Moderate -0.3 This is a standard C3 composite and inverse functions question with routine parts: stating range, composing functions (fg(x) = e^(ln x) + 2 = x + 2 simplifies nicely), solving f(2x+3) = 6 using logarithms, finding inverse by swapping and rearranging, and sketching with reflections in y=x. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
6. The functions \(f\) and \(g\) are defined by
$$\begin{aligned}
& \mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 , \quad x \in \mathbb { R } \\
& \mathrm {~g} : x \mapsto \ln x , \quad x > 0
\end{aligned}$$
State the range of f.
Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
Find \(\mathrm { f } ^ { - 1 }\), the inverse function of f , stating its domain.
On the same axes sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.
Range of \(f(x) > 2\). Accept \(y > 2, (2, \infty), f > 2\), as well as 'range is the set of numbers bigger than 2' but don't accept \(x > 2\)
(1)
(1)
(b) \(fg(x) = e^{\ln x + 2} = x + 2\)
M1, A1
For applying the correct order of operations. Look for \(e^{\ln x + 2}\). Note that \(\ln e^x + 2\) is M0. Simplifies \(e^{\ln x + 2}\) to \(x + 2\). Just the answer is acceptable for both marks.
Remember to isw any incorrect working after a correct answer.
(4)
(4)
(d) Let \(y = e^x + 2 \Rightarrow y - 2 = e^x \Rightarrow \ln(y-2) = x\)
M1
Starts with \(y = e^x + 2\) or \(x = e^y + 2\) and attempts to change the subject. All ln work must be correct. The 2 must be dealt with first. Eg \(y = e^x + 2 \Rightarrow \ln y = x + \ln 2 \Rightarrow x = \ln y - \ln 2\) is M0.
\(f^{-1}(x) = \ln(x - 2), \quad x > 2.\)
A1, B1ft
\(f^{-1}(x) = \ln(x-2)\) or \(y = \ln(x-2)\) or \(y = \ln\
(3)
(14 marks)
Notes:
- (a) B1: Range of \(f(x) > 2\). Accept \(y > 2, (2, \infty), f > 2\), as well as 'range is the set of numbers bigger than 2' but don't accept \(x > 2\).
- (b) M1: For applying the correct order of operations. Look for \(e^{\ln x + 2}\). Note that \(\ln e^x + 2\) is M0. A1: Simplifies \(e^{\ln x + 2}\) to \(x + 2\). Just the answer is acceptable for both marks.
- (c) M1: Starts with \(e^{2x+3} + 2 = 6\) and proceeds to \(e^{2x+3} = ...\); A1: \(e^{2x+3} = 4\); M1: Takes \(\ln\)'s both sides, \(2x + 3 = \ln...\) and proceeds to \(x = ...\); A1: \(x = \frac{\ln 4 - 3}{2}\) oe eg \(\ln 2 - \frac{3}{2}\). Remember to isw any incorrect working after a correct answer.
Answer
Marks
Guidance
- (d) M1: Starts with \(y = e^x + 2\) or \(x = e^y + 2\) and attempts to change the subject. All ln work must be correct. The 2 must be dealt with first. Eg \(y = e^x + 2 \Rightarrow \ln y = x + \ln 2 \Rightarrow x = \ln y - \ln 2\) is M0; A1: \(f^{-1}(x) = \ln(x-2)\) or \(y = \ln(x-2)\) or \(y = \ln\
x-2\
\) There must be some form of bracket; B1ft: Either \(x > 2\), or follow through on their answer to part (a), provided that it wasn't \(y \in \mathbb{R}\). Do not accept \(y > 2\) or \(f^{-1}(x) > 2\).
- (e) B1: Shape for \(y = e^x\). The graph should only lie in quadrants 1 and 2. It should start out with a gradient that is approx. 0 above the \(x\) axis in quadrant 2 and increase in gradient as it moves into quadrant 1. You should not see a minimum point on the graph.
- B1: \((0, 3)\) lies on the curve. Accept 3 written on the \(y\) axis as long as the point lies on the curve.
- B1: Shape for \(y = \ln x\). The graph should only lie in quadrants 4 and 1. It should start out with gradient that is approx. infinite to the right of the \(y\) axis in quadrant 4 and decrease in gradient as it moves into quadrant 1. You should not see a maximum point. Also with hold this mark if it intersects \(y = e^x\).
- B1: \((3, 0)\) lies on the curve. Accept 3 written on the \(x\) axis as long as the point lies on the curve.
- Condone lack of labels in this part
- Examples:
- Scores 1,0,1,0: Both shapes are fine, do not be concerned about asymptotes appearing at \(x = 2, y = 2\). Both co-ordinates are incorrect.
- Scores 0,1,1,1: Shape for \(y = e^x\) is incorrect, there is a minimum point on the graph. All other marks can be awarded.
| (a) $f(x) > 2$ | B1 | Range of $f(x) > 2$. Accept $y > 2, (2, \infty), f > 2$, as well as 'range is the set of numbers bigger than 2' but **don't accept** $x > 2$ |
|---|---|---|
| **(1)** | **(1)** | |
| (b) $fg(x) = e^{\ln x + 2} = x + 2$ | M1, A1 | For applying the correct order of operations. Look for $e^{\ln x + 2}$. Note that $\ln e^x + 2$ is M0. Simplifies $e^{\ln x + 2}$ to $x + 2$. Just the answer is acceptable for both marks. |
| **(2)** | **(2)** | |
| (c) $e^{2x+3} + 2 = 6 \Rightarrow e^{2x+3} = ...$ | M1A1 | Starts with $e^{2x+3} + 2 = 6$ and proceeds to $e^{2x+3} = ...$ |
| $e^{2x+3} = 4$ | A1 | |
| Takes $\ln$'s both sides, $2x + 3 = \ln...$ and proceeds to $x = ...$ | M1 | |
| $x = \frac{\ln 4 - 3}{2}$ oe. eg $\ln 2 - \frac{3}{2}$ | A1 | Remember to isw any incorrect working after a correct answer. |
| **(4)** | **(4)** | |
| (d) Let $y = e^x + 2 \Rightarrow y - 2 = e^x \Rightarrow \ln(y-2) = x$ | M1 | Starts with $y = e^x + 2$ or $x = e^y + 2$ and attempts to change the subject. All ln work must be correct. The 2 must be dealt with first. Eg $y = e^x + 2 \Rightarrow \ln y = x + \ln 2 \Rightarrow x = \ln y - \ln 2$ is M0. |
| $f^{-1}(x) = \ln(x - 2), \quad x > 2.$ | A1, B1ft | $f^{-1}(x) = \ln(x-2)$ or $y = \ln(x-2)$ or $y = \ln\|x-2\|$ There must be some form of bracket. Either $x > 2$, or follow through on their answer to part (a), provided that it wasn't $y \in \mathbb{R}$. Do not accept $y > 2$ or $f^{-1}(x) > 2$. |
| **(3)** | **(14 marks)** | |
**Notes:**
- (a) B1: Range of $f(x) > 2$. Accept $y > 2, (2, \infty), f > 2$, as well as 'range is the set of numbers bigger than 2' but **don't accept** $x > 2$.
- (b) M1: For applying the correct order of operations. Look for $e^{\ln x + 2}$. Note that $\ln e^x + 2$ is M0. A1: Simplifies $e^{\ln x + 2}$ to $x + 2$. Just the answer is acceptable for both marks.
- (c) M1: Starts with $e^{2x+3} + 2 = 6$ and proceeds to $e^{2x+3} = ...$; A1: $e^{2x+3} = 4$; M1: Takes $\ln$'s both sides, $2x + 3 = \ln...$ and proceeds to $x = ...$; A1: $x = \frac{\ln 4 - 3}{2}$ oe eg $\ln 2 - \frac{3}{2}$. Remember to isw any incorrect working after a correct answer.
- (d) M1: Starts with $y = e^x + 2$ or $x = e^y + 2$ and attempts to change the subject. All ln work must be correct. The 2 must be dealt with first. Eg $y = e^x + 2 \Rightarrow \ln y = x + \ln 2 \Rightarrow x = \ln y - \ln 2$ is M0; A1: $f^{-1}(x) = \ln(x-2)$ or $y = \ln(x-2)$ or $y = \ln\|x-2\|$ There must be some form of bracket; B1ft: Either $x > 2$, or follow through on their answer to part (a), provided that it wasn't $y \in \mathbb{R}$. Do not accept $y > 2$ or $f^{-1}(x) > 2$.
- (e) B1: Shape for $y = e^x$. The graph should only lie in quadrants 1 and 2. It should start out with a gradient that is approx. 0 above the $x$ axis in quadrant 2 and increase in gradient as it moves into quadrant 1. You should not see a minimum point on the graph.
- B1: $(0, 3)$ lies on the curve. Accept 3 written on the $y$ axis as long as the point lies on the curve.
- B1: Shape for $y = \ln x$. The graph should only lie in quadrants 4 and 1. It should start out with gradient that is approx. infinite to the right of the $y$ axis in quadrant 4 and decrease in gradient as it moves into quadrant 1. You should not see a maximum point. Also with hold this mark if it intersects $y = e^x$.
- B1: $(3, 0)$ lies on the curve. Accept 3 written on the $x$ axis as long as the point lies on the curve.
- **Condone lack of labels in this part**
- **Examples:**
- Scores 1,0,1,0: Both shapes are fine, do not be concerned about asymptotes appearing at $x = 2, y = 2$. Both co-ordinates are incorrect.
- Scores 0,1,1,1: Shape for $y = e^x$ is incorrect, there is a minimum point on the graph. All other marks can be awarded.
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6. The functions $f$ and $g$ are defined by
$$\begin{aligned}
& \mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 , \quad x \in \mathbb { R } \\
& \mathrm {~g} : x \mapsto \ln x , \quad x > 0
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item Find $\mathrm { fg } ( x )$, giving your answer in its simplest form.
\item Find the exact value of $x$ for which $\mathrm { f } ( 2 x + 3 ) = 6$
\item Find $\mathrm { f } ^ { - 1 }$, the inverse function of f , stating its domain.
\item On the same axes sketch the curves with equation $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, giving the coordinates of all the points where the curves cross the axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2012 Q6 [14]}}