| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing algebraic simplification and quotient rule application. Part (a) is routine algebraic manipulation, part (b) is a standard quotient rule differentiation with exponentials (showing a given result), and part (c) solves a simple equation. All techniques are standard C3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.06a Exponential function: a^x and e^x graphs and properties1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{(x-2)(x+4) - 2(x-2) + x - 8}{(x-2)(x+4)}\) | M1 | An attempt to combine to one fraction |
| A1 | Correct result of combining all three fractions | |
| \(= \frac{x^2 + 2x - 8 - 2x + 4 + x - 8}{(x-2)(x+4)}\) | ||
| \(= \frac{x^2 + x - 12}{[(x+4)(x-2)]}\) | A1 | Simplifies to give correct numerator. Ignore omission of denominator |
| \(= \frac{(x+4)(x-3)}{[(x+4)(x-2)]}\) | dM1 | An attempt to factorise the numerator |
| \(= \frac{x-3}{x-2}\) | A1 cso AG | Correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g'(x) = \frac{e^x(e^x-2) - e^x(e^x-3)}{(e^x-2)^2}\) | M1 | Applying \(\frac{vu'-uv'}{v^2}\) |
| A1 | Correct differentiation | |
| \(= \frac{e^{2x} - 2e^x - e^{2x} + 3e^x}{(e^x-2)^2} = \frac{e^x}{(e^x-2)^2}\) | A1 AG cso | Correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^x = (e^x - 2)^2\) | M1 | Puts their differentiated numerator equal to their denominator |
| \(e^{2x} - 5e^x + 4 = 0\) | A1 | \(e^{2x} - 5e^x + 4\) |
| \((e^x - 4)(e^x - 1) = 0\) | M1 | Attempt to factorise or solve quadratic in \(e^x\) |
| \(x = \ln 4\) or \(x = 0\) | A1 | Both \(x = 0,\ \ln 4\) |
# Question 7:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{(x-2)(x+4) - 2(x-2) + x - 8}{(x-2)(x+4)}$ | M1 | An attempt to combine to one fraction |
| | A1 | Correct result of combining all three fractions |
| $= \frac{x^2 + 2x - 8 - 2x + 4 + x - 8}{(x-2)(x+4)}$ | | |
| $= \frac{x^2 + x - 12}{[(x+4)(x-2)]}$ | A1 | Simplifies to give correct numerator. Ignore omission of denominator |
| $= \frac{(x+4)(x-3)}{[(x+4)(x-2)]}$ | dM1 | An attempt to factorise the numerator |
| $= \frac{x-3}{x-2}$ | A1 cso AG | Correct result |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g'(x) = \frac{e^x(e^x-2) - e^x(e^x-3)}{(e^x-2)^2}$ | M1 | Applying $\frac{vu'-uv'}{v^2}$ |
| | A1 | Correct differentiation |
| $= \frac{e^{2x} - 2e^x - e^{2x} + 3e^x}{(e^x-2)^2} = \frac{e^x}{(e^x-2)^2}$ | A1 AG cso | Correct result |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^x = (e^x - 2)^2$ | M1 | Puts their differentiated numerator equal to their denominator |
| $e^{2x} - 5e^x + 4 = 0$ | A1 | $e^{2x} - 5e^x + 4$ |
| $(e^x - 4)(e^x - 1) = 0$ | M1 | Attempt to factorise or solve quadratic in $e^x$ |
| $x = \ln 4$ or $x = 0$ | A1 | Both $x = 0,\ \ln 4$ |7. The function f is defined by
$$\mathrm { f } ( x ) = 1 - \frac { 2 } { ( x + 4 ) } + \frac { x - 8 } { ( x - 2 ) ( x + 4 ) } , \quad x \in \mathbb { R } , x \neq - 4 , x \neq 2$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = \frac { x - 3 } { x - 2 }$
The function g is defined by
$$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } - 3 } { \mathrm { e } ^ { x } - 2 } , \quad x \in \mathbb { R } , x \neq \ln 2$$
\item Differentiate $\mathrm { g } ( x )$ to show that $\mathrm { g } ^ { \prime } ( x ) = \frac { \mathrm { e } ^ { x } } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } }$
\item Find the exact values of $x$ for which $\mathrm { g } ^ { \prime } ( x ) = 1$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2009 Q7 [12]}}