Date | November 2021 | Marks available | 4 | Reference code | 21N.2.AHL.TZ0.10 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
Consider the function f(x)=x2-x-122x-15, x∈ℝ, x≠152.
Find the coordinates where the graph of f crosses the
x-axis.
y-axis.
Write down the equation of the vertical asymptote of the graph of f.
The oblique asymptote of the graph of f can be written as y=ax+b where a, b∈ℚ.
Find the value of a and the value of b.
Sketch the graph of f for -30≤x≤30, clearly indicating the points of intersection with each axis and any asymptotes.
Express 1f(x) in partial fractions.
Hence find the exact value of 3∫01f(x)dx, expressing your answer as a single logarithm.
Markscheme
Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.
attempts to solve x2-x-12=0 (M1)
(-3,0) and (4,0) A1
[2 marks]
Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.
(0,45) A1
[1 mark]
x=152 A1
Note: Award A0 for x≠152.
Award A1 in part (b), if x=152 is seen on their graph in part (d).
[1 mark]
METHOD 1
(ax+b)(2x-15)≡x2-x-12
attempts to expand (ax+b)(2x-15) (M1)
2ax2-15ax+2bx-15b≡x2-x-12
a=12 A1
equates coefficients of x (M1)
-1=-152+2b
b=134 A1
(y=x2+134)
METHOD 2
attempts division on x2-x-122x-15 M1
x2+134+… M1
a=12 A1
b=134 A1
(y=x2+134)
METHOD 3
a=12 A1
x2-x-122x-15≡x2+b+c2x-15 M1
x2-x-12≡(2x-15)x2+(2x-15)b+c
equates coefficients of x : (M1)
-1=-152+2b
b=134 A1
(y=x2+134)
METHOD 4
attempts division on x2-x-122x-15 M1
x2-x-122x-15=x2+13x2-122x-15
a=12 A1
13x2-122x-15=134+… M1
b=134 A1
(y=x2+134)
[4 marks]
two branches with approximately correct shape (for -30≤x≤30) A1
their vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1
their axes intercepts in approximately the correct positions A1
Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.
[3 marks]
attempts to split into partial fractions: (M1)
2x-15(x+3)(x-4)≡Ax+3+Bx-4
2x-15≡A(x-4)+B(x+3)
A=3 A1
B=-1 A1
(3x+3-1x-4)
[3 marks]
3∫0(3x+3-1x-4)dx
attempts to integrate and obtains two terms involving ‘ln’ (M1)
=[3 A1
A1
A1
Note: The final A1 is dependent on the previous two A marks.
[4 marks]