OCR MEI C3 2008 June — Question 6 8 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeExponential model with shifted asymptote
DifficultyModerate -0.8 This is a straightforward exponential decay question requiring only direct substitution (t=0), recognition of horizontal asymptote (as t→∞), solving a simple exponential equation by taking logarithms, and sketching a standard shifted exponential curve. All steps are routine applications of C3 techniques with no problem-solving or novel insight required.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b

6 In a chemical reaction, the mass \(m\) grams of a chemical after \(t\) minutes is modelled by the equation $$m = 20 + 30 \mathrm { e } ^ { - 0.1 t }$$
  1. Find the initial mass of the chemical. What is the mass of chemical in the long term?
  2. Find the time when the mass is 30 grams.
  3. Sketch the graph of \(m\) against \(t\).

Question 6:
Part (i):
AnswerMarks Guidance
AnswerMark Guidance
Initial mass \(= 20 + 30e^0 = 50\) gramsM1A1
Long term mass \(= 20\) gramsB1
[3]
Part (ii):
AnswerMarks Guidance
AnswerMark Guidance
\(30 = 20 + 30e^{-0.1t}\)M1
\(e^{-0.1t} = \frac{1}{3}\)M1 anti-logging correctly
\(-0.1t = \ln(\frac{1}{3}) = -1.0986...\)A1 11, 11.0, 10.99, 10.986 (not more than 3 d.p)
\(t = 11.0\) mins[3]
Part (iii):
AnswerMarks Guidance
AnswerMark Guidance
Correct shape through \((0, 50)\)B1 correct shape through \((0,50)\) – ignore negative values of \(t\)
\(\rightarrow 20\) as \(t \rightarrow \infty\)B1
[2]
# Question 6:

## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Initial mass $= 20 + 30e^0 = 50$ grams | M1A1 | |
| Long term mass $= 20$ grams | B1 | |
| | [3] | |

## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $30 = 20 + 30e^{-0.1t}$ | M1 | |
| $e^{-0.1t} = \frac{1}{3}$ | M1 | anti-logging correctly |
| $-0.1t = \ln(\frac{1}{3}) = -1.0986...$ | A1 | 11, 11.0, 10.99, 10.986 (not more than 3 d.p) |
| $t = 11.0$ mins | [3] | |

## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct shape through $(0, 50)$ | B1 | correct shape through $(0,50)$ – ignore negative values of $t$ |
| $\rightarrow 20$ as $t \rightarrow \infty$ | B1 | |
| | [2] | |

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6 In a chemical reaction, the mass $m$ grams of a chemical after $t$ minutes is modelled by the equation

$$m = 20 + 30 \mathrm { e } ^ { - 0.1 t }$$

(i) Find the initial mass of the chemical.

What is the mass of chemical in the long term?\\
(ii) Find the time when the mass is 30 grams.\\
(iii) Sketch the graph of $m$ against $t$.

\hfill \mbox{\textit{OCR MEI C3 2008 Q6 [8]}}