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# ELECTRONICS & COMMUNICATION ENGINEERING (ECE

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II Year B.E/B.Tech ELECTRONICS & COMMUNICATION ENGINEERING (ECE) MODEL QUESTION PAPERS (Effective from 2005 admitted batch) SCHOOL OF DISTANCE EDUCATION ANDHRA UNIVERSITY VISAKHAPATNAM - 530 003
• power amplifier circuit
• c. shunt motor
• squate wave at 10khz
• circuit diagram
• phase induction motor
• voltage
• time

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##### Amplifier

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 Publié par Nombre de lectures 8 Langue English

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MATH 1050 FINAL SOLUTIONS
Overview
The exam went ok. There were some very good scores and some
that were too low. 19 people took the exam at the scheduled time.
The average was 84, a disappointing result.
Here is the score distribution.
Score Number of Students
118 1
113 1
110 1
104 1
100 1
98 1
95 1
92 1
89 1
82 1
81 1
78 1
77 1
70 1
68 1
62 1
60 1
54 1
45 1
Every one who took the ﬂnal avoided getting an E. I wish I could
say that every one passed, but that did not happen. Here is the grade
distribution.
12 MATH 1050 FINAL SOLUTIONS
90-94 A- 2
85-89 B+ 2
80-84 B 2
75-79 B- 3
70-74 C+ 2
65-69 C 3
60-64 C- 2
55-59 D+ 3
This means that 14 out of 19 people passed the class. Congratula-
tions! This is not an easy class. Well done to those of you who did
well. For those who did not pass, I am sorry that it worked out this
way and wish you better luck next time.
Solutions and Commentary
1. Problem 1
Solve the inequality
2(1.1) 3x +2x¡1•0:
Solution 1.1. Notice that
2(1.2) P(x)=3x +2x¡1=(3x¡1)(x+1):
1Hence, the roots of P are x = and x = ¡1. Moreover, P is a
3
quadratic polynomial with a positive coe–cient. Hence, the graph of
y = P(x) opens upwards and hence P(x)• 0 between the roots of P;h i
1that is on the interval ¡1; .
3
1Remark 1.2. A lot of people wrote < x < ¡1 despite the fact that
3
there is no number x which satisﬂes this inequality.
A similar problem appeared on the ﬂrst midterm.
2. Problem 2
Solve the inequality
(2.1) 3<jx+5j:
Solution 2.1. Since 3<jx+5j, either
(2.2) 3<x+5;
or
(2.3) x+5<¡3:MATH 1050 FINAL SOLUTIONS 3
y
y=g(x)
37
(2,5)
x
Figure 1. Illustration for Problem 3
In the ﬂrst instance, we solve for x to get¡2<x. In the second case,
solving for x yields x <¡8. Hence, the solution set of our inequality
is the set
(2.4) (¡1;¡8)[(¡2;1):
Remark 2.2. It does not matter much to me how you present the solu-
tion set provided the presentationis correct. For example, the inequal-
ity
(2.5) ¡2<x<¡8
is not acceptable since x cannot be simultaneously bigger than¡2 and
less than¡8.
There was a similar problem on the ﬂrst midterm.
3. Problem 3
2Let f(x)=4x . Sketch the graph of
(3.1) y =2f(x¡2)+5:
Be sure to ﬂnd the vertex.
Solution 3.1. Let g(x)=2f(x¡2)+5. Then
2 2(3.2) g(x)=2[4(x¡2) ]+5=8(x¡2) +5:6
6
4 MATH 1050 FINAL SOLUTIONS
We see at once that g is a quadratic polynomial. Hence, the graph of
y =g(x) is a parabola. Notice that g(x) is written in (to coin a term)
completed square form. We see at once that the vertex of our parabola
is (2;5). Since the leading coe–cient of g is positive, theola
opens upwards. Therefore, g(x) ‚ 5 for every x inR and thus there
are no x-intercepts. Finally, we observe that
2(3.3) g(0)=8(¡2) +5=37:
We now have enough information to sketch the graph of y = g(x)
which we do in ﬂgure 1.
Remark 3.2. Several people multiplied g(x) out and then completed
the square. The above calculation shows that this is redundant. More
people should have found the y-intercept.
There was a similar problem on the ﬂrst midterm, although in that
problem the graph turned out to be a straight line.
4. Problem 4
Find the domain and range of
5x
(4.1) f(x)= :
x¡1
Solution 4.1. Notice that f is deﬂned for every number except x = 1.
Hence, the domain is the set fx j x = 1:g . To ﬂnd the range, we
¡1compute f . Let y =f(x).
Then
5x
y =
x¡1
y(x¡1)=5x
yx¡y =5x
(4.2)
yx¡5x=y
(y¡5)x=y
y
x= :
y¡5
Hence,
y¡1(4.3) f (y)= :
y¡5
¡1The domain of f is the range of f. Hence, the range of f is the set
fyjy =5g .MATH 1050 FINAL SOLUTIONS 5
Remark 4.2. I wrote the solution sets in such a way that x and y could
be interpreted as complex numbers if necessary. However, the usual
answer would interpret x and y as real numbers.
In any case, it is clear that 0 is in the range of f since f(0) = 0.
There was a similar problem on the ﬂrst midterm.
5. Problem 5
Farmer Palmer has recently acquired some land through Big Vern
of the London maﬂa. He wants to enclose some of that land into a
rectangular ﬂeld in which he can use (at night) his Crop-Circle O-
Matic to mess with hippies heads. He has 200 meters of electric fence.
Ifheusesallofthefence,whatisthelargesttheﬂeldwillbe? Carefully
Solution 5.1. Let l be the length of the ﬂeld in meters. Let w be the
width of the ﬂeld in meters. Let A be the area of the ﬂeld in square
meters. Let P be the perimeter of the ﬂeld in meters.
Then
(5.1) 2l+2w =P =200:
Solving for w, we ﬂnd:
2(l+w)=200
l+w =100(5.2)
w =100¡l:
Now
(5.3) A=lw =l(100¡l):
Hence, the area can be expressed as a quadratic in l. We complete the
square:
2A=100l¡l
2=¡(l ¡100l)
2(5.4) =¡((l¡50) ¡2500)
2=¡(l¡50) +2500
•2500;
with equality holding when l = 50. Hence, the maximum area of the
ﬂeld is 2500 square meters.
Remark 5.2. You cannot just assume that the rectangle is a square; in
the course of your justiﬂcation it may become clear that the maximum
area occurs when the rectangle is a square.6 MATH 1050 FINAL SOLUTIONS
y
y=f(x)
7/2
3
x
Figure 2. Illustration for Problem 6
You cannot assume that l and w have integer values. There is abso-
lutely no reason to assume that.
The source for this problem is a British comic called Viz. Farmer
Palmer and Big Vern are both characters in Viz, although it is not
clear that Vern is a member of the maﬂa. The comic is an adult comic
in the sense that it is full of childish humour. I’ll say no more at this
time.
The source for the date of the exam (10 Febtober) is a well known
Celebrity Jeopardy sketch from Saturday Night Live.
There was a similar problem on the ﬂrst midterm.
6. Problem 6
Let
x¡1(6.1) f(x)=2 +3:
Sketch the graph of y =f(x).
xSolution 6.1. The graph of f is the graph of y =2 shifted to the right
xby one unit and then up by three units. the graph of y = 2 has the
negativereal axis as a horizontal asymptote, is increasing, and concave
up. Therefore, the graph of y =f(x) has the line y =3 as a horizontalMATH 1050 FINAL SOLUTIONS 7
y
y=h(x)
1
y=g(x)
x1
y=f(x)
Figure 3. Illustration for Problem 7
asymptote, is increasing, and concave up. The y-intercept is
1 7¡1(6.2) f(0)=2 +3= +3= :
2 2
We sketch the graph of f in ﬂgure 2.
Remark 6.2. The asymptote needed to be clearly indicated.
There was a similar problem on the ﬂrst midterm.
7. Problem 7
Sketch the graph of y =log1 x.
2
Solution 7.1. Let f(x)=log1 x. By the deﬂnition of the logarithm,
2
‡ ·x1 1¡1 ¡x(7.1) f (x)= = =2 :
x2 2
¡1 xHence, the graph of y = f (x) is the graph of y = 2 (which was
discussed in the previous problem) re ected through the y-axis. The
¡1graph of y = f(x) is the graph of y = f (x) re ected through the
x-axis.
x ¡xLet h(x) = 2 and g(x) = 2 . We sketch the graphs of y = f(x),
y = g(x), and y = h(x) in ﬂgure 3. (The only one that we really need
is y =f(x); we include the other two for illustration).8 MATH 1050 FINAL SOLUTIONS
Remark 7.2. There was a similar problem on the second midterm.
8. Problem 8
3Solve log(x +5)=4 for x.
Solution 8.1. Writing the equation in exponential form, we obtain:
3 4x +5=10
3x +5=10000
3(8.1) x =9995
1‡ ·
3
x= 9995 :
Remark 8.2. Since we are taking cube roots, there is no § in our an-
swer. I suppose we could have taken complex cube roots (by the Fun-
damental Theorem of Algebra there are three solutions), but we have
not discussed how one would do that in this class.
There was a similar problem on the second midterm.
9. Problem 9
Use synthetic division to compute
3 2x +x +x¡1
(9.1) :
x¡5
x¡5
Solution 9.1. We synthetically divide by 5:
5 j 1 1 1 ¡1
5 30 155
(9.2)
1 6 31 154:
Hence,
3 2x +x +x¡1 1542(9.3) =x +6x+31+ :
x¡2 x¡5
Remark 9.2. This problem went fairly well.
There was a similar problem on the second midterm.6
6
MATH 1050 FINAL SOLUTIONS 9
10. Problem 10
Show that the polynomial
5(10.1) P(x)=x ¡4x+1
has no rational roots.
Solution 10.1. Bytherationalrootstheorem,theonlypossiblerational
roots of P are§1. But
(10.2) P(1)=1¡4+1 =0
and
(10.3) P(¡1)=¡1+4+1 =0:
Hence, P has no rational roots.
Remark 10.2. I do not know how to factor P to explicitly ﬂnd the ﬂve
roots. The whole point of the question was to use the rational roots
there were only two possible rational roots.
Do not confuse the term ‘rational’ with the term ‘integer’.
Note that since P(0) > 0 > P(1), there exists a (real) root between
zero and one by the intermediate value theorem.
Problem 7 on the second midterm is essentially the same, although
the statements appear to be diﬁerent.
11. Problem 11
Find a polynomial with real coe–cients whose roots include i, 1+i,
and 5.
Solution 11.1. Let P be a polynomial that satisﬂes the required con-
ditions. Since P must have real coe–cents, we observe that by the
conjugate pairs theorem,¡i and 1¡i must also be roots of P.
Hence, byan application ofthe factor theorem, a good candidate for
P would be:
P(x)=(x¡i)(x+i)(x¡(1+i))(x¡(1¡i))(x¡5)
2=(x +1)((x¡1)¡i)((x¡1)+i)(x¡5)
2 2=(x +1)((x¡1) +1)(x¡5)(11.1)
2 2=(x +1)(x ¡2x+1+1)(x¡5)
2 2=(x +1)(x ¡2x+2)(x¡5):
NoticethatP isaproductofpolynomialswithrealcoe–cients;hence
P also has real coe–cients.
We conclude that our candidate for P has the desired properties.6
10 MATH 1050 FINAL SOLUTIONS
Remark 11.2. I see no real need to multiply P out.
There was a similar problem on the second practice midterm.
12. Problem 12
Let A, B, and C be arbitrary 2£2 matrices. Let I be the identity2
2£2 matrix. Let 0 be the zero 2£2 matrix.2
Exactly one of the following statements is false. Identify the false
statement and give a counterexample.
(i) (A+B)+C =(A+B)+C
(ii) A£0 =0 .2 2
¡1(iii) If AB =I then B =A .2
(iv)AB =BA.
(v) A(B+C)=AB+AC.
Solution 12.1. Statement (i) is trivially true since the left and right
1hand sides say exactly the same thing . It can easily be shown that
statement (ii) is correct and it turns out that statements (iii) and (v)
are as well. (See sections 6.3 and 6.4 for details.) In any case, (iv) is
deﬂnitely false as we see in the example below.
• ‚ • ‚
1 2 6 2
Example 12.2. Let A= . Let B = :
0 ¡1 ¡2 0
Then
• ‚• ‚ • ‚
1 2 6 2 8 2
(12.1) AB = =
0 ¡1 ¡2 0 2 0
while
• ‚• ‚ • ‚
6 2 1 2 6 10
(12.2) BA= = =AB:
¡2 0 0 ¡1 ¡12 ¡4
Remark 12.3. Ithinkthatistheonlyquestionontheexamthatshould
have been remotely surprising. I wrote it in the hope that every one
would remember that matrix multiplication is not commutative.
13. Problem 13
Solve the following system:
x+y =1
y+z =2
(13.1)
z+w =¡1
x+w =0:
1I meant to write the associative law here.

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