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Date November 2021 Marks available 2 Reference code 21N.1.SL.TZ0.5
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number 5 Adapted from N/A

Question

Let the function h(x) represent the height in centimetres of a cylindrical tin can with diameter x cm.

hx=640x2+0.5 for 4x14.

The function h1 is the inverse function of h.

Find the range of h.

[3]
a.

Find h-110.

[2]
b.i.

In the context of the question, interpret your answer to part (b)(i).

[1]
b.ii.

Write down the range of h1.

[1]
b.iii.

Markscheme

h4=64042+0.5  OR  h14=640142+0.5                    (M1)


Note: Award (M1) for substituting 4 or 14 into h. This can be implicit from seeing 3.77 (3.76530) or 40.5.


3.77hx40.5   3.76530hx40.5              A1A1


Note: Award A1 for both correct endpoints seen, A1 for the endpoints in a correct interval.

 

[3 marks]

a.

hx=10  OR  h-1x=640x-0.5  OR  h-110=64010-0.5                    (M1)

x= 8.21cm 8.20782              A1

 

[2 marks]

b.i.

a tin that is 10cm high will have a diameter of 8.21cm (8.20782...)            A1


Note: Condone a correct answer expressed as the converse.

 

[1 mark]

b.ii.

4h-114           A1


Note: Accept 4y14. Do not FT in this part.

 

[1 mark]

b.iii.

Examiners report

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as 40.5y3.77 while some others wrote 40.5y3.77. A few candidates used integer x values from 4 to 14 to find corresponding values for hx and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was 6.9 with weaker candidates simply finding h10. Several candidates equated hx to 10 but missed out +0.5 in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting x by 10 and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

a.

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as 40.5y3.77 while some others wrote 40.5y3.77. A few candidates used integer x values from 4 to 14 to find corresponding values for hx and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was 6.9 with weaker candidates simply finding h10. Several candidates equated hx to 10 but missed out +0.5 in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting x by 10 and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

b.i.

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as 40.5y3.77 while some others wrote 40.5y3.77. A few candidates used integer x values from 4 to 14 to find corresponding values for hx and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was 6.9 with weaker candidates simply finding h10. Several candidates equated hx to 10 but missed out +0.5 in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting x by 10 and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

b.ii.

Part (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as 40.5y3.77 while some others wrote 40.5y3.77. A few candidates used integer x values from 4 to 14 to find corresponding values for hx and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was 6.9 with weaker candidates simply finding h10. Several candidates equated hx to 10 but missed out +0.5 in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting x by 10 and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).

b.iii.

Syllabus sections

Topic 2—Functions » SL 2.2—Functions, notation domain, range and inverse as reflection
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Topic 2—Functions

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