Edexcel C4 2013 June — Question 2 7 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeShow specific gradient value
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring standard techniques (product rule, chain rule for exponential) followed by substitution and algebraic manipulation into a specific logarithm form. While it has multiple steps, each is routine for C4 level, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.07s Parametric and implicit differentiation
2. The curve \(C\) has equation $$3 ^ { x - 1 } + x y - y ^ { 2 } + 5 = 0$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(( 1,3 )\) on the curve \(C\) can be written in the form \(\frac { 1 } { \lambda } \ln \left( \mu \mathrm { e } ^ { 3 } \right)\), where \(\lambda\) and \(\mu\) are integers to be found.
Question 2:
Implicit Differentiation of \(3^{x-1} + xy - y^2 + 5 = 0\)
AnswerMarks Guidance
Working/AnswerMark Guidance
\(3^{x-1} \rightarrow 3^{x-1}\ln 3\)B1 oe Correct differentiation of \(3^{x-1}\)
Differentiates implicitly to include either \(\pm\lambda x\frac{dy}{dx}\) or \(\pm ky\frac{dy}{dx}\)M1*
\(3^{x-1}\ln 3 + \left(y + x\frac{dy}{dx}\right) - 2y\frac{dy}{dx} = 0\)
\(xy \rightarrow +y + x\frac{dy}{dx}\)B1
\(\ldots + y + x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0\)A1
Substitutes \(x=1\), \(y=3\) into differentiated equation: \(3^{(1-1)}\ln 3 + 3 + (1)\frac{dy}{dx} - 2(3)\frac{dy}{dx} = 0\)dM1* Dependent on M1*
\(\ln 3 + 3 + \frac{dy}{dx} - 6\frac{dy}{dx} = 0 \Rightarrow 3 + \ln 3 = 5\frac{dy}{dx}\)
\(\frac{dy}{dx} = \frac{3+\ln 3}{5}\)dM1* Rearranges to make \(\frac{dy}{dx}\) subject
\(\frac{dy}{dx} = \frac{1}{5}(\ln e^3 + \ln 3) = \frac{1}{5}\ln(3e^3)\)A1 cso Uses \(3 = \ln e^3\) to achieve \(\frac{1}{5}\ln(3e^3)\)
Total: [7]
# Question 2:

## Implicit Differentiation of $3^{x-1} + xy - y^2 + 5 = 0$

| Working/Answer | Mark | Guidance |
|---|---|---|
| $3^{x-1} \rightarrow 3^{x-1}\ln 3$ | B1 oe | Correct differentiation of $3^{x-1}$ |
| Differentiates implicitly to include either $\pm\lambda x\frac{dy}{dx}$ or $\pm ky\frac{dy}{dx}$ | M1* | |
| $3^{x-1}\ln 3 + \left(y + x\frac{dy}{dx}\right) - 2y\frac{dy}{dx} = 0$ | | |
| $xy \rightarrow +y + x\frac{dy}{dx}$ | B1 | |
| $\ldots + y + x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$ | A1 | |
| Substitutes $x=1$, $y=3$ into differentiated equation: $3^{(1-1)}\ln 3 + 3 + (1)\frac{dy}{dx} - 2(3)\frac{dy}{dx} = 0$ | dM1* | Dependent on M1* |
| $\ln 3 + 3 + \frac{dy}{dx} - 6\frac{dy}{dx} = 0 \Rightarrow 3 + \ln 3 = 5\frac{dy}{dx}$ | | |
| $\frac{dy}{dx} = \frac{3+\ln 3}{5}$ | dM1* | Rearranges to make $\frac{dy}{dx}$ subject |
| $\frac{dy}{dx} = \frac{1}{5}(\ln e^3 + \ln 3) = \frac{1}{5}\ln(3e^3)$ | A1 cso | Uses $3 = \ln e^3$ to achieve $\frac{1}{5}\ln(3e^3)$ |

**Total: [7]**

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2. The curve $C$ has equation

$$3 ^ { x - 1 } + x y - y ^ { 2 } + 5 = 0$$

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point $( 1,3 )$ on the curve $C$ can be written in the form $\frac { 1 } { \lambda } \ln \left( \mu \mathrm { e } ^ { 3 } \right)$, where $\lambda$ and $\mu$ are integers to be found.\\

\hfill \mbox{\textit{Edexcel C4 2013 Q2 [7]}}
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