Date | May 2022 | Marks available | 5 | Reference code | 22M.1.SL.TZ2.8 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Consider the functions f(x)=1x-4+1, for x≠4, and g(x)=x-3 for x∈ℝ.
The following diagram shows the graphs of f and g.
The graphs of f and g intersect at points A and B. The coordinates of A are (3, 0).
In the following diagram, the shaded region is enclosed by the graph of f, the graph of g, the x-axis, and the line x=k, where k∈ℤ.
The area of the shaded region can be written as ln(p)+8, where p∈ℤ.
Find the coordinates of B.
Find the value of k and the value of p.
Markscheme
1x-4+1=x-3 (M1)
x2-8x+15=0 OR (x-4)2=1 (A1)
valid attempt to solve their quadratic (M1)
(x-3)(x-5)=0 OR x=8±√82-4(1)(15)2(1) OR (x-4)=±1
x=5 (x=3, x=5) (may be seen in answer) A1
B(5, 2) (accept x=5, y=2) A1
[5 marks]
recognizing two correct regions from x=3 to x=5 and from x=5 to x=k (R1)
triangle +k∫5f(x)d OR OR
area of triangle is OR OR (A1)
correct integration (A1)(A1)
Note: Award A1 for and A1 for .
Note: The first three A marks may be awarded independently of the R mark.
substitution of their limits (for ) into their integrated function (in terms of ) (M1)
A1
adding their two areas (in terms of ) and equating to (M1)
equating their non-log terms to (equation must be in terms of ) (M1)
A1
A1
[10 marks]
Examiners report
Nearly all candidates knew to set up an equation with in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from to . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of and were unattainable.