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Math 156 Applied Honors Calculus II Final Exam Review Sheet Solutions Fall 20171. True or False
a) TRUE. 1
n
+ 2
n
+ 3
n
+ · · ·+ n
n
=nP
i=1
i
n
= 1
n
·nP
i=1
i = 1
n
· n(n+1)
2
= n+1
2
b) TRUE. limn!1
nPi=1
f 0(xi
)�x =Rb
a
f 0(x)dx = f(b)� f(a) by FTC
c) FALSE. If n is doubled, then �x decreases by a factor of 1
2
and the error in the right-hand Riemann sum decreases bya factor of 1
2
, not 1
4
, as we saw in examples in class.
d) TRUE. Apply integration by parts. TO BE COMPLETED
e) FALSE. SinceR1
0
dx
x
2 diverges by p�test )R10
dx
x
2 diverges.
f) FALSE. When the spring is stretched from length 20 cm to 30 cm, the work done isR30�L
20�L
kxdx =R30�20
20�20
kxdx =R10
0
kxdx = 1
2
kx2
��100
= 50k = 2Joule. Then when the spring is stretched from length 30 cm to 40 cm, the work done isR40�L
30�L
kxdx =R40�20
30�20
kxdx =R20
10
kxdx = 1
2
kx2
��2010
= 1
2
k(400� 100) = 150k = 3 · 50k = 6Joule.
g) FALSE. TO BE COMPLETED
h) FALSE. A counterexample is an exponential distribution, f(x) attains its maximum value at x = 0 rather than µ = 1
c
.(However the statement is true for a normal distribution.)
i) TRUE.R1�1(x� µ)f(x)dx =
R1�1 xf(x)dx�
R1�1 µf(x)dx = µ� µ · 1 = 0
j) FALSE. After 100 years the sample has mass 1
2
kg, and after 400 years the sample has mass�1
2
�4
= 1
16
kg.
k) TRUE. continuous compounding ) y(t) = y0
ert = 1000e0.05t ) y(2) = 1000e0.05·2 = 1000e0.1; hence we need to showthat 1105 < 1000e0.1 < 11121st part : we need to show that 1105 < 1000e0.1; recall that ex = 1 + x + 1
2
x2 + · · · , so e0.1 = 1 + 0.1 + 1
2
(0.1)2 + · · · =1 + 0.1 + 0.005 + · · · ; then summing the first three terms yields e0.1 = 1.105 + · · · , and since the remainder is positive,we have e0.1 > 1.105, thus 1000e0.1 > 1000 · 1.105 = 11052nd part : we need to show that 1000e0.1 < 1112; this is equivalent to showing that e0.1 < 1.112; in this case letus consider e�x; it is fairly easy to see that e�x > 1 � x for x 6= 0; for example, consider the graph of e�x and1 � x; then let us set x = 0.1, so that we have e�0.1 > 1 � 0.1; this implies that e�0.1 > 0.9; and this implies thate0.1 < 1
0.9
= 1
910
= 10
9
= 1.1111 · · · < 1.112; so we’re done
l) FALSE. y(t) = 0 is a constant solution of the di↵erential equation, but it is unstable by looking at the phase plane.
m) FALSE. This is only true if the constant solution is stable.
n) FALSE. If the step size �t decreases, then the error also decreases.
o) FALSE. A counterexample is an
= 1
n
, bn
= n2.
p) FALSE. A counterexample is an
= 1
n
2 , bn = 1
n
, so that1P
n=0
an
converges, but1P
n=0
bn
diverges.
q) FALSE.
Method 1: draw a graph of y = 1
x
2 for x � 1 and note that1P
n=1
1
n
2 is a left-hand Riemann sum forR11
dx
x
2 , and hence the
sum is larger than the integral.
Method 2:1P
n=1
1
n
2 = ⇡
2
6
,R11
dx
x
2 = �1
x
��11
= 1
r) FALSE.Method 1: The error bound for a convergent alternating series says that |s � s
n
| < an+1
. If we set n = 1, then|s� s
1
| < a2
) |s� 1| < 1
2
) 1
2
< s < 3
2
, so s 6= 0.Method 2: In class we showed that 1� 1
2
+ 1
3
� 1
4
+ · · · = ln 2 6= 0.
s) FALSE.1P
n=1
1
n(n+1)
=1P
n�1
( 1n
� 1
n+1
) =1P
n=1
1
n
�1P
n=1
1
n+1
= 1�1 = 0
The 1st step is correct, but the 2nd step is incorrect; the original sum is equal to 1 (as shown on homework), so it isincorrect to write 1 = 1�1.
t) FALSE. The AST cannot be used to show that a series diverges.
u) FALSE. limn!1
���an+1
a
n
��� = limn!1
n
2
(n+1)
2 = 1 ) the ratio test is inconclusive
v) FALSE. A counterexample isP
n=1
(�1)nxn, which converges for x = 1 by the AST, but diverges for x = �1 since itis the harmonic series in that case.
1
w1) TRUE. The radius of convergence is at least R = 1, so the interval of convergence is at least 0 < x 2, whichcontains x = 1
2
.
w2) FALSE. geometric series, r = 2(x�1), converges for�1 < r < 1 , �1 < 2(x�1) < 1 , � 1
2
< x�1 < 1
2
, 1
2
< x < 3
2
,diverges at end points x = 1
2
, 3
2
x) TRUE. 1
1+x
= 1
1�(�x)
=1P
n=0
(�1)nxn, then di↵erentiating both sides yields � 1
(1+x)
2 =1P
n=1
(�1)nnxn�1 ) 1
(1+x)
2 =
1Pn=1
(�1)n+1nxn�1 =1P
n=0
(�1)n(n+ 1)xn, and we can check that the ioc is �1 < x < 1
y) FALSE. Don’t try to find f (3)(0), f (6)(0) directly. Instead use the Taylor series to derive them, since the general form
of a Taylor series is f(x) =1P
n=0
cn
xn, where cn
= f
(n)(0)
n!
) f (n)(0) = n! · cn
. Since ex = 1 + x + 1
2
x2 + 1
3!
x3 + · · · , then
e�x
2
= 1 � x2 + 1
2
x4 � 1
6
x6 + · · · . Then c3
= 0, since there is no x3 term, and hence f (3)(0) = 0. Then c6
= � 1
6
, thus
f (6)(0) = 6! · c6
= 720 ·�� 1
6
�= �120. Thus the statement is false.
z) TRUE.part 1 : Since ex = 1 + x + 1
2
x2 + 1
3!
x3 + · · · , then setting x = 1, we have e = 1 + 1 + 1
2
+ 1
3!
+ · · · , and it is clear thate > 2.part 2 : Setting x = �1 in the power series for ex, we have e�1 = 1� 1 + 1
2
� 1
3!
+ · · · , then using the error bound for aconvergent alternating series we have |e�1 � 1
2
| < 1
6
) 1
3
< e�1 < 2
3
, then e�1 > 1
3
) e < 3.
aa) TRUE. ex = 1 + x + 1
2
x2 + · · · ) e�x
2
= 1 + (�x2) + 1
2
(�x2)2 + · · · = 1 � x2 + 1
2
x4 � · · · Next we note that
e�x
2
> 1 � x2 for x > 0; to show this we can sketch the graph of e�x
2
and 1 � x2. To confirm the sketch, consider thefunction f(x) = e�x
2 � (1� x2), note that f(0) = 0 and f 0(x) = �2xe�x
2
+ 2x = 2x(1� e�x
2
) > 0 for x > 0, so f(x) is
an increasing function, and this implies f(x) > 0 for x > 0, and hence we have e�x
2
> 1 � x2 for x > 0. Then we haveR1
0
e�x
2
dx >R1
0
(1� x2)dx = (x� x
3
3
)��10
= 1� 1
3
= 2
3
.
bb) TRUE. Since the series is sin ⇡
2
(note that sinx = x� 1
3!
x3 + 1
5!
x5 � 1
7!
x7 + · · · ) and sin ⇡
2
= 1.
cc) TRUE. cosh2x� sinh2x =�e
x
+e
�x
2
�2 �
�e
x�e
�x
2
�2
= e
2x+2+e
�2x
4
� e
2x�2+e
�2x
4
= 1.
dd) FALSE.Rtanhxdx =
Rsinh x
cosh x
dx =R
1
cosh x
d coshx = ln(coshx) 6= sech2h. Note that (tanhx)0 = sech2x.
ee) TRUE. Note that T1
(x) = f(a) + f 0(a)(x� a), thus T 01
(a) = f 0(a).
↵) FALSE. In class we showed that the function defined by f(x) = e�x
2
for x 6= 0 and f(0) = 0 has the property thatf (n)(0) = 0 for all n, yet f(x) 6= 0 for x 6= 0.
gg) TRUE. ex = 1+x+ 1
2
x2+ · · · , so e�0.1 = 1+(�0.1)+ 1
2
(�0.1)2+ · · · = 1� 0.1+0.005� · · · , so |e�0.1� 0.9| 0.005,so 0.895 e�0.1 0.905.
hh) FALSE. Using the binomial series (1 + x)k = 1 + kx + · · · , replace x with x2, then (1 + x2)k = 1 + kx2 + · · · , andthen setting k = 1
2
, we havep1 + x2 = (1 + x2)
12 = 1 + 1
2
x2 + · · · .
ii) TRUE. Since cosh ix = e
ix
+e
�ix
2
= cos x+i sin x+cos x�i sin x
2
= cosx
jj) TRUE.
part 1 :R⇡/2
0
sin2 ✓d✓ =R⇡/2
0
( 12
� 1
2
cos 2✓)d✓ = ( 12
✓ � 1
2
· 1
2
sin 2✓)��⇡/20
= ⇡
4
part 2 :R⇡/2
0
cos2 ✓d✓ =R⇡/2
0
( 12
+ 1
2
cos 2✓)d✓ = ( 12
✓ + 1
2
· 1
2
sin 2✓)��⇡/20
= ⇡
4
kk) TRUE. Since e⇡i = cos⇡ + i sin⇡ = �1 ) ⇡i = log(�1). (Actually, more rigorously, log(�1) = (2k + 1)⇡i, wherek is an integer.)
ll) TRUE.�8
3
�= 8!
3!5!
= 8·7·63·2·1 = 8 · 7 = 56
mm) FALSE.�6
3
�= 6!
3!3!
= 6·5·43·2·1 = 20
nn) TRUE.�10
2
�= 10!
2!·8! =10!
8!·2! =�10
8
�
oo) TRUE.�7
3
�+�7
4
�= 7!
3!·4! +7!
4!·3! =2·7!3!·4! =
8·7!4·3!·4! =
8!
4!·4! =�8
4
�
pp) TRUE. (a+ b)k =kP
n=0
�k
n
�ak�nbn, then setting a = 1, b = �1 yields
kPn=0
�k
n
�(�1)n = (1 + (�1))k = 0k = 0
Question 2 Solutiona) geometric series : sum = 1
1� 20162017
= 2017
2017�2016
= 2017
b) It equalsR1
0
(1 + x)dx =⇣x+ x
2
2
⌘���1
0
= 3
2
(change i
n
! x, 1
n
! dx)
c) It equalsR1
0
1
1+x
dx = ln(1 + x)|10
= ln 2
2
d) By L’Hospital’s rule, limx!0
sin x
x
= limx!0
(sin x)
0
(x)
0 = limx!0
cos x
1
= cos 0 = 1.
e) By L’Hospital’s rule, limx!0
1�cos x
x
2 = limx!0
(1�cos x)
0
(x
2)
0 = limx!0
sin x
2x
= 1
2
.
f) Note that limn!1
�1 + x
n
�2n
= ( limn!1
�1 + x
n
�n
)2 = (ex)2 = e2x
g) By L’Hospital’s rule limx!0
p1+x�1
x
= limx!0
(
p1+x�1)
0
x
0 = limx!0
12p
1+x
1
= 1
2
h) limh!0
(x+h)
4�x
4
h
= limh!0
x
4+4x
3h+6x
2h
2+4xh
3+h
4�x
4
h
= limh!0
4x
3h+6x
2h
2+4xh
3+h
4
h
= limh!0
(4x3 + 6x2h+ 4xh2 + h3) = 4x3 (one
can use L’hospital rule, note that ‘h’ is variable here, regard x as constant)
i) By L’Hospital’s rule, limh!0
f(x+h)�f(x)
h
= limh!0
f
0(x+h)
1
= f 0(x).
j) Using L’Hospital’s rule twice, limh!0
f(x+h)�2f(x)+f(x�h)
h
2 = limh!0
f
0(x+h)�f
0(x�h)
2h
= limt!0
f
00(x+h)+f
00(x�h)
2
= f 00(x).
k) limh!0
Rh
0 f(x)dx
h
L’Hospital Rule������������! limh!0
(
Rh
0 f(x)dx)
0
(h)
0 = limh!0
f(h)
1
= f(0)
l) limh!0
Rh
0 xf(x)dx
h
2
L’Hospital Rule������������! limh!0
(
Rh
0 xf(x)dx)
0
(h
2)
0 = limh!0
hf(h)
2h
= limh!0
f(h)
2
= f(0)
2
integration
Question 3 Solution
a) TO BE COMPLETED
b) TO BE COMPLETED
c) ex =1P
n=0
1
n!
xn ) e�x
2
=1P
n=0
1
n!
(�x2)n =1P
n=0
1
n!
(�1)nx2n )Re�x
2
dx =R 1P
n=0
1
n!
(�1)nx2ndx =1P
n=0
1
n!
(�1)n 1
2n+1
x2n+1 =
x� 1
3
x3 + 1
10
x5 � 1
42
x7 + · · · .d) TO BE COMPLETED
e) integration by parts :Rx sinxdx =
Rx(�1)d cosx = �
Rxd cosx = �x cosx +
Rcosxdx = �x cosx + sinx =
sinx� x cosx
f) use integration by parts twiceonce :
Re�x sinxdx =
Re�x(�1)d cosx = �e�x cosx+
Rcosxde�x = �e�x cosx�
Re�x cosxdx
twice :Re�x cosxdx =
Re�xd sinx = e�x sinx�
Rsinxde�x = e�x sinx+
Re�x sinxdx
)Re�x sinxdx = �e�x cosx� e�x sinx�
Re�x sinxdx ) 2
Re�x sinx = �e�x cosx� e�x sinx
)Re�x sinx = 1
2
(�e�x cosx� e�x sinx) = � 1
2
e�x(cosx+ sinx)
g)R
dx
4x
2 = � 1
4x
h)R
x
4+x
2 dx =R 1
24+x
2 dx2 =
R 12
4+x
2 d(4 + x2) = 1
2
ln(4 + x2)
If you do not like the above way, instead use a change of variable u = 4 + x2, du = 2xdx ) dx = 1
2x
du.Rx
4+x
2 dx =R
x
u
· 1
2x
du =R
1
2u
du = 1
2
lnu+ C = 1
2
ln(4 + x2)
i) For this type of problem, one needs to use a trig substituion, x = 2 tan ✓, dx = 2(1 + tan2 ✓)d✓Rdx
4+x
2 =R
2(1+tan
2✓)
4+4 tan
2✓
d✓ =R
1
2
d✓ = 1
2
✓ = 1
2
arctan x
2
j) One can use the formula 1 + sinh2 ✓ = cosh2 ✓, and the change of variable x = 2 sinh ✓, dx = 2 cosh ✓d✓.Rdxp4+x
2 =R
2 cosh ✓p4+4 sinh
2✓
d✓ =R
2 cosh ✓p4 cosh
2✓
d✓ =R
2 cosh ✓
2 cosh ✓
d✓ =Rd✓ = ✓ = arcsinhx
2
= sinh�1
x
2
k) Using partial fractionsR
dx
4�x
2 =R
dx
(2+x)(2�x)
=R ⇣
1/4
2+x
+ 1/4
2�x
⌘dx =
R1/4
2+x
dx+R
1/4
2�x
dx = 1
4
ln |2 + x|� 1
4
ln |2� x|
l) partial fractions againR
dx
4x�x
2 =R
dx
x(4�x)
=R ⇣
1/4
x
+ 1/4
4�x
⌘dx =
R1/4
x
dx+R
1/4
4�x
dx = 1
4
ln |x|� 1
4
ln |4� x|
m) TO BE COMPLETED
n) using integration by parts once and using the equality sin2 x+ cos2 x = 1Rsin2 xdx =
Rsinx · sinxdx =
Rsinx(�1)d cosx = � sinx · cosx+
Rcosxd sinx = � sinx · cosx+
Rcos2 xdx = � sinx ·
cosx +R(1 � sin2 x)dx = � sinx · cosx +
R1dx �
Rsin2 xdx ) 2
Rsin2 xdx = � sinx · cosx + x )
Rsin2 xdx =
� 1
2
sinx cosx+ x
2
= x
2
� 1
4
sin 2xA simple way, using sin2 x = 1�cos 2x
2
, sin 2x = 2 cosx sinx, (cos 2x)2 = (1 + cos 4x)/2thus
Rsin2 xdx =
R1�cos 2x
2
dx = x
2
� sin 2x
4
o) using the equality sin2 x+ cos2 x = 1Rsin3 xdx =
Rsin2 x · sinxdx =
R(1 � cos2 x)(�1)d cosx = �
R(1 � cos2 x)d cosx = �
R1d cosx +
Rcos2 xd cosx =
� cosx+ 1
3
cos3 x
3
p) using integration by parts will be too complicated, using sin2 x = 1�cos 2x
2
, sin 2x = 2 cosx sinx, (cos 2x)2 = (1 +cos 4x)/2
ThusRsin4 x =
R �1�cos 2x
2
�2
dx =R
1�2 cos 2x+cos
22x
4
dx =R
1�2 cos 2x+1�sin
22x
4
dx
=R �
1
2
� 1
2
cos 2x� 1
4
sin2 2x�dx = x
2
� 1
4
sin 2x� 1
8
Rsin2 2xd(2x)
using 3 i)x!2x����������! x
2
� 1
4
sin 2x� 1
8
�2x
2
� sin 4x
4
�= 3
8
x� 1
4
sin 2x+ 1
32
sin 4x
Question 4 Solution
a)R2
p3
0
x
3p16�x
2 dx TO BE COMPLETED
b)R1�1 x2
1p2⇡
e�x
2
2 dx TO BE COMPLETED
c)R1�1(x� 1)2 1
2
p2⇡
e�(x�1)2
8 dx TO BE COMPLETED
Question 5 SolutionUsing variable substitution u = ⇡
2
� x, du = �dxR0
⇡/2
sin(
⇡
2 �u)
sin(
⇡
2 �u)+cos(
⇡
2 �u)
(�1)du = �R0
⇡/2
cosu
cosu+sinu
du =R⇡/2
0
cosu
cosu+sinu
du
both u and x are integral variables changing from 0 to 1
2
, one may replace them with ✓, thusR⇡/2
0
sin ✓
sin ✓+cos ✓
d✓ =R⇡/2
0
cos ✓
sin ✓+cos ✓
d✓ on one hand, one the other handR⇡/2
0
sin ✓
sin ✓+cos ✓
d✓+R⇡/2
0
cos ✓
sin ✓+cos ✓
d✓ =R⇡/2
0
⇣sin ✓
sin ✓+cos ✓
+ cos ✓
sin ✓+cos ✓
⌘d✓ =
R⇡/2
0
1d✓ = ⇡
2
.
ThusR⇡/2
0
sin ✓
sin ✓+cos ✓
d✓ =R⇡/2
0
cos ✓
sin ✓+cos ✓
d✓ = ⇡
4
Question 6 Solutiona) convergent by p�test
R11
1
x
2 dx =R10
�� 1
x
�0dx = � 1
x
��11
= 1
b) divergent by p�testR11
dx
x
= lnx|11
= 1c)
R11
dx
x�1
=R10
dy
y
divergent by p�test
d) divergent by p-testR1
0
dx
x
2 = � 1
x
��10
= 1e) convergent by p-test
R1
0
dxpx
= 2px|1
0
= 2
f) divergent by p-test Since bothR0
�1
dx
x
andR1
0
dx
x
diverges.
Question 7 Solutionsubstitute u = r2 � 2rx+ a2, so du = �2rdxlimits x = �a ) u = r2 � 2r(�a) + a2 = r2 + 2ra+ a2 = (r + a)2, x = a ) u = r2 � 2ra+ a2 = (r � a)2
V (r) = q
2a
Ra
�a
dxpr
2�2rx+a
2 = q
2a
R(r�a)
2
(r+a)
2
� 12r dupu
= � q
4ar
· 2pu��(r�a)
2
(r+a)
2 = � q
2ar
(|r � a|� |r + a|) =(
q
r
if r � aq
a
if 0 r a
Question 8a Solution
Assume the aquarium has length l, width w, and height h, let x be the vertical coordinate with x = 0 at the bottom ofthe aquarium (positive x is up), divide the water into layers of width �x = h
n
, each layer is a rectangular box with volumelw�x, the force on a layer is ⇢glw�x, the layer at level x
i
= i�x is raised a distance h�xi
, after letting n ! 1,�x ! 0,
the work done is W =Rh
0
⇢glw(h� x)dx = ⇢glw(hx� x
2
2
)��h0
= 1
2
⇢glwh2, substituting the given values, the work done is
W = 1
2
⇢g · 2 · 0.5 · 12 = 1
2
⇢g Joule. From the formula W = 1
2
glwh2, we see that if the width w is doubled, then the workis also doubled, and if the height h is doubled, then the work is multiplied by 4.
Question 8b Solution TO BE COMPLETED
Question 9 Solution
a) On x-axis, since one ion is held fixed at x = 0, the distance is x, replace r in F = � q
2
r
2 with x, Work =RF (x)dx =
R2
3
� q
2
x
2 dx = q
2
x
���2
3
= q
2
2
� q
2
3
= q
2
6
b) On x-axis, since one ion is held fixed at x = 1, the distance becomes x � 1, replace r in F = � q
2
r
2 with x � 1,
Work =RF (x)dx =
R2
3
� q
2
(x�1)
2 dx = q
2
x�1
���2
3
= q
2
2�1
� q
2
3�1
= q
2
2
c) The work can be calculated with respect to A(x = 0) and B(x = 1) separately, then put together. Add the results in
a) and b) together Work= q
2
6
+ q
2
2
= 2
3
q2.d) Divide the rod into many small pieces, each with width �w; here we use w to denote the position of small pieces(w changes from 0 to 1, it overlaps x = 0 to x = 1), for each piece the charge is q�w, and the force F (x) = � q·q�w
(x�w)
2 ,
4
work contributed by each piece =R2
3
F (x)dx =R2
3
� q
2�W
(x�w)
2 dx = q
2�w
x�w
���2
3
=⇣
1
2�w
� 1
3�w
⌘q2�w. Then we need a second
integral for w from 0 to 1 to sum all the pieces, Total Work=R1
0
⇣1
2�w
� 1
3�w
⌘q2�w
�w!dw�����!R1
0
⇣1
2�w
� 1
3�w
⌘q2dw =
q2 [� ln(2� w) + ln(3� w)]|10
= q2 ln 4
3
⇡ 0.28q2 (this result is reasonable since it is larger than q
2
6
⇡ 0.16q2 (case a) and
smaller than q
2
2
⇡ 0.5q2 (case b), cases a and b are two extreme cases (if we put all charge to one end of the rod), giventhat in the three cases (a,b,d) the total charge is the same.
Question 10 Solution
f(x) = coshx ) f 0(x) = sinhx ) 1 + (f 0(x))2 = 1 + sinh2 x = cosh2 x )p
1 + (f 0(x))2 = coshx
a) arclength =R1
�1
p1 + (f 0(x))2dx =
R1
�1
coshxdx = sinhx|1�1
= sinh 1� sinh(�1) = 2 sinh 1
b) surface area =R1
�1
2⇡f(x)p1 + (f 0(x))2dx = 2⇡
R1
�1
cosh2 xdx = 2⇡R1
�1
⇣e
x
+e
�x
2
⌘2
dx = 2⇡R1
�1
⇣e
2x+2+e
�2x
4
⌘dx
= 2⇡
4
⇣e
2x
2
+ 2x+ e
�2x
�2
⌘ ��1�1
= ⇡
2
⇣e
2
2
+ 2 + e
�2
�2
�⇣
e
�2
2
� 2 + e
2
�2
⌘⌘= ⇡
2
(e2+4�e�2) = ⇡
2
(2 sinh 2+4) = ⇡(sinh 2+2)
You can save some time using the fact that coshx is an even function, soR1
�1
· · · = 2R1
0
· · · .
Question 11 Solution
x = M
y
m
, y = M
x
m
case 1: m =Rb
a
f(x)dx, Mx
= 1
2
Rb
a
f2(x)dx, My
=Rb
a
xf(x)dx
case 2: m =Rb
a
(f(x)� g(x))dx, Mx
= 1
2
Rb
a
(f2(x)� g2(x))dx, My
=Rb
a
x(f(x)� g(x))dx
answers : (x, y) = (a)�3
2
, 6
5
�, (b)
�3
4
, 12
5
�, (c)
�0, 2
5
�, (d)
�1, 1
4
�
(b) m =R2
0
(4� x2)dx = (4x� x
3
3
)��20
= 8� 8
3
= 16
3
Mx
= 1
2
R2
0
(16� x4)dx = 1
2
(16x� x
5
5
)��20
= 1
2
(32� 32
5
) = 64
5
) y = M
x
m
= 64
5
· 3
16
= 12
5
My
=R2
0
(4x� x3)dx = (4x
2
2
� x
4
4
)��20
= 8� 16
4
= 4 ) x = M
y
m
= 4 · 3
16
= 3
4
(d) m =R10
1
1+x
2 dx ( = arctanx��10
= ⇡
2
)
change variable: x = tan ✓, dx = (1 + tan2 ✓)d✓, x = 0 ) ✓ = 0, x = 1 ) ✓ = ⇡
2
, since tan 0 = 0, tan ⇡
2
= 1m =
R10
1
1+x
2 dx =R ⇡
2
0
1+tan
2✓
1+tan
2✓
d✓ =R ⇡
2
0
1d✓ = arctanx��10
= ⇡
2
x =R 10 xf(x)dx
m
=R 10
x
1+x
2 dx
m
=R 10
12dx
2
1+x
2⇡
2=
12 ln(1+x
2)
��10
⇡
2= 1 !!!
The area is finite, but it has an infinitely large x.
y =R 10
12 f
2(x)dx
m
= 1
m
·R10
1
2
f2(x)dx = 1
⇡
2
R10
1
2
⇣1
1+x
2
⌘2
dx = 2
⇡
R10
12
(1+x
2)
2 dx = 2
⇡
R10
12+
12x
2� 12x
2
(1+x
2)
2 dx
= 2
⇡
⇣R10
12 (1+x
2)
(1+x
2)
2 dx�R10
12x
2
(1+x
2)
2 dx⌘= 2
⇡
⇣R10
12
1+x
2 dx�R10
12x·x
(1+x)
2 dx⌘
= 2
⇡
⇣1
2
arctanx��10
�R10
12x·
12 dx
2
(1+x
2)
2
⌘= 2
⇡
h1
2
· ⇡
2
+R10
x
4
d⇣
1
1+x
2
⌘i= 2
⇡
⇣⇡
4
+ x
4
· 1
1+x
2
���1
0
� 1
4
R10
1
1+x
2 dx⌘
= 2
⇡
⇣⇡
4
+ x
4
· 1
1+x
2
���1
0
� 1
4
R10
1
1+x
2 dx⌘= 2
⇡
⇣⇡
4
+ x
4
· 1
1+x
2
���1
0
� 1
4
arctanx��10
⌘
= 2
⇡
�⇡
4
+ 0� 1
4
· ⇡
2
�= 2
⇡
· ⇡
8
= 1
4
Question 12 Solutionf(t) = ce�ct, where c = 1
1000
and t � 0
a) Prob(0 t 200) =R200
0
ce�ctdt = �e�ct|2000
= 1� e�15 ⇡ 0.18
b) Prob(t � 800) =R1800
ce�ctdt = �e�ct|1800
= e�45 ⇡ 0.45
Question 13 SolutionFirst notice that f(x) � 0 for all x. Now we need only show that
R1
0
f(x)dx = 1R1
0
f(x)dx =R1
0
1
⇡
px(1�x)
dxx=u+
12�����!
R 12
� 12
1
⇡
p(u+
12 )(1�u� 1
2 )du =
R 12
� 12
1
⇡
p(
12+u)(
12�u)
du
=R 1
2
� 12
1
⇡
p14�u
2du
u=
12 sin ✓
������!R ⇡
2
�⇡
2
1
⇡· 12 cos ✓
· 1
2
cos ✓d✓ = 1
⇡
✓��⇡
2
�⇡
2= 1
An alternative approach to the integral is to make the substitution w =px, x = w2, dw = dx/(2
px). This givesR
1
0
f(x)dx =R1
0
1
⇡
px(1�x)
dx =R1
0
2dw
⇡
p1�w
2 = 2
⇡
arcsinw]10
= 2
⇡
(⇡/2� 0) = 1. The alternative is more e�cient but requires
some ingenuity to think of.
Di↵erential Equations
5
Question 14 Solutiona) y0 = �2y, y
0
= 1 (standard exponential decay) ) y = C · e�2t.y0
= 1 ) C = 1 ) y(t) = e�2t and limt!1
y(t) = 0
b) y0 = 1�2y ) dy
dt
= 1�2y separation of variables ) dy
1�2y
= dt integrate both sides ) � 1
2
ln |1� 2y| = t+C
) 1� 2y = C · e�2t ) y = 1�C·e�2t
2
y0
= 0 ) C = 1 ) y = 1�e
�2t
2
and limt!1
y(t) = 1
2
c) y0 = 1�y2 ) dy
dt
= (1+y)(1�y) separation of variables ) dy
(1+y)(1�y)
= dt partial fraction )12
1+y
dy+12
1�y
dy =
dt integrate both sides ) 1
2
ln |1 + y| � 1
2
ln |1� y| = t + C ) ln��� 1+y
1�y
��� = 2t + C ) 1+y
1�y
= C · e2t where C is
constant may be positive or negative ) y(t) = C·e2t�1
C·e2t+1
y0
= 0 ) C = 1 ) y(t) = e
2t�1
e
2t+1
. Furthermore y(t) = (e
2t�1)e
�t
(e
2t+1)e
�t
= e
t�e
�t
e
t
+e
�t
= tanh t
Check... y(t) = tanh t is the solution.limt!1
y(t) = 1
d) y0 = �ty ) dy
dt
= �ty separation of variables ) dy
y
= �tdt integrate both sides ) ln |y| = � 1
2
t2 + C )y = Ce�
12 t
2
y0
= 1 ) C = 1 ) y(t) = e�12 t
2
limt!1
y(t) = 0
Question 15 Solutiona) y = c
1
et + c2
e�t ) y0 = c1
et � c2
e�t ) y00 = c1
et + c2
e�t = y, thus it is a solution of y00 = y for any constantsc1
, c2
.b) y(0) = 1, y0(0) = 0 ) c
1
+ c2
= 1, c1
� c2
= 0 ) c1
= c2
= 1
2
) y(t) = 1
2
et + 1
2
e�t = coshxc) y(0) = 0, y0(0) = 1 ) c
1
+ c2
= 0, c1
� c2
= 1 ) c1
= 1
2
, c2
= � 1
2
) y(t) = 1
2
et � 1
2
e�t = sinhx
Question 16 Solution a) y0 = ky. Solve the equation, we have y(t) = y0
ekt. 200 = y(30) = y0
e30k and 800 = y(90) =y0
e90k. Therefore, y0
= 102 = 100 cells.b)200 = 100e30k, so k = ln 2
30
. Therefore, y(t) = 100 · 2t/30. Solve the equation 6400 = 100 · 2t/30. Then t =30 ln 64/ ln 2 = 30 · 6 = 180 hours.
Question 17 Solutiony(t) = y
0
e�kt
y(t) = 40 ·�1
2
� t
1.4⇥10�4
30 = 40 ·�1
2
� t
1.4⇥10�4
t = 1.4⇥ 10�4
ln
34
ln
12= 0.581⇥ 10�4s
Question 18 Solutiony(t) : tiger body mass (kg) as a function of time t (day)y0 = rate in� rate out = 2500 cal
day
· 1 kg
10000 cal
� 20 cal · y kg
10000 cal
· 1
day
= 2500�20y
10000
kg
day
y0 = � 1
500
(y � 125) Newton’s heating/cooling y0 = k(y � T )
y(t) = T + (y0
� T )ekt = 125 + (y0
� 125)e�t
500 ) limt!1
y(t) = 125 kg
Question 19 Solutiony(t) = T + (y
0
� T )e�kt , note that the patient’s temperature is T , y0
= 70�Fy(1) = 95 = T + (70� T )e�k
y(2) = 100 = T + (70� T )e�2k
) ( 95�T
70�T
)2 = 100�T
70�T
) (95� T )2 = (100� T )(70� T ) ) T 2 � 190T + 952 = T 2 � 170T + 70000 ) 20T = 2025) T = 101.25� F
Question 20 Solutiony0 = ky(M � y)y(t) = My0
y0+(M�y0)e�kMt
y0
= 10, M = 4000y(t) = 40000
10+(4000�10)e
�4000kt
measure time in days20 = 40000
10+(4000�10)e
�4000·7k
6
e�k =�199
399
� 128000
y(t) = 40000
10+3990( 199399 )
t
7
let y(t) = 1
2
· 4000 = 2000, solve t = 7 ln 399
ln 399�ln 199
⇡ 60 days.
Question 21 SolutionThis equation, y0 = f(y) with f(y) = 2y, is particularly simple to study with Euler’s Method. The recursion for Euler’s
Method in this case is uk+1
= uk
+ (�t)f(uk
) = (�t)(2uk
). After factoring, uk+1
= (1 + 2�t)(uk
). This means that, inthis case, we don’t actually need to do each iteration of the process; we can foresee that u
k
= (1+2�t)ku0
= (1+2�t)k.So u
n
= (1 + 2/n)n, where n = 1/�t is the number of steps.�t u
n
1 (1 + 2)1 = 31/2 (1 + 1)2 = 41/4 (1 + 1/2)4 = 5.0625Remark. The general solution of this di↵erential equation is y = Ce2t, and the solution which fits our initial condition
is y = e2t. Thus the true answer is y(1) = e2 ⇡ 7.389. We need to use smaller steps to get a satisfactory approximation.With 1000 steps, we would still only have u
n
⇡ 7.374.
Series
Question 22 Solution
a) divergent1P
n=1
1
2n
= 1
2
1Pn=1
1
n
by p-test of series, p = 1.
b) convergent since1P
n=1
1
2
n
=1P
n=1
�1
2
�n
= 1
2
1Pn=0
�1
2
�n
= 1
2
· 1
1� 12= 1 < 1.
c) convergent by p-test of series, p = 2.d) convergent by Alternating Series Test, lim
n!1an
= limn!1
1
n
2 = 0, an+1
< an
and the sign is alternating.
e) divergent by Ratio Test, L = limn!1
���an+1
a
n
��� = limn!1
2
n+1
(n+1)
2 · n
2
2
n
= 2 > 1, (L > 1 divergent)
Question 23 Solution
a) 0.111111.... = 0.1+0.01+0.001+0.0001+ · · · = 1
10
+ 1
100
+ 1
1000
+ 1
1000
+ · · · =1P
n=1
1
10
n
= 1
10
1Pn=0
�1
10
�n
= 1
10
· 1
1� 110
= 1
9
b) 0.1212121212... = 12
100
+ 12
10000
+ 12
1000000
+ · · · =1P
n=1
12
100
n
= 12
100
1Pn=0
1
100
n
= 12
100
· 1
1� 1100
= 12
99
c) 0.4999999... = 0.45 + 0.045 + 0.0045 + 0.00045 + · · · = 45
100
+ 45
1000
+ 45
10000
+ · · ·
= 45
100
�1 + 1
10
+ 1
100
+ 1
1000
+ · · ·�= 45
100
1Pn=0
1
10
n
= 45
100
· 1
1� 110
= 1
2
(ie, 0.49999999... = 0.5)
Question 24 Solution
a) recall :1P
n=0
x
n
n!
= ex for all x, so setting x = 2 we obtain1P
n=0
2
n
n!
= e2
b) recall :1P
n=0
xn = 1
1�x
for �1 < x < 1, so setting x = 1
3
we obtain1P
n=0
1
3
n
= 1
1� 13= 3
2
)1P
n=1
1
3
n
=1P
n=0
1
3
n
�1 = 3
2
�1 = 1
2
c) di↵erentiating the first equation in part (b) with respect to x we obtain1P
n=1
nxn�1 = 1
(1�x)
2 for �1 < x < 1, so setting
x = 1
3
we obtain1P
n=1
n( 13
)n�1 = 1
(1� 13 )
2 = 9
4
)1P
n=1
n( 13
)n · ( 13
)�1 = 9
4
)1P
n=1
n( 13
)n = 9
4
· 1
3
= 3
4
Question 25 Solution
Given that1P
n=1
1
n
2 = ⇡
2
6
, note that1P
n=0
1
(2n+1)
2 includes all the odd terms.1P
n=1
1
n
2 =odd terms+even terms )1P
n=1
1
n
2 =
1Pn=0
1
(2n+1)
2 +1P
n=1
1
(2n)
2 )1P
n=1
1
n
2 =1P
n=0
1
(2n+1)
2 + 1
4
1Pn=1
1
n
2
) ⇡
2
6
=1P
n=0
1
(2n+1)
2 + 1
4
⇡
2
6
)1P
n=0
1
(2n+1)
2 = ⇡
2
6
� ⇡
2
24
= ⇡
2
8
.
Question 26 Solutiona) Use |s� s
10
| R1n
f(x)dx since all terms are positive, where f(x) = 1
x
2
|s� s10
| R110
f(x)dx =R110
1
x
2 dx = � 1
x
��110
= 0.1
b) Use |s� s10
| an+1
where an+1
= 1
(n+1)
2 since the series is an alternating series.
|s� s10
| an+1
= 1
11
2 = 1
121
7
Question 27 SolutionAssume that the dog starts with student A and runs to student B; if the time interval has duration t
0
, then we have10t
0
+ 2t0
= 20 (the distance the dog runs plus the distance B walks is equal to the distance between them when theystart). This implies that 12t
0
= 20 or t0
= 5
3
.In the next time interval of duration t
1
, the dog runs back to A; then we have 10t1
+ 2t1
= 20� 2t0
� 2t0
(the distancethe dog runs plus the distance A walks is equal to the distance between them when they start). This implies that12t
1
= 20� 4 · 5
3
= 40
3
or t1
= 10
9
= 5
3
· 2
3
.Note that everything stayed the same except that the starting distance changed from 20 to 40
3
= 20 · 2
3
.In the next time interval of duration t
2
, the dog runs back to B; then we have 10t2
+ 2t2
= 40
3
� 2t1
� 2t1
(the distancethe dog runs plus the distance B walks is equal to the distance between them when they start). This implies that12t
2
= 40
3
� 4 · 5
3
· 2
3
= 80
9
or t2
= 80
9·12 = 5
3
· ( 23
)2.Note that everything stayed the same except that the starting distance changed from 40
3
to 80
9
= 40
3
· 2
3
.The pattern repeats.
The total time is T = t0
+ t1
+ t2
+ · · · = 5
3
+ 5
3
· 2
3
+ 5
3
·�2
3
�2
+ · · · = 5
3
·⇣1 + 2
3
+�2
3
�2
+ · · ·⌘= 5
3
· 1
1� 23= 5.
Hence the dog travels a distance D = 10 mph ⇥ 5 hours = 50 miles.The question asks us to express D as an infinite series, but we could also find the distance directly as follows. The studentsmeet in the middle after walking a distance of 10 miles; since they walk at 2 mph, this must have taken 5 hours. This isalso how long the dog runs, so the dog must have run a total distance of 10 mph ⇥ 5 hours = 50 miles.
Question 28 SolutionConsider the remaining points as happening in pairs. In any pair of points, you will win both with probability p2, loseboth with probability (1� p)2, or win one and lose one with probability 2p(1� p). In the first case, the game ends andyou win. In the second case, the games and you lose. In the third case, there is no result and we are back to the samesituation.
So any scenario in which you win has the following form: you split pairs of points with your opponent any number oftimes (0, 1, 2, ...), and then you win a pair. This the probability of winning is p2 + (2p(1� p))p2 + (2p(1� p))2p2 + · · · =1P
n=0
[2p(1� p)]np2 = p21P
n=0
[2p(1� p)]n = p2 · 1
1�2p(1�p)
= p
2
1�2p+2p
2 .
p = 1
2
) p
2
1�2p+2p
2 = 1
2
(no surprise)
p = 1
4
) p
2
1�2p+2p
2 = 1
10
p = 3
4
) p
2
1�2p+2p
2 = 9
10
(no surprise that this answer and the previous would sum to 1)
Question 29 Solution
a) The total length removed = 1
3
+ 2 · 1
3
· 1
3
+ 4 · 1
3
· 1
3
· 1
3
+ · · · = 1
3
h1 + 2
3
+�2
3
�2
+ · · ·i= 1
3
· 1
1� 23= 1
b) Notice that any point which is the endpoint of a remaining interval will never be removed. Every stage doubles thenumber of remaining intervals, so this accounts for infinitely many points.
Power Series, Taylor Series
Question 30 Solution
a) L = limn!1
���an+1
a
n
��� = limn!1
���xn+1
x
n
��� = |x| < 1 ) the radius of convergence is 1; since at two end points x = ±1, the
series diverges, the interval of convergence is �1 < x < 1. The sum is 1
1�x
for �1 < x < 1.
b) L = limn!1
���an+1
a
n
��� = limn!1
��� 2n
x
n+1
2
n+1x
n
��� =��x2
�� < 1 ) |x| < 2 ) the radius of convergence is 2; since at two end points
x = ±2, the series diverges, the interval of convergence is �2 < x < 2. The sum is1P
n=0
�x
2
�n
= 1
1� x
2= 2
2�x
for �2 < x < 2.
c) L = limn!1
���an+1
a
n
��� = limn!1
��� (x�1)
n+1
(x�1)
n
��� = |x� 1| < 1 ) � 1 < x � 1 < 1 ) 0 < x < 2 ) the radius of convergence
is 1 (the length of the interval divided by 2); since at two end points x = 0 and 2, the series diverges, the interval ofconvergence is 0 < x < 2. The sum is 1
1�(x�1)
= 1
2�x
for 0 < x < 2.
d) L = limn!1
���an+1
a
n
��� = limn!1
��� nx
n+1
(n+1)x
n
��� = |x| < 1 ) � 1 < x < 1 ) the radius of convergence is 1 ; since at x = 1, the
series is harmonic series thus diverges, while at x = �1, the series converges by AST (alternating series test), the interval
of convergence is �1 x < 1. Note that xn =Rnxn�1dx ) x
n
n
=Rxn�1dx the sum is
1Pn=1
Rxn�1dx =
R 1Pn=1
xn�1dx =
R 1Pn=0
xndx =R
1
1�x
dx = � ln(1� x) = ln 1
1�x
for �1 x < 1.
ln 1
1�x
=1P
n=1
x
n
n
8
e) L = limn!1
���an+1
a
n
��� = limn!1
��� (n+1)x
n+1
nx
n
��� = |x| < 1 ) � 1 < x < 1 ) the radius of convergence is 1; since at x = ±1,
the series diverges, the interval of convergence is �1 < x < 1.
Note that (xn)0 = nxn�1,1P
n=1
nxn = x1P
n=1
nxn�1 = x1P
n=1
(xn)0 = x ·✓ 1P
n=0
xn
◆0= x ·
⇣1
1�x
⌘0= x · 1
(1�x)
2 = x
(1�x)
2
x
(1�x)
2 =1P
n=1
nxn
Question 31 Solution
Namely find cn
, such that f(x) = 1
1�x
=1P
n=0
cn
(x� 1
2
)n.
1
1�x
= 1
1�(x� 12+
12 )
= 1
1�(x� 12 )�
12= 1
12�(x� 1
2 )= 1
2
· 1
1�2(x� 12 )
= 1
2
·1P
n=0
(2(x� 1
2
))n =1P
n=0
2n�1(x� 1
2
)n
Question 32 Solutionf(x) = sinhx ) f(0) = sinh 0 = 0f 0(x) = coshx ) f 0(0) = cosh 0 = 1f 00(x) = sinhx ) f 00(0) = sinh 0 = 0f 000(x) = coshx ) f 000(0) = cosh 0 = 1...
sinhx = f(0) + f 0(0)x+ f
00(0)
2
x2 + f
000(0)
3!
x3 + · · · = x+ 1
3!
x3 + 1
5!
x5 · · · =1P
n=0
x
2n+1
(2n+1)!
coshx = (sinhx)0 =
✓ 1Pn=0
x
2n+1
(2n+1)!
◆0=
1Pn=0
⇣x
2n+1
(2n+1)!
⌘0=
1Pn=0
x
2n
(2n)!
Question 33 Solution
a) sinx =1P
n=0
(�1)n x
2n+1
(2n+1)!
= x� x
3
3!
+ x
5
5!
� · · · , cosx =1P
n=0
(�1)n x
2n
(2n)!
= 1� x
2
2!
+ x
4
4!
� x
6
6!
+ · · ·
sin2 x+ cos2 x =⇣x� x
3
3!
+ x
5
5!
� · · ·⌘2
+⇣1� x
2
2!
+ x
4
4!
� x
6
6!
+ · · ·⌘2
=⇣x2 � 2x
4
3!
+ x
6
36
+ 2x
6
5!
+ · · ·⌘+⇣1� 2x
2
2
+ x
4
4
+ 2x
4
4!
� 2 x
6
2·4! � 2x
6
6!
+ · · ·⌘= · · · = 1 +O(x8)
b) In fact we know that all terms in the power series for sin2 x+ cos2 x vanish after the first term 1, though proving it isa nice fairly involved exercise.
Question 34 Solution
Because ex =1P
n=0
x
n
n!
, e�x
2
=1P
n=0
(�x
2)
n
n!
=1P
n=0
(�1)n x
2n
n!
= 1� x2 + x
4
2
� x
6
6
+ x
8
24
� x
10
120
+ · · ·
T1
(x) = T0
(x) = 1 and T2
(x) = 1� x2. See Figure below for sketch.
Figure 1: Graph for Problem 34.
Question 35 SolutionShow that 0 f(x) < 1; lim
x!1f(x) = 1; lim
x!0
+f (n)(x) = lim
x!0
+P ( 1
x
)e�1/x, where P ( 1x
) is a polynomial of 1
x
; when x ! 0+
e�1/x ! 0 exponentially (faster than any polynomial) thus f (n)(x) ! 0 regardless of the form of P ( 1x
).
9
Question 36 Solutionmethod 1 It can be shown using Taylor series for f(x) =
px about x = a, that
px =
pa+ 1
2
pa
(x�a)� 1
8
a�32 (x�a)2+· · · .
Setting a = 9 yieldspx = 3 + 1
6
(x� 9)� 1
216
(x� 9)2 + · · · .This is a convergent alternating series (why?), so |s� s
n
| < an+1
, i.e.��px�
�3 + 1
6
(x� 9)��� < 1
216
(x� 9)2.
Setting x = 10 yields��p10� 3.16666
�� < 1
216
< 1
200
= 0.005.
The approximate value isp10 ⇡ 3.16666.
method 2 Use the binomial series, (1 + x)k = 1 + kx+ k(k�1)
2
x2 + · · · for �1 < x < 1.p10 =
p9 + 1 =
q9(1 + 1
9
) =p9q1 + 1
9
= 3(1 + 1
9
)1/2; we have x = 1
9
and k = 1
2p10 = 3(1 + 1
2
· 1
9
� 1
8
· 1
9
2 + · · · ) = 3 + 1
6
� 1
216
+ · · · )��p10� 3.16666
�� < 1
216
< 1
200
= 0.005
Question 37 Solution
Since f(x) = ln(1+x) = x� x
2
2
+ x
3
3
� x
4
4
+ x
5
5
+ · · · =1P
n=1
(�1)n+1
x
n
n
is an alternating series, we can use the error bound
for alternating series: |s� sn
| an+1
where an+1
= x
n+1
n+1
. We are interested in estimating ln 3
2
= ln(1 + 1
2
); hence, we
take x = 1
2
. Let us find the smallest n such that an+1
= 1
(n+1)2
n+1 0.001. We find that n = 6 and s6
⇡ 0.4047; the
exact value is s = ln 3
2
⇡ 0.4055, so the error is less than 10�3, as desired.
Question 38 Solutiona) tanx = x+ x
3
3
+Remainderb) e�x sinx = x� x2 +Remainderc) 1�cos x
x
= 1
2
x� 1
24
x3 +Remainder
Question 39 SolutionB
0
= f(0) = 1; B1
= f 0(0) = � 1
2
; B2
= f 00(0) = 1
6
(using L’Hospital rule).
Question 40 Solutiona) f(x) = x, f(0) = 0, f 0(0) = 1b) f(x) = sinx, f(0) = 0, f 0(x) = cosx, f 0(0) = 1c) f(x) = ln(1 + x), f(0) = 0, f 0(x) = 1
1+x
, f 0(0) = 1b) f(x) = ex � 1, f(0) = 0, f 0(x) = ex, f 0(0) = 1If the functions are sketched in a neighborhood of x = 0, the order they appear (from top to bottom), consider their
Taylor approximationsx = x, sinx = x� 1
6
x3 + · · · , ln(1 + x) = x� 1
2
x2 + · · · , ex � 1 = x+ 1
2
x2
Thus on right hand side of 0, from top to bottom, ex � 1, x, sinx and ln(1 + x); on left hand side of 0, from top tobottom, ex � 1, sinx, x, and ln(1 + x).
Question 41 Solutiona) J
0
(x) = 1� x
2
4
+ x
4
64
� · · · it is alternating series.R1
0
⇣1� x
2
4
⌘dx = 11
12
error boundR1
0
x
4
64
dx = x
5
5·64 = 1
320
b) J0
(x)0 =1P
n=1
(�1)
n
2nx
2n�1
2
2n(n!)
2 =1P
n=0
(�1)
n+1(2n+2)x
2n+1
2
2n+2((n+1)!)
2
J0
(x)00 =1P
n=1
(�1)
n
2n(2n�1)x
2n�2
2
2n(n!)
2
xJ0
(x)00 =1P
n=1
(�1)
n
2n(2n�1)x
2n�1
2
2n(n!)
2 =1P
n=0
(�1)
n+1(2n+2)(2n+1)x
2n+1
2
2n+2((n+1)!)
2
xJ0
(x) =1P
n=0
(�1)
n
x
2n+1
2
2n(n!)
2
xJ0
(x)00 + J0
(x)0 + xJ0
(x) =1P
n=0
(�1)
n+1x
2n+1
2
2n(n!)
2
h2n+2
2
2(n+1)
2 + (2n+1)(2n+2)
2
2(n+1)
2 � 1i= 0
Thus J0
(x) satisfies xy00 + y0 + xy = 0
Question 42 Solution
f(t) =1P
n=0
tn = 1 + t+ t2 + t3 + t4 + · · ·
a) step 1 : di↵erentiate the series, f 0(t) = 1 + 2t+ 3t2 + 4t3 + · · ·
10
step 2 : square the series, f2(t) = (1 + t+ t2 + t3 + t4 + · · · ) · (1 + t+ t2 + t3 + t4 + · · · ) = 1 + 2t+ 3t2 + 4t3 + · · ·Thus f 0(t) = f2(t), so f(t) satisfies the di↵erential equation y0 = y2 and we can check that the initial condition is f(0) = 1.
b) y0 = y2 ) dy
dt
= y2 ) dy
y
2 = dt )R
dy
y
2 =Rdt ) � 1
y
= t + C ) y = 1
�C�t
, now apply initial condition
t = 0, y = 1 ) C = �1 ) y = 1
1�t
Since we showed in part (a) that f(t) satisfes the di↵erential equation and initial condition, we obtain f(t) = 1
1�t
. This
is a round-about way of deriving the sum of a geomertic series, f(t) =1P
n=0
tn = 1 + t+ t2 + t3 ++t4 + · · · = 1
1�t
.
Question 43 Solutiona1)
R10
sin x
x
dx converges
pf:R10
sin x
x
dx =1P
n=0
R(n+1)⇡
n⇡
sin x
x
dx =R⇡
0
sin x
x
dx +1P
n=1
R(n+1)⇡
n⇡
sin x
x
dx R⇡
0
sin x
x
dx +1P
n=1
R(n+1)⇡
n⇡
sin x
n⇡
dx; the last step
follows because x � n⇡ in each interval; then we haveR10
sin x
x
dx R⇡
0
sin x
x
dx + 1
⇡
1Pn=1
1
n
R(n+1)⇡
n⇡
sinxdx =R⇡
0
sin x
x
dx +
1
⇡
1Pn=1
� cos x
n
��(n+1)⇡
n⇡
=R⇡
0
sin x
x
dx + 1
⇡
1Pn=1
cosn⇡�cos(n+1)⇡
n
=R⇡
0
sin x
x
dx + 1
⇡
1Pn=1
(�1)n 2
n
=R⇡
0
sin x
x
dx + 2
⇡
1Pn=1
(�1)
n
n
, which
converges since the integral is proper (note that limx!0
sin x
x
= 1) and the series converges by the AST ok- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -a2)
R10
�� sin x
x
�� dx diverges
pf:R10
�� sin x
x
�� dx =1P
n=0
R(n+1)⇡
n⇡
|sin x|x
dx =R⇡
0
|sin x|x
dx+1P
n=1
R(n+1)⇡
n⇡
|sin x|x
dx �R⇡
0
sin x
x
dx+1P
n=1
R(n+1)⇡
n⇡
|sin x|(n+1)⇡
dx; the last
step follows because x n(+1)⇡ in each interval; then we haveR10
sin x
x
dx �R⇡
0
sin x
x
dx + 1
⇡
1Pn=1
1
n+1
R(n+1)⇡
n⇡
|sinx| dx =
R⇡
0
sin x
x
dx+ 1
⇡
1Pn=1
2
n+1
=R⇡
0
sin x
x
dx+ 2
⇡
1Pn=1
1
n+1
, which diverges because the series is the harmonic series ok
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -b)
R10
sin x
x
dx = ⇡
2
pf: set f(a) =R10
sin x
x
e�axdx; then note that f(0) =R10
sin x
x
dx; so we need to evaluate f(0); to do that we will evaluatef(a) and then set a = 0; to solve for f(a), we will find an expression for f 0(a) and then integrate; hence we havef 0(a) =
R10
sin x
x
d
da
[e�ax]dx =R10
sin x
x
·�xe�axdx = �R10
sinxe�axdx
substitute: u = sinx, dv = e�axdx ) du = cosxdx, v = e
�ax
�a
) f 0(a) = �hsinx · e
�ax
�a
��10
�R10
cosx e
�ax
�a
dxi= � 1
a
R10
cosxe�axdx
substitute: u = cosx, dv = e�axdx ) du = � sinxdx, v = e
�ax
�a
) f 0(a) = � 1
a
hcosx · e
�ax
�a
��10
�R10
sinx e
�ax
a
dxi= � 1
a
⇥1
a
+ 1
a
f 0(a)⇤) f 0(a) = � 1
a
⇥1
a
+ 1
a
f 0(a)⇤) f 0(a)
⇥1 + 1
a
2
⇤=
� 1
a
2 ) f 0(a)⇥a2 + 1
⇤= �1 ) f 0(a) = � 1
1+a
2 ) f(a) = � tan�1(a) + C; to evaluate the constant note that lima!1
f(a) =
lima!1
R10
sin x
x
e�axdx =R10
sin x
x
lima!1
[e�ax]dx =R10
sin x
x
·0dx =R10
0dx = 0; hence we have lima!1
f(a) = lima!1
� tan�1(a)+
C = � tan�1(1) + C = �⇡
2
+ C = 0 ) C = ⇡
2
) f(a) = � tan�1(a) + ⇡
2
) f(0) = � tan�1(0) + ⇡
2
= ⇡
2
ok
Question 44 Solutioncosx = 1� 1
2
x2 + 1
4!
x4 � · · · ; it is a convergent alternating series, so |cosx� 1| 1
2
x2, and��cosx� (1� 1
2
x2)�� 1
4!
x4;setting x = ⇡
5
gives the result
Question 45 Solutionerf(x) = 2p
⇡
Rx
0
e�t
2
dt = 2p⇡
Rx
0
(1� t2 + t
4
2
+ ...)dt = 2p⇡
(x� x
3
3
+ x
5
10
+ ...)
Question 46 Solution
a) a
a+b
= a
b
· 1
1+
a
b
= a
b
· 1
1�(� a
b
)= a
b
·1P
n=0
��a
b
�n
a
a+b
= a
b
⇣1� a
b
+ a
2
b
2 + · · ·⌘= a
b
� a
2
b
2 + a
3
b
3 + · · ·b) Using the Binomial Series Theorem
(1 + x)k = 1 + kx+ k(k�1)
2
x2 + · · · for �1 < x < 1pR2 � r2 = R
q1� r
2
R
2 = R⇣1� r
2
R
2
⌘ 12= R
1� 1
2
· r
2
R
2 +(
12 )(
12�1)
2
⇣� r
2
R
2
⌘2
+ · · ·�= R� r
2
· r
R
� r
8
· r
3
R
3 + · · ·
Question 47 Solution
11
Starting from the Taylor series of f(x) about x = a:
f(x) = f(a) + f 0(a)(x� a) +f 00(a)
2!(x� a)2 + · · · ,
replace x ! x+ h, a ! x. It follows that x� a ! (x+ h)� x = h, and the Taylor series becomes
f(x+ h) = f(x) + f 0(x)h+f 00(x)
2!h2 + · · ·
Question 48 Solutiona) let y = 0 ) x = ±(1 + ✏), let x = 0 ) y = ±1
b) Solve for y: y = f(x) = ±r1�
⇣x
1+✏
⌘2
A(✏) = 2R1+✏
�(1+✏)
r1�
⇣x
1+✏
⌘2
dx = 4R1+✏
0
r1�
⇣x
1+✏
⌘2
dx
c)
A(✏) = 4R1+✏
0
r1�
⇣x
1+✏
⌘2
dx = 4(1 + ✏)R1+✏
0
r1�
⇣x
1+✏
⌘2
d x
1+✏
= 4(1 + ✏)R1
0
p1� u2du = 4(1 + ✏)⇡
4
= (1 + ✏)⇡
The first two nonzero terms are ⇡ + ⇡✏.
Question 49 SolutionV (x) = Gm1
|x�x1| +Gm2|x�x2| for x ! 1 ie., x > x
1
and x > x2
) V (x) = Gm1x�x1
+ Gm2x�x2
Using the hint set y = 1/x, ie., x = 1/y and expand the potential in powers of yV (1/y) = Gm1
1/y�x1+ Gm2
1/y�x2= Gm1y
1�x1y+ Gm2y
1�x2y
Using Geometric Series Formula 1
1�x
i
y
=1P
n=0
(xi
y)n where i = 1, 2
V (1/y) = Gm1
y1P
n=0
(x1
y)n +Gm2
y1P
n=0
(x2
y)n = Gm1
1Pn=0
xn
1
yn+1 +Gm2
1Pn=0
xn
2
yn+1
= (Gm1
+Gm2
)y + (Gm1
x1
+Gm2
x2
)y2 + (Gm1
x2
2
+Gm2
x2
2
)y3 + · · ·change y = 1
x
backV (x) = (Gm
1
+Gm2
) 1x
+ (Gm1
x1
+Gm2
x2
) 1
x
2 + (Gm1
x2
2
+Gm2
x2
2
) 1
x
3 + · · ·Thus a = (Gm
1
+Gm2
), b = (Gm1
x1
+Gm2
x2
), and c = (Gm1
x2
2
+Gm2
x2
2
)
Question 50 Solution
a) limr!0
V (r) = limr!0
V0
(( r0r
)12 � 2( r0r
)6) = V0
limr!0
r
120 �2r
60r
6
r
12 = V0
r
1200
+ = 1
limr!1
V (r) = limr!1
V0
(( r0r
)12 � 2( r0r
)6) = V0
limr!1
r
120 �2r
60r
6
r
12 = V0
· 0 = 0
b) V 0(r) = V0
((�12) r120
r
13 � 2(�6) r60
r
7 ) = V0
(�12r
120
r
13 + 12r
60
r
7 ) = V0
�12r
120 +12r
60r
6
r
13
V 0(r) = 0 , �12r120
+ 12r60
r6 = 0 , 12r60
r6 = 12r120
, r6 = r60
, r = r0
.d) Find the quadratic Taylor approximation at x = x
0
, ie., c0
+ c1
(x� x0
) + c2
(x� x0
)2
using the Theorem(1 + x)k = 1 + kx+ k(k�1)
2
x2 + · · · for �1 < x < 1
V (x) = V0
h�x0x
�12 � 2
�x0x
�6
i= V
0
⇣x
x0
⌘�12
� 2⇣
x
x0
⌘�6
�= V
0
⇣x�x0+x0
x0
⌘�12
� 2⇣
x�x0+x0x0
⌘�6
�
= V0
⇣1 + x�x0
x0
⌘�12
� 2⇣1 + x�x0
x0
⌘�6
�
using the above Theorem⇣1 + x�x0
x0
⌘�12
= 1� 12x�x0x0
+ (�12)·(�12�1)
2
�x�x0
x
�2
+ · · · = 1� 12
x0(x� x
0
) + 78
x
20(x� x
0
)2 + · · ·⇣1 + x�x0
x0
⌘�6
= 1� 6x�x0x0
+ (�6)·(�6�1)
2
�x�x0
x
�2
+ · · · = 1� 6
x0(x� x
0
) + 21
x
20(x� x
0
)2 + · · ·
V (x) = V0
h⇣1� 12
x0(x� x
0
) + 78
x
20(x� x
0
)2 + · · ·⌘� 2
⇣1� 6
x0(x� x
0
) + 21
x
20(x� x
0
)2⌘i
+ · · ·
= V0
h�1 + 36
x
20(x� x
0
)2i+ · · · = �V
0
+ 36V0
x
20(x� x
0
)2 + · · ·T2
(x) = �V0
+ 36V0
x
20(x� x
0
)2
e) We know that work equals the integral of the force:
W =1Rr0
f(r)dr =1Rr0
�V 0(r)dr = �V (r)|1r0
= �0� (�V (r0
)) = V (r0
) = V0
(1� 2) = �V0
12
binomial series
Question 51 SolutionUsing the Binomial Series Theorem,(1 + x)k = 1 + kx+ k(k�1)
2
x2 + k(k�1)(k�2)
3!
x3 + k(k�1)(k�2)(k�3)
4!
x4 + · · · for �1 < x < 1
(1 + x2)k = 1 + kx2 + k(k�1)
2
x4 + k(k�1)(k�2)
3!
x6 + k(k�1)(k�2)(k�3)
4!
x8 + · · ·(1 + x2)
12 = 1 + 1
2
x2 +12 (
12�1)
2
x4 +12 (
12�1)(
12�2)
3!
x6 +12 (
12�1)(
12�2)(
12�3)
4!
x8 + · · · = 1 + 1
2
x2 � 1
8
x4 + 1
16
x6 � 5
128
x8 + · · ·This is an alternating series and the assumptions in the AST apply.Using the second order Taylor approximation, T
2
(x) = 1 + 1
2
x2, we have |s� T2
| < 1
8
x4, where s is the exact value.R1
0
p1 + x2dx ⇡
R1
0
(1 + 1
2
x2)dx = x+ 1
6
x3
��10
= 7
6
and the error is less thanR
1
8
x4dx = 1
40
x5
��10
= 1
40
Question 52 SolutionConsider the expansion 1p
1�2ax+x
2 = c0
+ c1
x+ c2
x2 + · · · , where a is a constant. Find c0
, c1
, c2
in terms of a.
We can apply the binomial expansion.1p
1�2ax+x
2 = (1� 2ax+ x2)�1/2 = (1 + (�2ax+ x2))�1/2 = 1 + (� 1
2
)(�2ax+ x2) +(� 1
2 )(�32 )
2
(�2ax+ x2)2 + · · ·= 1 + ax� 1
2
x2 + 3
8
(4a2x2 � 4ax3 + x4) + · · · = 1 + ax+ (� 1
2
+ 3
2
a2)x2 + · · · , hence c0
= 1, c1
= a, c2
= 1
2
(3a2 � 1)
Question 53 Solutiona) Show that
�k+1
n+1
�=
�k
n
�+�
k
n+1
�
This is true since the left hand side is number of ways of choosing n+ 1 objects from a set of k+ 1 objects (disregardingthe order in which the objects are chosen).The right hand side means that: assume all k + 1 objects are white, one may randomly pick one object from the k + 1objects, coloring it red, then put it back. Now choose n + 1 objects from these k + 1 objects, there are two di↵erentsituations: one situation is that the red one is chosen, the number of ways is
�k
n
�(it is equivalent to choosing n from k
objects); the other situation is the red one is not chosen, the number of ways is�
k
n+1
�(it is equivalent to choosing n+ 1
objects from k objects).The left hand side equals the right hand side, since it is the same thing, choosing n+ 1 objects from k + 1 objects.�k+1
n+1
�= (k+1)!
(n+1)!(k�n)!
= k!·(k+1)
(n+1)!(k�n)!
= k!·(k�n+n+1)
(n+1)!(k�n)!
== k!·(k�n)+k!·(n+1)
(n+1)!(k�n)!
= k!·(k�n)
(n+1)!(k�n)!
+ k!·(n+1)
(n+1)!(k�n)!
= k!
(n+1)!(k�n�1)!
+k!
n!(k�n)!
=�
k
n+1
�+
�k
n
�
b)�0
0
�1�
1
0
� �1
1
�1 1�
2
0
� �2
1
� �2
2
�1 2 1�
3
0
� �3
1
� �3
2
� �3
3
�1 3 3 1�
4
0
� �4
1
� �4
2
� �4
3
� �4
4
�1 4 6 4 1�
5
0
� �5
1
� �5
2
� �5
3
� �5
4
� �5
5
�1 5 10 10 5 1�
6
0
� �6
1
� �6
2
� �6
3
� �6
4
� �6
5
� �6
6
�1 6 15 20 15 6 1
Denote elements in the triangle by ak,n
, nth element on kth row. Each row is obtained by adding the two entries diagonallyabove, a
k+1,n+1
= ak,n
+ ak,n+1
which is equivalent to�k+1
n+1
�=
�k
n
�+�
k
n+1
�.
c) (a+b)6 =�6
0
�a6+
�6
1
�a5b+
�6
2
�a4b2+
�6
3
�a3b3+
�6
4
�a2b4+
�6
5
�ab5+
�6
6
�b6 = a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6
complex numbers
Question 54 Solutiona) 1 + i =
p2e
⇡
4 i (x = 1, y = 1 already in Cartesian form)b) (1 + i)2 = 1 + 2i+ i2 = 1 + 2i� 1 = 2i = 2e
⇡
2 i (x = 0, y = 2)c) (1 + i)3 = (1 + i)2(1 + i) = 2i(1 + i) = �2 + 2i = 2
p2e
3⇡4 i (x = �2, y = 2)
d) 1
1+i
= 1·(1�i)
(1+i)·(1�i)
= 1�i
1�i
2 = 1�i
1�(�1)
= 1�i
2
= 1
2
� 1
2
i = 1p2
e�⇡
4 i (x = 1
2
, y = � 1
2
)
e)p1 + i =
�p2e
⇡
4 i
� 12 = 2
14 e
⇡
8 i = 214
�cos ⇡
8
+ i sin ⇡
8
�= 2
14 cos ⇡
8
+ i 214 sin ⇡
8
(x = 214 cos ⇡
8
, y = 214 sin ⇡
8
)
Question 55 Solutiona) See 49 c) (1 + i)6 = 1 + 6 · i+ 15 · i2 + 20 · i3 + 15 · i4 + 6 · i5 + i6 = 1 + 6i� 15� 20i+ 15 + 6i� 1 = �8ib) 1 + i =
p2e
⇡
4 i, since x = 1, y = 1, r =px2 + y2 =
p2, ✓ = arctan y
x
= arctan 1 = ⇡
4
, re✓i =p2e
⇡
4 i
(1 + i)6 = (p2e
⇡
4 i)6 = 262 e
64⇡i = 23e
32⇡i = 8(cos 3
2
⇡ + i sin 3
2
⇡) = �8i
Question 56 Solution
a) z2 + 2z � 2 = 0 ) a = 1, b = 2, c = �2, z1,2
= �b±pb
2�4ac
2a
= �2±p4+8
2
= �1±p3 two real roots
13
b) z2 +2z+2 = 0 ) a = 1, b = 2, c = 2, z1,2
= �b±pb
2�4ac
2a
= �2±p4�8
2
= �2±p�4
2
= �2±2
p�1
2
= �1±p�1 = �1± i
c) z2 = (re✓i)2 = r2e2✓i = 1 ) r = 1, ✓ = ⇡k, where k is any integer ) z1,2
= ±1
d) z3 = r3e3✓i = 1 ) r = 1, ✓ = 2⇡
3
k, where k is any integer ) three roots: z1
= 1 and z2,3
= � 1
2
±p3
2
ie) z4 = r4e4✓i = 1 ) z2 = ±1 ) four roots: z
1,2
= ±1, z3,4
= ±if) ez = 1 on real axis there is on root z = 0, but in complex plane there are infinite roots, let z = x + yi, ez = ex+yi =ex(cos y + i sin y) = 1 ) x = 0 and y = 2k⇡, where k is any integer, roots are z
k
= 2k⇡ i.
Question 57 Solution(cos ✓ + i sin ✓)n =
�ei✓
�n
= ein✓ = cosn✓ + i sinn✓
Question 58 Solutionei(a+b) = eia ·eib ) cos(a+b)+i sin(a+b) = (cos a+i sin a)·(cos b+i sin b) = cos a cos b�sin a sin b+i(sin a cos b+cos a sin b)
take real and imaginary parts : cos(a+ b) = cos a cos b� sin a sin b , sin(a+ b) = sin a cos b+ cos a sin b
Question 59 Solution
a1)Reax cos bx dx =
Reax 1
b
d sin bx = 1
b
eax · sin bx �R
1
b
sin bx deax = 1
b
eax · sin bx �R
a
b
eax sin bx dx = 1
b
eax · sin bx �Ra
b
eax�� 1
b
�d cos bx = 1
b
eax ·sin bx+R
a
b
2 eax d cos bx = 1
b
eax ·sin bx+ a
b
2 eax ·cos bx� a
b
2
Rcos bx deax = 1
b
eax ·sin bx+ a
b
2 eax ·
cos bx�a
2
b
2
Reax cos bx dx )
Reax cos bx dx= 1
b
eax·sin bx+ a
b
2 eax·cos bx�a
2
b
2
Reax cos bx dx )
⇣1 + a
2
b
2
⌘ Reax cos bx dx =
1
b
eax · sin bx+ a
b
2 eax · cos bx )
Reax cos bx dx = e
ax
a
2+b
2 (b · sin bx+ a · cos bx)- - - - - - - - - - - - - - - - - - - -
a2)Reax sin bx dx =
Reax
�� 1
b
�d cos bx = � 1
b
eax · cos bx +R
1
b
cos bx deax = � 1
b
eax · cos bx +R
a
b
eax cos bx dx = � 1
b
eax ·cos bx+
Ra
b
eax · 1
b
d sin bx = � 1
b
eax · cos bx+R
a
b
2 eax d sin bx = � 1
b
eax · cos bx+ a
b
2 eax · sin bx� a
b
2
Rsin bx deax = � 1
b
eax ·cos bx + a
b
2 eax · sin bx � a
2
b
2
Reax sin bx dx )
Reax sin bx dx = � 1
b
eax · cos bx + a
b
2 eax · sin bx � a
2
b
2
Reax sin bx dx )⇣
1 + a
2
b
2
⌘ Reax sin bx dx = � 1
b
eax · cos bx+ a
b
2 eax · sin bx )
Reax sin bx dx = e
ax
a
2+b
2 (a · sin bx� b · sin bx)- - - - - - - - - - - - - - - - - - - -
b) e(a+ib)x = eax+ibx = eax · eibx = eax (cos bx+ i sin bx)Re(a+ib)xdx = 1
a+ib
e(a+ib)x = 1·(a�ib)
(a+ib)·(a�ib)
e(a+ib)x = a�ib
a
2+b
2 e(a+ib)x
- - - - - - - - - - - - - - - - - - - -
c) SinceRe(a+ib)xdx =
R(eax cos bx+ i eax sin bx) dx =
Reax cos bx dx+ i
Reax sin bx dx = a�ib
a
2+b
2 e(a+ib)x, and
a�ib
a
2+b
2 e(a+ib)x = a�ib
a
2+b
2 (eax cos bx+ i eax sin bx) = e
ax
a
2+b
2 (a cos bx+ b sin bx) + ih
e
ax
a
2+b
2 (a sin bx� b sin bx)i, we have
Reax cos bx dx = e
ax
a
2+b
2 (a cos bx+ b sin bx) andR
eax sin bx dx = e
ax
a
2+b
2 (a sin bx� b sin bx)
Question 60 Solution
a) since eix = cosx+ i sinx and e�ix = cosx� i sinx ) eix + e�ix = 2 cosx ) cosx = e
ix
+e
�ix
2
b) eix � e�ix = 2i sinx ) sinx = e
ix�e
�ix
2i
c) d
dx
cosx = d
dx
⇣e
ix
+e
�ix
2
⌘= ie
ix�ie
�ix
2
=(ieix�ie
�ix)i2i
= �e
ix
+e
�ix
2i
= � sinx
d) d
dx
sinx = d
dx
⇣e
ix�e
�ix
2i
⌘= ie
ix
+ie
�ix
2i
= e
ix
+e
�ix
2
= cosx
e) TO BE COMPLETEDf) TO BE COMPLETEDg) TO BE COMPLETEDh) TO BE COMPLETED
14
