Edexcel C3 2015 June — Question 2 10 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2015
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSketch y=|f(x)| for non-linear f(x)
DifficultyModerate -0.8 This is a straightforward modulus function question requiring standard techniques: sketching an exponential function, reflecting negative parts for the modulus, finding intercepts (ln(5/2)), and solving |f(x)|=2 by cases. All steps are routine C3 material with no novel problem-solving required, making it easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02n Sketch curves: simple equations including polynomials1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b

2. Given that $$\mathrm { f } ( x ) = 2 \mathrm { e } ^ { x } - 5 , \quad x \in \mathbb { R }$$
  1. sketch, on separate diagrams, the curve with equation
    1. \(y = \mathrm { f } ( x )\)
    2. \(y = | \mathrm { f } ( x ) |\) On each diagram, show the coordinates of each point at which the curve meets or cuts the axes. On each diagram state the equation of the asymptote.
  2. Deduce the set of values of \(x\) for which \(\mathrm { f } ( x ) = | \mathrm { f } ( x ) |\)
  3. Find the exact solutions of the equation \(| \mathrm { f } ( x ) | = 2\)

Question 2:
Part (ai)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct shape of \(y = 2e^x - 5\)B1 Exponential curve shape
\(\left(\ln\left(\frac{5}{2}\right), 0\right)\) and \((0,-3)\) markedB1 Both intercepts required
Asymptote \(y=-5\) shownB1
Part (aii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct shape of \(y =2e^x-5 \) including cusp
\(\left(\ln\left(\frac{5}{2}\right), 0\right)\) and \((0,3)\) markedB1ft Both intercepts, follow through
Asymptote \(y=5\) shownB1ft Follow through
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x \ldots \ln\left(\frac{5}{2}\right)\)B1ft Follow through
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(2e^x - 5 = -2 \Rightarrow x = \ln\left(\frac{3}{2}\right)\)M1A1
\(x = \ln\left(\frac{7}{2}\right)\)B1
## Question 2:

### Part (ai)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape of $y = 2e^x - 5$ | B1 | Exponential curve shape |
| $\left(\ln\left(\frac{5}{2}\right), 0\right)$ and $(0,-3)$ marked | B1 | Both intercepts required |
| Asymptote $y=-5$ shown | B1 | |

### Part (aii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape of $y = |2e^x-5|$ including cusp | B1ft | Shape including cusp, follow through |
| $\left(\ln\left(\frac{5}{2}\right), 0\right)$ and $(0,3)$ marked | B1ft | Both intercepts, follow through |
| Asymptote $y=5$ shown | B1ft | Follow through |

### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x \ldots \ln\left(\frac{5}{2}\right)$ | B1ft | Follow through |

### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2e^x - 5 = -2 \Rightarrow x = \ln\left(\frac{3}{2}\right)$ | M1A1 | |
| $x = \ln\left(\frac{7}{2}\right)$ | B1 | |
2. Given that

$$\mathrm { f } ( x ) = 2 \mathrm { e } ^ { x } - 5 , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item sketch, on separate diagrams, the curve with equation
\begin{enumerate}[label=(\roman*)]
\item $y = \mathrm { f } ( x )$
\item $y = | \mathrm { f } ( x ) |$

On each diagram, show the coordinates of each point at which the curve meets or cuts the axes.

On each diagram state the equation of the asymptote.
\end{enumerate}\item Deduce the set of values of $x$ for which $\mathrm { f } ( x ) = | \mathrm { f } ( x ) |$
\item Find the exact solutions of the equation $| \mathrm { f } ( x ) | = 2$
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3 2015 Q2 [10]}}