rule; could expand first |
| \(e^{2x}(6x^2 + 6x + 4)\) | A1 | Obtain any fully correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{2x} > 0\) for all \(x\) | B1 | B0 if clearly considering \(f(x)\) or \(f''(x)\) and not \(f'(x)\) |
| \(6x^2 + 6x + 4 = 6\left(x + \frac{1}{2}\right)^2 + \frac{5}{2}\); minimum value is \(\frac{5}{2}\) so \(> 0\) for all \(x\) | M1 | Attempt to show 3-term quadratic factor is \(> 0\) for all \(x\); complete the square or consider discriminant; could be multiple or fraction of their quadratic |
| Gradient \(e^{2x}(6x^2 + 6x + 4) > 0\) for all \(x\) so it is increasing for all \(x\) | A1 | Full justification that quadratic factor is always positive |
| A1 | Justify increasing function as \(f'(x) > 0\) for all \(x\) |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(e^{2x}) = 2e^{2x}$ | B1 | Seen anywhere in solution |
| $6xe^{2x} + (2 + 3x^2)(2e^{2x})$ | M1 | Attempt product rule; could expand first |
| $e^{2x}(6x^2 + 6x + 4)$ | A1 | Obtain any fully correct expression |
**[3 marks]**
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{2x} > 0$ for all $x$ | B1 | B0 if clearly considering $f(x)$ or $f''(x)$ and not $f'(x)$ |
| $6x^2 + 6x + 4 = 6\left(x + \frac{1}{2}\right)^2 + \frac{5}{2}$; minimum value is $\frac{5}{2}$ so $> 0$ for all $x$ | M1 | Attempt to show 3-term quadratic factor is $> 0$ for all $x$; complete the square or consider discriminant; could be multiple or fraction of their quadratic |
| Gradient $e^{2x}(6x^2 + 6x + 4) > 0$ for all $x$ so it is increasing for all $x$ | A1 | Full justification that quadratic factor is always positive |
| | A1 | Justify increasing function as $f'(x) > 0$ for all $x$ |
**[4 marks]**
---8
\begin{enumerate}[label=(\alph*)]
\item Differentiate $\left( 2 + 3 x ^ { 2 } \right) \mathrm { e } ^ { 2 x }$ with respect to $x$.
\item Hence show that $\left( 2 + 3 x ^ { 2 } \right) \mathrm { e } ^ { 2 x }$ is increasing for all values of $x$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2020 Q8 [7]}}