OCR MEI C3 2010 June — Question 5 7 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyStandard +0.3 This is a straightforward implicit differentiation question with two standard parts: (i) differentiate implicitly using product rule and chain rule, then rearrange to show the given result, and (ii) set dy/dx = 0 to find stationary points. The algebra is routine and the question provides the derivative formula to verify, making it slightly easier than average but still requiring proper technique.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation
5 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).
Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y^3 = xy - x^2\)
\(3y^2 \, dy/dx = x \, dy/dx + y - 2x\)B1 \(3y^2 dy/dx\)
B1\(x \, dy/dx + y - 2x\); must show '\(x \, dy/dx + y\)' on one side
\((3y^2 - x)\, dy/dx = y - 2x\)M1 collecting terms in \(dy/dx\) only
\(dy/dx = \frac{y-2x}{3y^2-x}\) *E1
TP when \(dy/dx = 0 \Rightarrow y - 2x = 0\)
\(y = 2x\)M1 or \(x = 1/8\) and \(dy/dx = 0 \Rightarrow y = \frac{1}{4}\)
\((2x)^3 = x \cdot 2x - x^2\), so \(8x^3 = x^2\)M1 or \((1/4)^3 = (1/8)(1/4)-(1/8)^2\); or \(x=1/8 \Rightarrow y^3 = (1/8)y - 1/64\) M1
\(x = \frac{1}{8}\) *(or 0)E1 [7] verifying \(y=\frac{1}{4}\) is solution (must show evidence) M1; *just stating \(y=\frac{1}{4}\) is M1 M0 E0 
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^3 = xy - x^2$ | | |
| $3y^2 \, dy/dx = x \, dy/dx + y - 2x$ | B1 | $3y^2 dy/dx$ |
| | B1 | $x \, dy/dx + y - 2x$; must show '$x \, dy/dx + y$' on one side |
| $(3y^2 - x)\, dy/dx = y - 2x$ | M1 | collecting terms in $dy/dx$ only |
| $dy/dx = \frac{y-2x}{3y^2-x}$ * | E1 | |
| TP when $dy/dx = 0 \Rightarrow y - 2x = 0$ | | |
| $y = 2x$ | M1 | or $x = 1/8$ and $dy/dx = 0 \Rightarrow y = \frac{1}{4}$ |
| $(2x)^3 = x \cdot 2x - x^2$, so $8x^3 = x^2$ | M1 | or $(1/4)^3 = (1/8)(1/4)-(1/8)^2$; or $x=1/8 \Rightarrow y^3 = (1/8)y - 1/64$ M1 |
| $x = \frac{1}{8}$ *(or 0) | E1 [7] | verifying $y=\frac{1}{4}$ is solution (must show evidence) M1; *just stating $y=\frac{1}{4}$ is M1 M0 E0 |

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5 Given that $y ^ { 3 } = x y - x ^ { 2 }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }$.\\
Hence show that the curve $y ^ { 3 } = x y - x ^ { 2 }$ has a stationary point when $x = \frac { 1 } { 8 }$.

\hfill \mbox{\textit{OCR MEI C3 2010 Q5 [7]}}
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