Edexcel C3 2006 January — Question 8 10 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2006
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeVerify composite identity
DifficultyModerate -0.8 This is a straightforward multi-part question testing basic composite function mechanics, routine differentiation, and simple algebraic manipulation. Part (a) requires substituting one function into another with basic exponential/logarithm properties (e^ln2 = 2). Parts (b)-(d) are standard bookwork: sketching an exponential, stating its range, and solving a simple equation involving the derivative. All steps are routine C3 techniques with no problem-solving insight required.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)

8. The functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } : x \rightarrow 2 x + \ln 2 , & x \in \mathbb { R } , \\ \mathrm {~g} : x \rightarrow \mathrm { e } ^ { 2 x } , & x \in \mathbb { R } . \end{array}$$
  1. Prove that the composite function gf is $$\operatorname { gf } : x \rightarrow 4 \mathrm { e } ^ { 4 x } , \quad x \in \mathbb { R }$$
  2. In the space provided on page 19, sketch the curve with equation \(y = \operatorname { gf } ( x )\), and show the coordinates of the point where the curve cuts the \(y\)-axis.
  3. Write down the range of gf.
  4. Find the value of \(x\) for which \(\frac { \mathrm { d } } { \mathrm { d } x } [ \operatorname { gf } ( x ) ] = 3\), giving your answer to 3 significant figures.

Question 8:
(a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(gf(x) = e^{2(2x+\ln 2)}\)M1
\(= e^{4x}e^{2\ln 2}\)M1
\(= e^{4x}e^{\ln 4}\)M1
\(= 4e^{4x}\)A1 Give mark at this point, cso
(Hence \(gf: x \mapsto 4e^{4x}, \quad x \in \mathbb{R}\)) (4)
(b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct exponential shape passing through \((0, 4)\)B1 Shape and point (1)
(c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Range is \(\mathbb{R}_+\)B1 Accept \(gf(x) > 0\), \(y > 0\) (1)
(d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d}{dx}[gf(x)] = 16e^{4x}\)M1 A1
\(e^{4x} = \frac{3}{16}\)M1
\(4x = \ln\frac{3}{16}\)
\(x \approx -0.418\)A1 (4)
[10]
## Question 8:

**(a)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $gf(x) = e^{2(2x+\ln 2)}$ | M1 | |
| $= e^{4x}e^{2\ln 2}$ | M1 | |
| $= e^{4x}e^{\ln 4}$ | M1 | |
| $= 4e^{4x}$ | A1 | Give mark at this point, cso |
| (Hence $gf: x \mapsto 4e^{4x}, \quad x \in \mathbb{R}$) | | **(4)** |

**(b)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct exponential shape passing through $(0, 4)$ | B1 | Shape and point **(1)** |

**(c)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Range is $\mathbb{R}_+$ | B1 | Accept $gf(x) > 0$, $y > 0$ **(1)** |

**(d)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}[gf(x)] = 16e^{4x}$ | M1 A1 | |
| $e^{4x} = \frac{3}{16}$ | M1 | |
| $4x = \ln\frac{3}{16}$ | | |
| $x \approx -0.418$ | A1 | **(4)** |
| | | **[10]** |
8. The functions $f$ and $g$ are defined by

$$\begin{array} { l l } 
\mathrm { f } : x \rightarrow 2 x + \ln 2 , & x \in \mathbb { R } , \\
\mathrm {~g} : x \rightarrow \mathrm { e } ^ { 2 x } , & x \in \mathbb { R } .
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Prove that the composite function gf is

$$\operatorname { gf } : x \rightarrow 4 \mathrm { e } ^ { 4 x } , \quad x \in \mathbb { R }$$
\item In the space provided on page 19, sketch the curve with equation $y = \operatorname { gf } ( x )$, and show the coordinates of the point where the curve cuts the $y$-axis.
\item Write down the range of gf.
\item Find the value of $x$ for which $\frac { \mathrm { d } } { \mathrm { d } x } [ \operatorname { gf } ( x ) ] = 3$, giving your answer to 3 significant figures.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3 2006 Q8 [10]}}