| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward application of the chain rule for differentiation. Part (a) requires finding dy/dx, setting it to zero, and solving a simple linear equation. Part (b) requires finding the second derivative and showing it's always positive, which follows directly since exponentials are always positive. The algebra is routine with no conceptual challenges beyond standard A-level techniques. |
| Spec | 1.07f Convexity/concavity: points of inflection1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = (2x+3)e^{x^2+3x}\) | M1 | Attempt to differentiate using the chain rule; obtain derivative of form \(f(x)e^{x^2+3x}\); M0 if attempt to split results in sum not product |
| \((2x+3)e^{x^2+3x}\) | A1 | Obtain correct derivative; brackets must be seen or implied by later work |
| \((2x+3)e^{x^2+3x} = 0\), \(2x+3=0\), \(x = -\frac{3}{2}\) | A1 | Equate correct derivative to 0 and solve to obtain \(x = -\frac{3}{2}\); A0 if any additional solutions for \(x\) |
| \(e^{x^2+3x} > 0\) for all \(x\), or \(e^{x^2+3x} \neq 0\), or \(x^2+3x = \ln 0\) not possible | B1FT | Indicate no solutions from the exponential term; must have some reason; A0 for 'math error' or 'doesn't work' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\ln y = x^2 + 3x\), \(\frac{1}{y}\frac{dy}{dx} = 2x+3\) | M1 | Take ln and attempt implicit differentiation; must deal correctly with \(\ln y\) |
| \(\frac{dy}{dx} = y(2x+3)\) | A1 | Obtain correct derivative |
| \(2x+3=0\), \(x = -\frac{3}{2}\) | A1 | Equate correct derivative to 0 and solve |
| \(e^{x^2+3x} \neq 0\) | B1 | Indicate no solutions from exponential term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = 2e^{x^2+3x} + (2x+3)^2 e^{x^2+3x}\) | M1 | Attempt to differentiate again using the product rule correctly; obtain derivative of form \((ax^2+bx+c)e^{x^2+3x}\) |
| \(\frac{d^2y}{dx^2} = \left(2+(2x+3)^2\right)e^{x^2+3x}\) | A1 | Obtain correct derivative |
| convex means \(\frac{d^2y}{dx^2} > 0\) | B1 | State or clearly imply correct condition at any point in proof; must be general statement |
| \((2x+3)^2 \geq 0\) hence \(2+(2x+3)^2 > 0\) | M1 | Explain why correct quadratic is always positive; could use completed square, expanded quadratic \(4x^2+12x+11\), or show no real roots (must also state positive quadratic) |
| \(e^{g(x)} > 0\) for all \(x\); quadratic \(> 0\) for all \(x\) hence curve is always convex | A1 | Full and convincing proof to show curve is convex for all \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = y(2x+3)\), \(\frac{d^2y}{dx^2} = 2y + (2x+3)\frac{dy}{dx}\) | M1 | Attempt second derivative using implicit differentiation and product rule |
| Correct derivative obtained | A1 | Will need to use \(\frac{dy}{dx} = y(2x+3)\) to make further progress; then B1 M1 A1 as above |
## Question 6(a):
**Differentiating $y = e^{x^2+3x}$**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = (2x+3)e^{x^2+3x}$ | M1 | Attempt to differentiate using the chain rule; obtain derivative of form $f(x)e^{x^2+3x}$; M0 if attempt to split results in sum not product |
| $(2x+3)e^{x^2+3x}$ | A1 | Obtain correct derivative; brackets must be seen or implied by later work |
| $(2x+3)e^{x^2+3x} = 0$, $2x+3=0$, $x = -\frac{3}{2}$ | A1 | Equate correct derivative to 0 and solve to obtain $x = -\frac{3}{2}$; A0 if any additional solutions for $x$ |
| $e^{x^2+3x} > 0$ for all $x$, or $e^{x^2+3x} \neq 0$, or $x^2+3x = \ln 0$ not possible | B1FT | Indicate no solutions from the exponential term; must have some reason; A0 for 'math error' or 'doesn't work' |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\ln y = x^2 + 3x$, $\frac{1}{y}\frac{dy}{dx} = 2x+3$ | M1 | Take ln and attempt implicit differentiation; must deal correctly with $\ln y$ |
| $\frac{dy}{dx} = y(2x+3)$ | A1 | Obtain correct derivative |
| $2x+3=0$, $x = -\frac{3}{2}$ | A1 | Equate correct derivative to 0 and solve |
| $e^{x^2+3x} \neq 0$ | B1 | Indicate no solutions from exponential term |
---
## Question 6(b):
**Showing curve is always convex**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2y}{dx^2} = 2e^{x^2+3x} + (2x+3)^2 e^{x^2+3x}$ | M1 | Attempt to differentiate again using the product rule correctly; obtain derivative of form $(ax^2+bx+c)e^{x^2+3x}$ |
| $\frac{d^2y}{dx^2} = \left(2+(2x+3)^2\right)e^{x^2+3x}$ | A1 | Obtain correct derivative |
| convex means $\frac{d^2y}{dx^2} > 0$ | B1 | State or clearly imply correct condition at any point in proof; must be general statement |
| $(2x+3)^2 \geq 0$ hence $2+(2x+3)^2 > 0$ | M1 | Explain why correct quadratic is always positive; could use completed square, expanded quadratic $4x^2+12x+11$, or show no real roots (must also state positive quadratic) |
| $e^{g(x)} > 0$ for all $x$; quadratic $> 0$ for all $x$ hence curve is always convex | A1 | Full and convincing proof to show curve is convex for all $x$ |
**Alternative method for first 2 marks:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = y(2x+3)$, $\frac{d^2y}{dx^2} = 2y + (2x+3)\frac{dy}{dx}$ | M1 | Attempt second derivative using implicit differentiation and product rule |
| Correct derivative obtained | A1 | Will need to use $\frac{dy}{dx} = y(2x+3)$ to make further progress; then B1 M1 A1 as above |6 A curve has equation $y = \mathrm { e } ^ { x ^ { 2 } + 3 x }$.
\begin{enumerate}[label=(\alph*)]
\item Determine the $x$-coordinates of any stationary points on the curve.
\item Show that the curve is convex for all values of $x$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q6 [9]}}