IBDP Maths SL 1.7 Laws of exponents with rational exponents AA HL Paper 2- Exam Style Questions- New Syllabus

Question

All living plants contain an isotope of carbon called carbon-14. When a plant dies, the isotope decays so that the amount of carbon-14 present in the remains of the plant decreases. The time since the death of a plant can be determined by measuring the amount of carbon-14 still present in the remains.

The amount, $ A $, of carbon-14 present in a plant $ t $ years after its death can be modelled by $ A = A_0 e^{-kt} $, where $ t \geq 0 $ and $ A_0, k $ are positive constants.

At the time of death, a plant is defined to have 100 units of carbon-14.

(a) Show that $ A_0 = 100 $. [1 mark]

The time taken for half the original amount of carbon-14 to decay is known to be 5730 years.

(b) Show that $ k = \frac{\ln 2}{5730} $. [3 marks]

(c) Find, correct to the nearest 10 years, the time taken after the plant’s death for 25% of the carbon-14 to decay. [3 marks]

▶️ Answer/Explanation

Markscheme

(a) At $ t = 0 $, $ A = 100 $. Substitute into $ A = A_0 e^{-kt} $: $ 100 = A_0 e^{0} = A_0 $. A1

Thus, $ A_0 = 100 $. AG

[1 mark]

(b) Half-life is 5730 years, so when $ t = 5730 $, $ A = \frac{100}{2} = 50 $. Substitute: $ 50 = 100 e^{-5730k} $. M1

$ e^{-5730k} = \frac{1}{2} $. A1

EITHER

Take natural log: $ -5730k = \ln \frac{1}{2} = -\ln 2 $. M1

$ k = \frac{\ln 2}{5730} $. A1

$ e^{5730k} = 2 $, so $ 5730k = \ln 2 $. M1

$ k = \frac{\ln 2}{5730} $. A1

[3 marks]

(c) If 25% of carbon-14 has decayed, 75% remains, so $ A = 75 $. Substitute: $ 75 = 100 e^{-\frac{\ln 2}{5730} t} $. M1

$ e^{-\frac{\ln 2}{5730} t} = 0.75 $. A1

Take natural log: $ -\frac{\ln 2}{5730} t = \ln 0.75 $. M1

$ t = -\frac{\ln 0.75}{\ln 2 / 5730} \approx 2388.164 $. A1

To the nearest 10 years: $ t \approx 2390 $. A1

[3 marks]

Total [7 marks]

Question

Let $ f(x) = \log_3 \frac{x}{2} + \log_3 16 – \log_3 4 $, for $ x > 0 $.

(a) Show that $ f(x) = \log_3 2x $. [2 marks]

(b) Find the value of $ f(0.5) $ and of $ f(4.5) $. [3 marks]

(c)

(i) The function $ f $ can also be written in the form $ f(x) = \frac{\ln ax}{\ln b} $. Write down the value of $ a $ and of $ b $. [2 marks]

(ii) On graph paper, sketch the graph of $ f $, for $ -5 \leq x \leq 5 $, $ -5 \leq y \leq 5 $, using a scale of 1 cm to 1 unit on each axis. [3 marks]

(iii) Write down the equation of the asymptote. [1 mark]

(d) Write down the value of $ f^{-1}(0) $. [1 mark]

(e) The point A lies on the graph of $ f $. At A, $ x = 4.5 $. On your diagram, sketch the graph of $ f^{-1} $, noting clearly the image of point A. [4 marks]

▶️ Answer/Explanation

Markscheme

(a) Simplify $ f(x) = \log_3 \frac{x}{2} + \log_3 16 – \log_3 4 $.

Combine terms: $ \log_3 \frac{x}{2} + \log_3 16 = \log_3 \left( \frac{x}{2} \cdot 16 \right) = \log_3 8x $. A1

Then: $ \log_3 8x – \log_3 4 = \log_3 \frac{8x}{4} = \log_3 2x $. A1

Thus, $ f(x) = \log_3 2x $. AG

[2 marks]

(b) Using $ f(x) = \log_3 2x $: M1

$ f(0.5) = \log_3 (2 \cdot 0.5) = \log_3 1 = 0 $. A1

$ f(4.5) = \log_3 (2 \cdot 4.5) = \log_3 9 = \log_3 3^2 = 2 $. A1

[3 marks]

[2 marks]

– Passes approximately through $ (0.5, 0) $. A1

– Correct logarithmic shape, increasing and concave down. A1

– Asymptotic to the y-axis ($ x = 0 $). A1

Graph description: The graph is defined for $ x > 0 $, passes through $ (0.5, 0) $, $ (1, \log_3 2 \approx 0.63) $, and $ (4.5, 2) $. It approaches the y-axis ($ x = 0 $) as $ x \to 0^+ $, with y-values tending to $-\infty$. Within $ -5 \leq x \leq 5 $, $ -5 \leq y \leq 5 $, the graph starts near the y-axis at large negative y-values and rises slowly through the specified points.

Sketch of f(x)

[3 marks]

[1 mark]

(d) For $ f^{-1}(0) $, solve $ f(x) = 0 $: $ \log_3 2x = 0 \Rightarrow 2x = 3^0 = 1 \Rightarrow x = 0.5 $. A1

[1 mark]

(e) Sketch of $ f^{-1} $:

– Passes approximately through $ (0, 0.5) $. A1

– Correct shape, reflected over $ y = x $, resembling an exponential curve. A1

– Asymptotic to the x-axis ($ y = 0 $). A1

– Image of point A $ (4.5, 2) $ is $ (2, 4.5) $, clearly marked on the curve. A1

Graph description: The inverse $ f^{-1}(x) = \frac{3^x}{2} $ is defined for all $ x $, passes through $ (0, 0.5) $ and $ (2, 4.5) $, and is asymptotic to $ y = 0 $ as $ x \to -\infty $. Within $ -5 \leq x \leq 5 $, $ -5 \leq y \leq 5 $, it rises from near the x-axis at negative $ x $, through $ (0, 0.5) $, to $ (2, 4.5) $, staying below $ y = 5 $.

$Sketch of f^{-1}(x)$

[4 marks]

Total [16 marks]

IBDP Maths SL 1.7 Laws of exponents with rational exponents AA HL Paper 2- Exam Style Questions

IBDP Maths SL 1.7 Laws of exponents with rational exponents AA HL Paper 2- Exam Style Questions- New Syllabus

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