Date | May 2021 | Marks available | 1 | Reference code | 21M.2.AHL.TZ2.12 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
A function f is defined by f(x)=arcsin(x2-1x2+1), x∈ℝ.
A function g is defined by g(x)=arcsin(x2-1x2+1), x∈ℝ, x≥0.
Show that f is an even function.
By considering limits, show that the graph of y=f(x) has a horizontal asymptote and state its equation.
Show that f'(x)=2x√x2(x2+1) for x∈ℝ, x≠0.
By using the expression for f'(x) and the result √x2=|x|, show that f is decreasing for x<0.
Find an expression for g-1(x), justifying your answer.
State the domain of g-1.
Sketch the graph of y=g-1(x), clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.
Markscheme
EITHER
f(-x)=arcsin((-x)2-1(-x)2+1)=arcsin(x2-1x2+1)=f(x) R1
OR
a sketch graph of y=f(x) with line symmetry in the y-axis indicated R1
THEN
so f(x) is an even function. AG
[1 mark]
as x→±∞, f(x)→arcsin 1(→π2) A1
so the horizontal asymptote is y=π2 A1
[2 marks]
attempting to use the quotient rule to find ddx(x2-1x2+1) M1
ddx(x2-1x2+1)=2x(x2+1)-2x(x2-1)(x2+1)2 (=4x(x2+1)2) A1
attempting to use the chain rule to find ddx(arcsin(x2-1x2+1)) M1
let u=x2-1x2+1 and so y=arcsin u and dydu=1√1-u2
f'(x)=1√1-(x2-1x2+1)2×4x(x2+1)2 M1
=4x√(x2+1)2-(x2-1)2×1(x2+1) A1
=4x√4x2×1(x2+1) A1
=2x√x2(x2+1) AG
[6 marks]
f'(x)=2x|x|(x2+1)
EITHER
for x<0, |x|=-x (A1)
so f'(x)=-2xx2+1 A1
OR
|x|>0 and x2+1>0 A1
2x<0, x<0 A1
THEN
f'(x)<0 R1
Note: Award R1 for stating that in f'(x), the numerator is negative, and the denominator is positive.
so f is decreasing for x<0 AG
Note: Do not accept a graphical solution
[3 marks]
x=arcsin(y2-1y2+1) M1
sin x=y2-1y2+1⇒y2 sin x+sin x=y2-1 A1
y2=1+sin x1-sin x A1
domain of g is x∈ℝ, x≥0 and so the range of g-1 must be y∈ℝ, y≥0
hence the positive root is taken (or the negative root is rejected) R1
Note: The R1 is dependent on the above A1.
so (g-1(x)=)√1+sin x1-sin x A1
Note: The final A1 is not dependent on R1 mark.
[5 marks]
domain is -π2≤x<π2 A1
Note: Accept correct alternative notations, for example, ⌊-π2, π2⌊ or ⌊-π2, π2).
Accept [-1.57, 1.57[ if correct to 3 s.f.
[1 mark]
A1A1A1
Note: A1 for correct domain and correct range and y-intercept at y=1
A1 for asymptotic behaviour x→π2
A1 for x=π2
Coordinates are not required.
Do not accept x=1.57 or other inexact values.
[3 marks]