MEI Core 3
Functions / Natural logarithms and exponentials
Chapter Assessment
1. The functions f, g and h are defined as follows:
f ( x)  e x
x
g( x)  x
x0
h( x)  2 x  1 x 
Find each of the following functions, giving the domain and range of each.
(i) fg(x)
(ii) gh(x)
(iii) hf(x)
-1
-1
(iv) f (x)
(v) h (x)
x  x3
is odd.
1  x2
(ii) Explain what this tells you about the graph of y  f ( x) .
2. (i)
Show algebraically that the function f ( x) 
3. (i) Sketch the graph of y  2 x  1 .
(ii) Hence, or otherwise, solve each of the following equations:
(a)
2x 1  3  x
(b)
2 x  1  3x  2
[9]
[6]
[3]
[1]
[2]
[3]
[3]
4. Solve the equations
(i) 2e x  3e  x  5
(ii) ln(2 x  1)  ln x  2
giving your answers in exact form.
[3]
[3]
x 1
for x  0.
x
The graph approaches the line y = 1 as x becomes very large.
5. The diagram below shows the graph y = f(x), where f ( x) 
1
1
(i)
(ii)
(iii)
(iv)
(v)
Write down the domain and range of f(x).
Find the inverse function f-1(x).
Write down the domain and range of f-1(x).
Sketch the graph of y = f-1(x) for the domain you gave in (iii).
What is the relationship between the graph of y = f(x) and the graph of
y = f-1(x)?
© MEI, 10/05/10
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[3]
[2]
[2]
[1]
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MEI C3 Functions Assessment solutions
6. The graph of a function y = f(x) is shown below. The graph has a local maximum
at ( 1, 1) and a local minimum at (2,  2).
y
(-1, 1)
x
(2, -2)
Sketch the graphs of:
(i) y  3f (2 x)
(ii) y  2f ( x  1)
(iii) y  f (2 x)  1
(iv) y  f ( x)  1
(v) y  f ( x)
(vi) giving the coordinates of the turning points in each case.
[3]
[3]
[3]
[3]
[3]
7. A function is defined by f ( x)  2cos 1 x  1 .
(i) Write down the domain and range of this function.
(ii) Find the value of f(-0.5).
(iii) Find the inverse function f 1 ( x) .
[2]
[2]
[3]
8. (i) Solve the inequality 3x  2  4 .
[3]
(ii) Write the inequality  2 < x < 7 in the form x  a  b .
[2]
Total 70 marks
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MEI C3 Functions Assessment solutions
Functions / Natural logarithms and exponentials
Solutions to Chapter Assessment
1. (i)
fg( x )  f( x )  e x
Domain is x  0
Range is fg( x )  1
(ii) gh( x )  g(2 x  1)  2 x  1
Domain is x   21
Range is gh( x )  0 .
Since g(x) is
defined as a
function, only the
positive square
root is used,
giving this range.
[3]
[3]
(iii) hf( x )  h(e )  2e  1
Domain is x 
Range is hf( x )  1
x
x
[3]
(iv) y  e
ln y  x
x
f 1( x )  ln x
Domain is x  0
Range is f 1( x ) 
[3]
(v) y  2 x  1
y  1  2x
y 1
x
2
h1( x ) 
x 1
2
Domain is x 
Range is h1( x ) 
[3]
2. (i)
x  x3
1  x2
(  x )  (  x )3  x  x 3 ( x  x 3 )
f(  x ) 


  f( x )
1  (  x )2
1  x2
1  x2
f( x ) 
so the function f is odd.
[3]
(ii) The graph of y  f( x ) has rotational symmetry of order 2 about the
origin.
[1]
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MEI C3 Functions Assessment solutions
3. (i) y  2 x  1
1
 21
[2]
(ii) (a)
2x  1  3  x
The graph shows that there are two solutions.
2x  1  3  x
(2 x  1)  3  x
3x  2
x
2 x  1  3  x
4  x
2
3
Solutions are x 
2
3
3
 21
and x  4
[3]
(b)
2 x  1  3x  2
The graph shows that there is just one solution
2 x  1  3x  2
 21
3 x
2
3
The solution is x = 3.
[3]
4. (i) 2e x  3e  x  5
Substituting y  e x :
2 y  3 y 1  5
Multiplying through by y: 2 y 2  3  5 y
2y2  5 y  3  0
(2 y  1)( y  3)  0
y   21 or 3
e x   21 or 3
Since e x cannot be negative, e x  3
x  ln 3
[3]
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MEI C3 Functions Assessment solutions
(ii) ln(2 x  1)  ln x  2
ln(2 x  1)  ln x  2
 2x  1 
ln 
2
 x 
2x  1
 e2
x
2 x  1  xe 2
1  xe 2  2 x
1  x(e 2  2)
x
1
e 2
2
[3]
5. (i) Domain is x  0
Range is f( x )  1
[2]
x 1
(ii) y 
x
xy  x  1
1  x  xy
1  x(1  y )
x
1
1y
f 1( x ) 
1
1x
[3]
(iii) Domain is x  1
Range is f( x )  0
[2]
(iv)
1
1
[2]
(v) They are reflections of each other in the line y = x.
[1]
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MEI C3 Functions Assessment solutions
6. (i)
y  3f(2 x )
The graph is stretched scale factor 3 in the y direction, and scale factor
in the x direction.
The turning points are (1, -6) and (  21 , 3).
1
2
(  21 , 3)
(1, -6)
[3]
(ii) y  2f( x  1)
The graph is translated 1 unit horizontally to the right, and stretched
scale factor 2 in the y direction.
The turning points are (3, -4) and (0, 2)
(0, 2)
(3, -4)
[3]
(iii) y  f(2 x )  1
The graph is stretched scale factor 21 parallel to the x axis, and translated
1 unit downwards.
The turning points are (1, -3) and ( 21 , 0)
y
( 21 , 0)
x
(1, -3)
[3]
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MEI C3 Functions Assessment solutions
(iv) y  f(  x )  1
The graph is reflected in the y-axis, and translated 1 unit upwards.
The turning points are (-2, -1) and (1, 2).
[3]
(v) y  f( x )
The negative parts of the graph are reflected in the x-axis.
The turning points are (2, 2) and (-1, 1).
[3]
1
7. f( x )  2 cos x  1
(i) Domain for f(x) is 1  x  1
Range for cos 1 x is 0  x   , so range for f(x) is 1  f( x )  2  1 .
[2]
1
(ii) f( x )  2 cos ( 0.5 )  1
 2

2
1
3
4
1
3
[2]
1
(iii) y  2 cos x  1
y  1  2 cos 1 x
y1
 cos 1 x
2
y 1

 2 
x  cos 
x 1
The inverse function is f 1( x )  cos 

 2 
[3]
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MEI C3 Functions Assessment solutions
8. (i)
3x  2  4
4  3 x  2  4
2  3 x  6
 23  x  2
(ii) 2  x  7
2  2.5  x  2.5  7  2.5
4.5  x  2.5  4.5
[3]
x  2.5  4.5
[2]
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