OCR MEI C3 — Question 6 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeShow dy/dx equals given expression
DifficultyModerate -0.3 This is a straightforward implicit differentiation question requiring application of the chain rule to (x+y)² and basic algebraic rearrangement to reach the given form. The verification of the stationary point is then trivial substitution. Slightly easier than average as it's a standard technique with clear steps and the target expression is provided.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation
6 A curve is defined by the equation \(( x + y ) ^ { 2 } = 4 x\). The point \(( 1,1 )\) lies on this curve.
By differentiating implicitly, show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x + y } - 1\).
Hence verify that the curve has a stationary point at \(( 1,1 )\).
Question 6:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((x+y)^2 = 4x \Rightarrow 2(x+y)\!\left(1+\frac{dy}{dx}\right) = 4\)M1 Implicit differentiation of LHS; allow one error but must include \(dy/dx\); ignore superfluous \(dy/dx = \ldots\) for M1 and both A1s if not pursued; condone missing brackets
\(1 + \frac{dy}{dx} = \frac{4}{2(x+y)} = \frac{2}{x+y}\)A1 Correct expression \(= 4\)
\(\frac{dy}{dx} = \frac{2}{x+y} - 1\)A1 www (AG); A0 if missing brackets in earlier working
*or* \(x^2 + 2xy + y^2 = 4x\)
\(2x + 2x\frac{dy}{dx} + 2y + 2y\frac{dy}{dx} = 4\)M1dep Implicit differentiation of LHS; dep correct expansion; correct expression \(= 4\) (oe after re-arrangement); allow 1 error provided \(2x\,dy/dx\) and \(2y\,dy/dx\) are correct, but must expand \((x+y)^2\) correctly for M1
\(\frac{dy}{dx}(2x+2y) = 4-2x-2y\)
\(\frac{dy}{dx} = \frac{4}{2x+2y} - 1 = \frac{2}{x+y} - 1\)A1 www (AG); A0 if missing brackets in earlier working
When \(x=1,\ y=1\): \(\frac{dy}{dx} = \frac{2}{1+1} - 1 = 0\)B1 (AG) oe (e.g. from \(x+y=2\)); or e.g. \(2/(x+y)-1=0 \Rightarrow x+y=2,\ \Rightarrow 4=4x,\ \Rightarrow x=1,\ y=1\) (oe)
[4] 
## Question 6:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+y)^2 = 4x \Rightarrow 2(x+y)\!\left(1+\frac{dy}{dx}\right) = 4$ | M1 | Implicit differentiation of LHS; allow one error but must include $dy/dx$; ignore superfluous $dy/dx = \ldots$ for M1 and both A1s if not pursued; condone missing brackets |
| $1 + \frac{dy}{dx} = \frac{4}{2(x+y)} = \frac{2}{x+y}$ | A1 | Correct expression $= 4$ |
| $\frac{dy}{dx} = \frac{2}{x+y} - 1$ | A1 | www **(AG)**; A0 if missing brackets in earlier working |
| *or* $x^2 + 2xy + y^2 = 4x$ | | |
| $2x + 2x\frac{dy}{dx} + 2y + 2y\frac{dy}{dx} = 4$ | M1dep | Implicit differentiation of LHS; dep correct expansion; correct expression $= 4$ (oe after re-arrangement); allow 1 error provided $2x\,dy/dx$ and $2y\,dy/dx$ are correct, but must expand $(x+y)^2$ correctly for M1 |
| $\frac{dy}{dx}(2x+2y) = 4-2x-2y$ | | |
| $\frac{dy}{dx} = \frac{4}{2x+2y} - 1 = \frac{2}{x+y} - 1$ | A1 | www **(AG)**; A0 if missing brackets in earlier working |
| When $x=1,\ y=1$: $\frac{dy}{dx} = \frac{2}{1+1} - 1 = 0$ | B1 | **(AG)** oe (e.g. from $x+y=2$); or e.g. $2/(x+y)-1=0 \Rightarrow x+y=2,\ \Rightarrow 4=4x,\ \Rightarrow x=1,\ y=1$ (oe) |
| **[4]** | | |

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6 A curve is defined by the equation $( x + y ) ^ { 2 } = 4 x$. The point $( 1,1 )$ lies on this curve.\\
By differentiating implicitly, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x + y } - 1$.\\
Hence verify that the curve has a stationary point at $( 1,1 )$.

\hfill \mbox{\textit{OCR MEI C3  Q6 [4]}}
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