| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Moderate -0.3 This is a straightforward application of the product rule to find a gradient, followed by routine tangent equation work. While it requires careful algebraic manipulation and the question specifies 'detailed reasoning', the techniques are standard AS-level calculus with no novel problem-solving required. The multi-step nature and potential for algebraic errors prevent it from being significantly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x^{-\frac{3}{2}} - 2x^{\frac{3}{2}}\) | M1 | Uses negative or fractional powers and laws of indices to rewrite the equation. May be implied by one correct term |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3}{2}x^{-\frac{5}{2}} - 2 \times \frac{3}{2}x^{\frac{1}{2}}\) | M1 | Differentiates their non-integer power(s) of \(x\). Need not be simplified |
| \(x = \frac{1}{4} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3}{2}\times\left(\frac{1}{4}\right)^{-\frac{5}{2}} - 2\times\frac{3}{2}\times\left(\frac{1}{4}\right)^{\frac{1}{2}}\) | M1 | Substitution seen in their expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \left[-\frac{3}{2}\times 32 - 3\times\frac{1}{2}\right] = -\frac{99}{2}\) | A1 [4] | AG Do not allow without clear expression seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tangent is \(y - \frac{31}{4} = -\frac{99}{2}\left(x - \frac{1}{4}\right)\) | M1 | Uses given \(x\)- and \(y\)-coordinates and gradient in a formula for straight line. Allow if their value for gradient used. Do not allow for gradient 2/99 used |
| \(396x + 8y - 161 = 0\) | A1 [2] | Must be in the form \(ax + by + c = 0\) where \(a, b, c\) are integers |
| Alternative: \(y = -\frac{99}{2}x + c\), so \(\frac{31}{4} = -\frac{99}{2}\times\frac{1}{4} + c \Rightarrow c = \frac{161}{8}\) | M1 | Uses \(y = -\frac{99}{2}x + c\) and attempts to evaluate \(c\). Allow if their gradient used. Do not allow gradient 2/99 |
| \(396x + 8y - 161 = 0\) | A1 | Must be in the form \(ax + by + c = 0\) where \(a, b, c\) are integers |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^{-\frac{3}{2}} - 2x^{\frac{3}{2}}$ | M1 | Uses negative or fractional powers and laws of indices to rewrite the equation. May be implied by one correct term |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3}{2}x^{-\frac{5}{2}} - 2 \times \frac{3}{2}x^{\frac{1}{2}}$ | M1 | Differentiates their non-integer power(s) of $x$. Need not be simplified |
| $x = \frac{1}{4} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3}{2}\times\left(\frac{1}{4}\right)^{-\frac{5}{2}} - 2\times\frac{3}{2}\times\left(\frac{1}{4}\right)^{\frac{1}{2}}$ | M1 | Substitution seen in their expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \left[-\frac{3}{2}\times 32 - 3\times\frac{1}{2}\right] = -\frac{99}{2}$ | A1 [4] | AG Do not allow without clear expression seen |
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## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tangent is $y - \frac{31}{4} = -\frac{99}{2}\left(x - \frac{1}{4}\right)$ | M1 | Uses given $x$- and $y$-coordinates and gradient in a formula for straight line. Allow if their value for gradient used. Do not allow for gradient 2/99 used |
| $396x + 8y - 161 = 0$ | A1 [2] | Must be in the form $ax + by + c = 0$ where $a, b, c$ are integers |
| **Alternative:** $y = -\frac{99}{2}x + c$, so $\frac{31}{4} = -\frac{99}{2}\times\frac{1}{4} + c \Rightarrow c = \frac{161}{8}$ | M1 | Uses $y = -\frac{99}{2}x + c$ and attempts to evaluate $c$. Allow if their gradient used. Do not allow gradient 2/99 |
| $396x + 8y - 161 = 0$ | A1 | Must be in the form $ax + by + c = 0$ where $a, b, c$ are integers |
---5 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the curve $\mathrm { y } = \sqrt { \mathrm { x } } \left( \frac { 1 } { \mathrm { x } ^ { 2 } } - 2 \mathrm { x } \right)$ at the point $\left( \frac { 1 } { 4 } , \frac { 31 } { 4 } \right)$ is $- \frac { 99 } { 2 }$.
\item Find the equation of the tangent to the curve at $\left( \frac { 1 } { 4 } , \frac { 31 } { 4 } \right)$ giving your answer in the form $\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q5 [6]}}