IB Mathematics AA SL - Questionbank
Topic 2 All - Functions
Question 96
[Maximum mark: 16]
↑
y
.
The graph of a function f has its vertex at the point (1, 7) and intersects the x-axis at x = −3.
&
<
= 0
ay12 7
(a) Find f in the form f (x) = a (x − h) + k .
=
f ( 3)
The function g is defined by g(x) = mx2 − 2nx +
2n2
where x ∈ R, m ∈ R+ and n ∈ R.
m
2
=
-
+
+(3)
+
0
-
& In the case where g(−2) = g(1) = 3,
(b)
&
when
(i) find the value of m and the value of n;
(2/2m
-
72)7 3
4 + 44 + 3
2n
=
-
m
When
X
=I
-
1) +7 = 0
[3]
16a + 7 =p
16a
=
a =
X = -2
&
ac 3
-
7
-T
7
f(x
(x)
=
-
74m + 4n +
E (m
-
-
2n
12 +
+) p
3m + 6n = 0
1m-2n
&
1
m-2n
(ii) Find the range of g .
.
-
m+ 2n = 0
+
-
The linear function h is defined by h(x) = nx + 1 where x ∈ R.
m=
2n2
2n 2n +
In
-
-
=
-
4n - n
=
3
!
-
fgx
=
-
2
g(x) Tgx
=
-
+
c
5)x
Ex
T(x2 x)
+
=
(Xz
=
+
[6]
⑮
m=2
.
c.
3
+
+
2n
[7]
3
(c) Show that the graphs of h and g have two distinct points of intersection for every possible value of m ∈
R+ and n ∈ R.
54 = 3
g(x)
-
5
+ X
mx2 2nx + m
MX2 3nX
=
nx +
+
=
0
S
Question 97
[Maximum mark: 8]
A population of endangered snow leopards, P , can be modelled by the equation
Pt = P0 ekt ,
where P0 is the initial population, and t is measured in years.
(a) After one year, it is estimated that
P1
= 0.93.
P0
(i) Find the value of k .
(ii) Interpret the meaning of the value of k .
(b) Find the least number of whole years for which
[3]
Pt
< 0.50.
P0
[5]
Question 98
Vo = VitEd
Up
=
Un
In 2
vs
Us = 14
[Maximum mark: 15]
=
U , <n 1)d
Vo
+
-
Vo
=
=
=
In 2
Vi + Sluz :
In2
2d
Vis-Mind
31nz-Inz 2d
d Ind
=
=
The arithmetic sequence has the sixth term of u6 = ln 2 and the eighth term of u8 = 32 ln 4.
(a) (i) Show the common difference is.
ln 2.
10
Sn =
·
-
(ii) Find u1 .
S
Su =
E(2ri + (n 1d)
-
Sn
E (2 ( 4142) + (n 1)(nz)
-
.
= c-2mmm2 Inz)
-
(iii) Hence, show that the sum of n terms is given by Sn = (n2 − 9n) ln 2. Where n ∈ Z+ .
&
[5]
=
<MIn2-912]
A new function, f (x), is formed using Sn with the discrete variable n being replaced with the continuous
variable x, such that x ∈ R. Hence f (x) = (x2 − 9x) ln 2.
Sn =
Ins2 x + /Su)xX =
(b) &
(i) Write down the equation of the axis of symmetry of f (x).
.
=
⑫
(ii) Sketch f (x) on the grid below, identify the axis of symmetry and any axis intercepts.
(x2
-
9X)In2
x2 qx
x =
=
=
0
0
0/9
f(z)
=
s
(iii) Hence or otherwise, find the minimum value of Sn in the form a ln b. Where a, b ∈ Z
=
=
=>
Enz .
*. 21n52 (n-9)
·
ncn-9)InS2
M (n) InS2
[5]
Consider the linear function g(x) = (6 − 3x) ln 2.
(c)
(i) Solve f (x) = g(x).
(ii) Hence, solve the inequality f (x) < g(x).
(iii) Hence or otherwise, find the range of values of n that satisfy Sn < g(x).
[5]
Question 99
[Maximum mark: 5]
Solve the equation 9x + 2 ⋅ 3x+1 = 1.
9x
=
132)x
3x
+
a =
1
134)2
=
34
=
.
3
3x
=
a
a+ 2 3
a
.
.
=
1
-
az + ba
-
1
=
0
-
Question 100
34
20
3 - To
<0
=S
.
162
> 8
⑮
[Maximum mark: 17]
fixsax
n = 3
The quadratic function f (x) has a vertex at (h, −2).
f(0)
=
acu
W
(a) The function can be written in the form of f (x) = a(x − h)2 + k .
=
3)2-2
-
99
-
2
-
(i) If the equation of the axis of symmetry of f (x) is x = 3, write down the values of h and k .
(ii) If the y−intercept of f (x) is −11, find a.
gaz--Il
A new function g(x) is defined such that g ′ (x) = f (x).
f(x)
(b) S
(i) Show that the function f (x) has no real roots.
=
- (X
-
[4]
.
3)22
(x 3(230
-
-
-
(x
-
3240
(x
no
-
31324-2
real
roots
(ii) Hence, or otherwise, justify that g(x) is a decreasing function.
O
&
[3]
S
The function f (x) is vertically translated such that it now has 2 distinct roots, x1 and x2 , where x2 > x1 . The
coordinates of the new vertex is (3, m). Where m > 0.
-
(c) Show that x1 = 3 −
m and x2 = 3 +
m.
[2]
The area enclosed by f (x) and the x−axis is 32
units.
3
(d)
(i) Use the substitution u = x − 3 to find ∫ f (x) dx in terms of m.
(ii) Hence find m.
(e) Hence, or otherwise, find the region(s) of x, for which g(x) is a decreasing function.
Question 101
[Maximum mark: 15]
[6]
[2]
Let f (x) = a(x + 1)(x + 5), for x ∈ R, where a ∈ Z. The following diagram shows part of the graph of f .
The graph of f has x-intercepts at (p, 0) and (q, 0), and a y -intercept at (0, −10).
(a) S
(i) Write down the value of p and the value of q .
(ii) Find the value of a. y
/
=
10
+
f(x)
-
2(X + 1)
(c) Find the coordinates of the vertex.
S
=
1
10 = a(0 + 1)(0 + S)
5 a = 10
S
(b) Find the equation of the axis of symmetry.
=
p = 59
X
=
3
[6]
[2]
[3]
(x + s) The graph of a function g is obtained from the graph of f by a reflection in the y -axis, followed by a translation
=
&
of 2 units in the direction of the positive y -axis. The point P(−2, 6) on the graph of f is mapped to point Q on
the graph of g .
2(X2
=
-
+
[x + X
(d) Find the coordinates of Q.
ts)
=
-
2x2
-
12x
10
6)
,
2) (2 8)
,
f( 3)
-
1) (2
=
= -
73
,
- =
2x( 3+
)c
8)
Question 102
-
S +5) 8
=
[4]
[Maximum mark: 8]
!
(a) Write down the domain and range of the logarithmic function y = loga x
&
where a > 0 and a =
 1.
[2]
(b) Given that logx2 y = 9 logy (x2 ), find all the possible expressions
of y as a function of x.
[6]
>
Question 103
[Maximum mark: 7]
Consider f (x) = logk (8x − 2x2 ), for 0 < x < 4, where k > 0.
The equation f (x) = 3 has exactly one solution. Find the value of k .
Question 104
[Maximum mark: 12]
The following table shows values of ln x and ln y .
1
The relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.
(a) Find the value of a and the value of b.
[3]
(b) Use the regression equation to estimate the value of y when x = 4.12.
[3]
The relationship between x and y can be modelled using the formula y = pxq , where p =
 0 and q =
 0, q =
 1.
(c) Expressing ln y in terms of ln x, find the value of p and the value of q .
Question 105
[6]
[Maximum mark: 19]
Consider the function
f (x) = 14 x2 − x + 3
(a) Write f (x) in the form a(x − h)2 + k . Where a, h, k ∈ Q.
[3]
(b) Hence, find ∫ f (x) dx.
[2]
A rectangle ABCD has vertices shown below.
f (x) cuts the rectangle into two parts.
The part above the curve is labelled U and the part below the curve is labelled L.
(c)
(i) Verify the point (4,3) lies on f (x).
(ii) Hence, or otherwise, sketch f (x) on the diagram above and identify the parts U and L.
(d)
[2]
(i) Write down the area of rectangle ABCD.
(ii) Hence, show that the area of U is twice the area of L.
[5]
The rectangle ABCD is modified such that point C(b, f (b)) is a variable point on the function f (x). Where
b > 2 and b ∈ R.
The point A(2, 2) is fixed in the same position as previously.
(e) Find the area of the new rectangle in terms of b.
b
[2]
(f) Find ∫ f (x) dx in terms of b.
[1]
(g) Hence, or otherwise, prove that the area of U is twice the area of L for any position of point C .
[4]
2