CURVE SKECTHING
1.
The velocity v m s–1 of a particle at time t seconds, is given by v = 2t + cos2t, for 0 ≤ t ≤ 2.
(a)
Write down the velocity of the particle when t = 0.
(1)
When t = k, the acceleration is zero.
(b)
π
.
4
(i)
Show that k =
(ii)
Find the exact velocity when t =
π
.
4
(8)
(c)
When t <
π dv
π dv
,
> 0 and when t > ,
> 0.
4 dt
4 dt
Sketch a graph of v against t.
(4)
(Total 13 marks)
2.
Let f(x) =
ax
x 1
2
, –8 ≤ x ≤ 8, a 
. The graph of f is shown below.
The region between x = 3 and x = 7 is shaded.
(a)
Show that f(–x) = –f(x).
(2)
(b) Given that f′′(x) =
2ax( x 2  3)
( x 2  1) 3
, find the coordinates of all points of inflexion.
(7)
(Total 9 marks)
IB Questionbank Maths SL
1
3.
Let f(x) = 3 +
20
x 4
2
, for x ≠ ±2. The graph of f is given below.
diagram not to scale
The y-intercept is at the point A.
(a)
(i)
Find the coordinates of A.
(ii)
Show that f′(x) = 0 at A.
(7)
(b) The second derivative f′′(x) =
40(3x 2  4)
( x 2  4) 3
. Use this to
(i)
justify that the graph of f has a local maximum at A;
(ii)
explain why the graph of f does not have a point of inflexion.
(6)
(c)
Describe the behaviour of the graph of f for large │x│.
(1)
(d) Write down the range of f.
(2)
(Total 16 marks)
IB Questionbank Maths SL
2
4.
The graph of a function g is given in the diagram below.
The gradient of the curve has its maximum value at point B and its minimum value at point D.
The tangent is horizontal at points C and E.
(a)
Complete the table below, by stating whether the first derivative g′ is positive or negative,
and whether the second derivative g′′ is positive or negative.
Interval
g′
g′′
axb
exƒ
(b)
Complete the table below by noting the points on the graph described by the following
conditions.
Conditions
Point
g′ (x) = 0, g′′ (x)  0
g′ (x)  0, g′′ (x) = 0
(Total 6 marks)
IB Questionbank Maths SL
3
5.
The diagram shows the graph of y = f (x).
y
0
x
On the grid below sketch the graph of y = f (x).
y
0
x
(Total 6 marks)
IB Questionbank Maths SL
4
ANSWERS:
1.
(a) v = 1
(b)
(i)
A1
d
2t   2
dt
N1
1
A1
d
cos 2t   2 sin 2t
dt
Note: Award A1 for coefficient 2 and A1 for –sin 2t.
evidence of considering acceleration = 0
A1A1
(M1)
dv
 0, 2  2 sin 2t  0
dt
e.g.
correct manipulation
A1
e.g. sin 2k  1, sin 2t  1
2k 
k
π
π
 accept 2t  
2
2
A1
π
4
AG
(ii) attempt to substitute t 

into v
4
N0
(M1)
 
 2 
e.g. 2   cos 
4
 4 
v

2
A1
N2
8
A1A1A2
Notes: Award A1 for y-intercept at (0, 1), A1 for curve having
π
zero gradient at t  , A2 for shape that is concave down to
4
π
π
the left of
and concave up to the right of . If a correct
4
4
π
curve is drawn without indicating t = , do not award the
4
second A1 for the zero gradient, but award the final A2 if
appropriate. Sketch need not be drawn to scale. Only essential
features need to be clear.
N4
4
(c)
IB Questionbank Maths SL
5
2.
(a)
METHOD 1
evidence of substituting –x for x
a ( x)
f(–x) =
( x) 2  1
 ax
f(–x) = 2
(= –f(x))
x 1
(M1)
A1
AG
N0
METHOD 2
y = –f(x) is reflection of y = f(x) in x axis
and y = f(–x) is reflection of y = f(x) in y axis
(M1)
sketch showing these are the same
f(–x) =
(b)
 ax
x 2 1
A1
(= –f(x))
AG
evidence of appropriate approach
e.g. f″(x) = 0
(M1)
to set the numerator equal to 0
e.g. 2ax(x2 – 3) = 0; (x2 – 3) = 0

a 3  
a 3 
(0, 0),  3 ,
(accept x = 0, y = 0 etc.)
,  3 ,

4  
4 

(A1)
A1A1A1A1A1
N0
N5
[9]
3.
(a)
(i)
coordinates of A are (0, –2)
(ii)
derivative of x2 – 4 = 2x (seen anywhere)
evidence of correct approach
e.g. quotient rule, chain rule
A1A1
finding f′(x)
e.g. f′(x) = 20 × (–1) × (x2 – 4)–2 × (2x),
(i)
(ii)
(A1)
(M1)
A2
( x 2  4)(0)  (20)(2 x)
( x 2  4) 2
substituting x = 0 into f′(x) (do not accept solving f′(x) = 0)
at A f′(x) = 0
(b)
M1
AG
reference to f′(x) = 0 (seen anywhere)
reference to f″(0) is negative (seen anywhere)
evidence of substituting x = 0 into f″(x)
40  4   5 
finding f″(0) =


(4) 3  2 
then the graph must have a local maximum
(R1)
R1
M1
reference to f″(x) = 0 at point of inflexion,
recognizing that the second derivative is never 0
(R1)
A1
IB Questionbank Maths SL
N2
N0
A1
AG
N2
6
e.g. 40(3x2 + 4) ≠ 0, 3x2 + 4 ≠ 0, x2 ≠ 
4
, the numerator is
3
always positive
Note: Do not accept the use of the first derivative in part (b).
(c)
correct (informal) statement, including reference to approaching y = 3
e.g. getting closer to the line y = 3, horizontal asymptote at y = 3
(d)
correct inequalities, y ≤ –2, y > 3, FT from (a)(i) and (c)
A1
N1
A1A1
N2
[16]
4.
(a)
Interval
g
g
a<x<b
positive
positive
e<x<f
negative
negative
A1A1
A1A1
N4
A1
N1
A1
N1
(b)
Conditions
Point
g (x) = 0, g (x) < 0
C
g (x) < 0, g (x) = 0
D
[6]
IB Questionbank Maths SL
7
5.
y
x
(A2)(A1)(A1)(A2) (C6)
Note: Award A2 for correct shape (approximately parabolic),
A1 A1 for intercepts at 0 and 4, A2 for minimum between
x = 1.5 and x = 2.5.
[6]
IB Questionbank Maths SL
8