STEP II, 1999
2
Section A:
Pure Mathematics
1
Let
x
= 10
100
,
y
= 10
x
,
z
= 10
y
, and let
a
1
=
x
!
,
a
2
=
x
y
,
a
3
=
y
x
,
a
4
=
z
x
,
a
5
=
e
xyz
,
a
6
=
z
1
/y
,
a
7
=
y
z/x
.
(i)
Use Stirling’s approximation
n
!
≈
√
2
π n
n
+
1
2
e

n
, which is valid for large
n
, to show
that log
10
(log
10
a
1
)
≈
102.
(ii)
Arrange the seven numbers
a
1
,
. . .
,
a
7
in ascending order of magnitude, justifying
your result.
2
Consider the quadratic equation
nx
2
+ 2
x
√
(
pn
2
+
q
) +
rn
+
s
= 0
,
(
*
)
where
p >
0,
p
±
=
r
and
n
= 1, 2, 3,
. . .
.
(i)
For the case where
p
= 3,
q
= 50,
r
= 2,
s
= 15, ﬁnd the set of values of
n
for which
equation (
*
) has no real roots.
(ii)
Prove that if
p < r
and 4
q
(
p

r
)
> s
2
, then (
*
) has no real roots for any value of
n
.
(iii)
If
n
= 1,
p

r
= 1 and
q
=
s
2
/
8, show that (
*
) has real roots if, and only if,
s
6
4

2
√
2 or
s
>
4 + 2
√
2 .
3
Let
S
n
(
x
) =
e
x
3
d
n
d
x
n
‡
e

x
3
·
.
Show that S
2
(
x
) = 9
x
4

6
x
and ﬁnd S
3
(
x
).
Prove by induction on
n
that S
n
(
x
) is a polynomial. By means of your induction argument,
determine the order of this polynomial and the coeﬃcient of the highest power of
x
.
Show also that if
d
S
n
d
x
= 0 for some value
a
of
x
, then
S
n
(
a
)
S
n
+1
(
a
)
6
0.
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3
4
By considering the expansions in powers of
x
of both sides of the identity
(1 +
x
)
n
(1 +
x
)
n
≡
(1 +
x
)
2
n
,
show that
n
X
s
=0
±
n
s
¶
2
=
±
2
n
n
¶
,
where
±
n
s
¶
=
n
!
s
! (
n

s
)!
.
By considering similar identities, or otherwise, show also that:
(i)
if
n
is an even integer, then
n
X
s
=0
(

1)
s
±
n
s
¶
2
= (

1)
n/
2
±
n
n/
2
¶
;
(ii)
n
X
t
=1
2
t
±
n
t
¶
2
=
n
±
2
n
n
¶
.
5
Show that if
α
is a solution of the equation
5cos
x
+ 12sin
x
= 7
,
then either
cos
α
=
35

12
√
120
169
or cos
α
has one other value which you should ﬁnd.
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 Math, Probability, Probability distribution, Probability theory, Quadratic equation, real roots, randomly selected MB666

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