Functions
1.
Figure 3
y
(0, c)
O
(d, 0)
x
Figure 3 shows a sketch of the curve with equation y = f(x), x  0. The curve meets the coordinate
axes at the points (0, c) and (d, 0).
In separate diagrams sketch the curve with equation
(a) y = f1(x),
(2)
(b) y = 3f(2x).
(3)
Indicate clearly on each sketch the coordinates, in terms of c or d, of any point where the curve
meets the coordinate axes.
Given that f is defined by
f : x  3(2x )  1, x  ℝ, x  0,
(c) state
(i)
the value of c,
(ii) the range of f.
(3)
(d) Find the value of d, giving your answer to 3 decimal places.
(3)
The function g is defined by
g : x  log2 x, x  ℝ, x  1.
(e) Find fg(x), giving your answer in its simplest form.
(3)
2.
The function f is defined by
f: x 
(a) Show that f(x) =
5x  1
3
–
, x > 1.
x  x2 x2
2
2
, x > 1.
x 1
(4)
(b) Find f –1(x).
(3)
The function g is defined by
g: x  x2 + 5, x  ℝ.
(c) Solve fg(x) =
1
4
.
(3)
3.
The functions f and g are defined by
f : x  2x + ln 2,
x  ℝ,
g : x  e2x,
x  ℝ.
(a) Prove that the composite function gf is
gf : x  4e4x,
x  ℝ.
(4)
(b) Sketch the curve with equation y = gf(x), and show the coordinates of the point where the curve
cuts the y-axis.
(1)
(c) Write down the range of gf.
(1)
(d) Find the value of x for which
d
[gf(x)] = 3, giving your answer to 3 significant figures.
dx
(4)
4.
For the constant k, where k > 1, the functions f and g are defined by
f: x  ln (x + k),
x > –k,
g: x  2x – k,
x  ℝ.
(a) On separate axes, sketch the graph of f and the graph of g.
On each sketch state, in terms of k, the coordinates of points where the graph meets the coordinate
axes.
(5)
(b) Write down the range of f.
(1)
k
4
(c) Find fg   in terms of k, giving your answer in its simplest form.
(2)
The curve C has equation y = f(x). The tangent to C at the point with x-coordinate 3 is parallel to the
line with equation 9y = 2x + 1.
(d) Find the value of k.
Functions Answers
Question
Scheme
Marks
number
y
1.
(a)
shape B1
intersections with axes (c, 0), (0, d) B1
(0, d)
O
(b)
(c, 0)
(2)
x
y
shape B1
x intersection ( 12 d, 0) B1
(0, 3c)
y intersection (0, 3c) B1
O
( 12 d, 0)
x
(c)(i) c = 2
B1
(ii) 1 < f(x)  (candidate’s) c value
(d) 3(2x) = 1  2x =
(3)
1
3
B1 B1 ft
and take logs; x =
ln
1
3
M1; A1
ln 2
d (or x) = 1.585 (3 decimal places)
 log x
(e) fg(x) = f[log2 x] = [ 3(2 2 )  1 ]; = [ 3(2
=
3
1
x
log2 1x
)  1 ] or
(3)
3
2
log 2 x
A1
1
(3)
M1; A1
A1
(3)
(14 marks)
2
(a)
5x  1
3

( x  2)( x  1) x  2
=
B1
5 x  1  3( x  1)
( x  2)( x  1)
M1
M1 for combining fractions even if the denominator is not lowest common
=
2x  4
=
( x  2)( x  1)
2( x  2)
2
=
( x  2)( x  1) x  1
M1 A1 cso
*
(4)
M1 must have linear numerator
(b)
y=
2

x 1
f –1(x) =
2 x
x
fg(x) =
2
x 4
Setting
2
A1
o.e.
(attempt)
2
=
x 4
2
M1A1
xy  y  2  xy = 2 + y
[
2
]
" g"  1
1
and finding x2 = …;
4
x = 2
(3)
M1
M1; A1
(3)
[10]
Question
Scheme
Marks
Number
3.
gf  x   e2 2 x ln 2
(a)
M1
 e4 x e2ln 2
M1
 e 4 x eln 4
M1
 4 e4 x
Give mark at this
point, cso
 Hence gf : x
4 e4 x ,
A1
(4)
B1
(1)
B1
(1)
x  R
(b)
y
Shape and
point
4
O
(c)
Range is R
gf  x   0, y  0
x
Accept
(d)
d
 gf  x    16 e 4 x
dx
e4 x 
3
16
4 x  ln
3
16
x  0.418
M1 A1
M1
A1
(4)
[10]
Question
Number
Scheme
Marks
4.
Log graph: Shape
B1
Intersection with –ve x-axis
dB1
(0, ln k), (1 – k, 0)
B1
(a)
Mod graph :V shape, vertex
on +ve x-axis
k
2


B1
(0, k) and  , 0 
(b) f(x)  R
,
k
4
(c) fg   = ln{k + |
= ln (
(d)
–   f ( x)   ,
2k
 k |} or
4


–  y  
k


f |  |
2
3k
)
2
(5)
B1
(1)
M1
A1
dy
1

dx x  k
Equating (with x = 3) to grad. of line;
B1
(2)
B1
1
2

3 k 9
M1; A1
k = 1½
A1
(4)
(12 marks)