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Solutions to Math-106(Fall/2004) Final Exam Problems
1- a) D =Ñ − {− 1 }.
b) ( ) =−∞
→+
f x x 11
lim , ( ) =+∞
→−
f x x 1
lim , lim ( ) = 0
→ +∞
f x x
, lim ( ) = 0
→ −∞
f x x
c) ( )
( ) ( )
3 2 3
3
3 2
3 2
′ = x
x
x
x
x x x f x.
f ′ ( ) x < 0 for − < ⇒ < ⇒
3 3
1 2 x 0 x 3 2
< x. So f is decreasing on
3
3
3
f ′^ x > 0 ⇒ > x ⇒ x <. So f is increasing on ( − ∞,− 1 )and
3 2
d)
e)
( )
dx
x
x V y dx ⋅
∫ ∫
∞ ∞
0
3 2
2
0
2
3 u = x + ⇒ du = x ⋅ dx
2 (^3)
→∞
∞
∫ 1 1
2
lim 3 3
b
u b^ u
du x dx du V
π π
lim 3
b → ∞ b
cubic units.
Side lengths are 2 x and y.
So Area; A = 2 x ⋅ y
2 2 2 2 x + y = 3 = 9 ⇒ y = 9 − x
2 ⇒ A = 2 x ⋅ 9 − x
2
2
x
x x x dx
dA
2 2
2 2
2
2 2 = ⇒ = ⇒ = −
= ⋅ − ⋅ − x x x
x x
x
x x
and
2
2
y = −.
Hence side lengths are
3 2 , and largest area is 9
2
2 ⋅ ⋅ = sq.units.
3-a) df = f ′⋅ dx and dg = g ′⋅ dx.
( ) ( f g ) dx ( f g g f ) dx
dx
d f g f g g f f g dx
d ⋅ = ′⋅ + ′⋅ ⇒ ⋅ = ⋅ ⋅ = ′⋅ + ′⋅ ⋅
= g ⋅ df + f ⋅ dg ⇒ f ⋅ dg = f ⋅ g − g ⋅ df.
b) If we put u = − x , then we have
x
dx du
⋅
and dx = 2 u ⋅ du.
I e dx e u du
x u = ⋅ = ⋅ ⋅ ⋅
−
- In order to evaluate this integral, we will use integration
by parts; Let f = u &
u g = e. Then
I [ u e e du ] ( u e e C ) ( u ) e C ( x ) e C
u u u u u x = ⋅ ⋅ − ⋅ = ⋅ ⋅ − − = ⋅ − ⋅ + =− ⋅ + ⋅ +
−
where C is a constant.
c) lim[ 2 ( 1 ) ] lim[ 2 ( 1 ) 2 ]
0
−
→∞
−
→∞
∞ −
b
b a
b x
b
x e dx x e b e
say a : s (^) n a n
→∞
lim. Then, s (^) n a n
+^ =
→ ∞
lim 1 , and so ( )
→∞ →∞
→∞ →∞
n
n n
n n n n
n n
lim a lim s 1 s lim s 1 lim s
= a − a = 0.
b) If lim = 0 → ∞
n n
a , we can not say that the series (^) ∑
∞
n = 0
an is convergent. To see that, consider
the Harmonic Series (^) ∑
∞
= 1
n n^
. Here n
a (^) n
= and lim = 0 → ∞
n n
a , but one knows that the Harmonic
Series is divergent.
6-a)
( ) ( )
( ) ( ) 3
lim 2
2 / 3
lim
(^1 )
=
⋅ +
→∞
→∞
x
n
n x
x n
x n
n n n
n n
n
Therefore, the series converges if 1 ( 2 3 ) 3
x
x and diverges if
1 ( 2 3 ) 3
x
x
. Hence the radius of convergence is: R = 3.
b) Endpoints of the interval of convergence are − 2 m 3 :-5 and 1.
When
∑ ∑
∞
=
∞
=
1 1
n
n
n
n
n
n n
x. This new series is convergent by the
Alternating series Test, because
n n
is decreasing and 0
lim = →∞ n
n
When
∑ ∑
∞
=
∞
=
1 1
n n
n
n
n n
x. This series is a divergent p-series (^)
p.
Therefore the interval of convergence is[ − 5 , 1 ).
7) Taylor series for a function f ( ) x about x = 1 is ( )
( )
n n
n
f x n
f x 1 1 !
0
= (^) ∑ ⋅ ⋅ −
∞
=
. So
( )
n n
n
x n
x ln 1 1 !
ln
0
= (^) ∑ ⋅ ⋅ −
∞
=
( ) ( ) ln( ) 1 1
ln 1 = 0 , ln′ = ⇒ ′ = x
x
( ) ln( ) 1 1
ln 2
x
x
( ) ln ( ) 1 2
ln 3
x
x
ln
( )
( )
ln ( ) 1 3!
4
4 = − ⇒ =− x
x
( )
( )
ln ( ) 1 4!
ln
5 5
5 = ⇒ = x
x , (inductively)
( )
x n n
n n ln 1 1 !,
1 Ù
and
( )
( )
ln 1 ln 1 1 1!
1 1
n n
n
n n n n
.
Hence
ln
2 3 4
1
1
±
= (^) ∑
∞
=
x x x x n
x x
n
n n
Alternative solution:
Let y = x − 1.
0
1
1 ln ln 1
y n y
n n
n n n
y
n
u u du u
du y x x y
∞ +
=
∞
=
∫ ∫∑^ ∑
∑ ∑ ∑
∞
=
∞ +
=
∞ +
=
− ⋅ −
1
1
0
1
0
1 1 1
m
m m
n
n n
n
n n
m
x
n
x
n
y .
