1201/1602SCE midsemester exam 2005 answers
1
1201SCE/1602SCE
Answers for midsemester exam 2005
Question 1
(a)
The first few lines of the Pascal triangle are easy to write down.
Then use the
rule that new entry = value just above + value above and to left
1
1
1
2
1
Å
(
x
+
y
)
2
1
3
3
1
Å
(
x
+
y
)
3
1
4
6
4
1
Å
(
x
+
y
)
4
So the expansion is
4
4
3
2
2
3
4
(
)
4
6
4
x
y
x
x y
x y
xy
y
+
=
+
+
+
+
The expansion of
4
4
3
2
2
3
4
4
3
2
2
3
4
(
2 )
4
(2 )
6
(2 )
4 (2 )
(2 )
8
24
32
16
x
y
x
x
y
x
y
x
y
y
x
x y
x y
xy
y
+
=
+
+
+
+
=
+
+
+
+
(b)
See below – it is important to get the general shape right
Because I drew this using a computer package this graph has scales on the
x
and
y
axes – this is not essential.
The key points are:
The square root function starts at 0 and goes upwards
but has a concave shape;
The 1/
x
function decreases from large values
towards zero; the log(
x
) graph is negative for
x
< 1 and positive for
x
> 1 and
also has a concave shape (concave is another word for convex downwards)
(c)
y
=
f
(
x
)
has an inverse function if this equation has just one solution for
x
for
each
y
value (in the range of
f
)
The notation is
1
( )
x
f
y
−
=
The inverse function for
x
y
e
=
is
log ( )
e
x
y
=
1201/1602SCE midsemester exam 2005 answers
2
(d)
The horizontal line test says that if each horizontal line meets the graph of
y
=
f
(
x
)
at just one
x
value then the function
f
(
x
) has an inverse function.
For the cos(
x
) function we need to restrict our attention to one part of the
graph where the graph goes just once from -1 to +1
The region where
x
goes from 0 to
π
is marked in bold. Each
y
value between -
1 and +1 has just one solution for cos(
x
) =
y
between
x
= 0
and
x
=
π
When you calculate an inverse cos on your calculator you will get an answer
between 0 and
π
radians.
(e)
NOT DONE IN 2007
Definitions are
1
1
cosh( )
{
}
sinh( )
{
}
2
2
x
x
x
x
x
e
e
and
x
e
e
−
−
=
+
=
−
Use these definitions to get
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
cosh ( )
sinh ( )
{
}
{
}
4
4
1
1
{
2
.
}
{
2
.
}
4
4
1
1
{
2
}
{
2
}
4
4
2
1
{
}
{
}
4
2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
e
e
e
e
e
e e
e
e
e e
e
e
e
e
e
e
e
e
e
−
−
−
−
−
−
−
−
−
−
+
=
+
+
−
=
+
+
+
−
+
=
+
+
+
−
+
=
+
=
+
This matches what we would get from the definition of cosh(2
x
)
So we have checked that
2
2
cosh ( )
sinh ( )
cosh(2 )
x
x
x
+
=

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