CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
SMUAPR15
ASSESSMENT_CODE BT0063_SMUAPR15
QUESTION_TYPE DESCRIPTIVE_QUESTION
QUESTION_ID
QUESTION_TEXT
4984
a.Explain the concept of Venn diagram.
b.Let A = {1, 2, 3} and B = {4, 5}. Find A × B and B × A and show
that A × B ≠ B × A.
c.Discuss the convergence of the series
SCHEME OF
EVALUATION
a.Most of the relationship between sets can be represented by means
of diagrams. Figures representing sets in the form of enclosed region
in the planes are called Venn diagrams named after British logician
John Venn. The universal set U is represented by the interior of a
rectangle. Other sets are represented by the interior of circles.
b.A X B= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
And B X A={(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}
Note that (1, 4) belongs to A X B but it does not belong to B X A.
Hence the proof.
c. The series is of the form:
1+(1/2)+(1/4)+(1/8)+…+[(1/2)]^(n-1)
Therefore the partial sums,
S1=1, S2=(3/2) S3=(7/4)…..
Here the sequence of partial sums converges. Hence the sequence
given also converges. (Using the principle of partial sums)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
4986
QUESTION_TEXT
Define a group and give one example.
SCHEME OF
EVALUATION
Definition of Group:
A non-empty set G is said to be a group with respect to the binary
operation * if the following axioms are satisfied:
a.Closure law: For every a, b in G, a*b belongs to G
b.Associative law: For every a, b, c in G,
a*(b*c)=(a*b)*c
c.Existence of identity element: There exists an element e in G such
that a*e=e*a=a for every a in G.
d.Existence of inverse: For every a in G, there exists an element b in
G such that,
a*b=b*a=e. here b is called the inverse of a.
a group G with respect to binary operation * is denoted by(G, *)
Example:
The set Z of integers is a group with respect to the usual addition as
the binary operation.
a.Closure law: We know that the sum of two integers is also an
integer. Hence for every m, n in Z, m+n belongs to Z.
b.Associative law: It is well known that the addition of integers is
associative
c.Existence of identity element: There exists 0 in Z such that,
m+0=0+m=m for every m in Z. Hence 0 is the additive identity.
d.Existence of inverse: For every m in Z, there exists –m in Z such
that m+(–m)=(–m)+m=0. Here –m is called the additive inverse of
m or simply the negative of m.
Therefore (Z, +) is a group.
QUESTION_TYPE DESCRIPTIVE_QUESTION
QUESTION_ID
4987
a.Find the probability that at least one head appears in a throw of 3
unbiased coins.
b.An article consists of two parts A and B. The probabilities of
QUESTION_TEXT defects in A and B are 0.08 and 0.04. What is the probability that the
assembled part will not have any defect?
c.The mean of marks scored by 30 girls of a class is 44%. The mean
of 50 boys is 42%. Find the mean for the whole class.
SCHEME OF
EVALUATION
a.The possible outcomes are: HHH, HHT, THH, HTH, HTT, TTH,
THT, TTT
The probability that at least one head appears=7/8
b.Let x and y be the events that A and B do not have any defect
respectively.
P(x)=0.92 and P(y)=0.96
Assuming independence,
Probability that the assembled part will not have any
defect=P(x)*P(y)=0.92*0.96=0.8832
c.Here n1=30, n2=50
Mean marks of boys=42%
Mean marks of girls=44%
Therefore the mean for the whole class
=[(30*44)+(50+42)]/(30+50)=(1230+2100)/80=42.75%
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
4988
QUESTION_TEXT
Define:
i.Conjunction
ii.Disjunction
iii.Negation
iv.Tautology
v.contradiction
i . Conjunction: If two statements p and q are connected by the
word “and”, then the resulting compound statement “p and q” is
called conjunction of p and q.
i ii. Disjunction: If two simple statements p and q are connected by
the word “or” then the resulting compound statement “p or q” is
called disjunction of p and q.
iii. Negation: An assertion that, a statement fails of denial of a
statement is called the negation of the statement.
iv. Tautology: A statement is said to be a tautology if it is true for
all logical possibilities.
v . contradiction: A statement is called contradiction if it is false
for all logical possibilities.
(Each definition carries 2 marks)
SCHEME OF
EVALUATION
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
72355
QUESTION_TEXT
a.
b.
SCHEME OF
EVALUATION
a.
Let e and e1 be the two identity elements of a group G. then for
every a in G, ae = ea = a and
Ae1 = e1a = a
Substitute a = e1 in (i) and a = e in (ii).
Then, e1e = ee1 = e1
Hence e1 = ee1 = e
Therefore identity element in a group G is unique.
b.
Let b and c be the two inverse of an element a in G.
Then, ab = ba = e
Ac = ca = e
Now consider, b = be
= b (ac)
= (ba) c
= ec
=c
Therefore inverse of every element in a group G is unique.
Prove that the identity element in a group is unique.
Prove that in a group G the inverse of an element is unique.
QUESTION_T
DESCRIPTIVE_QUESTION
YPE
QUESTION_I
D
118222
QUESTION_T Prove that the set Z4 = {0, 1, 2, 3} is an abelian group w.r.t addition
EXT
modulo 4.
The composition table w.r.t addition modulo 4 is an as below:
Since 1+3=40(mod 4), 3+3=62(mod 4), 2+3=51 (mod 4) etc. (2
MARK)
1. Closure law: From the above composition table for all a, bG, a+4b also
belongs to Z4
(1 MARK)
SCHEME OF
EVALUATIO
N
2. Associative law: Since a+(b+c) and (a+b)+c leave the same remainder
when divided by 4, we have
a+4(b+4c)=(a+4b)+4c
(2 MARK)
3. Existence of identity element: From the above table, we observe that 0
Z4 satisfies a+40=0+4a=a for every a Z4
0 is the identity element. (2 MARK)
4. Existence of inverse: From the above table, the inverse of 0, 1, 2 are
respectively 0, 3, 2, 1 because 0+40=0, 1+43=0, 2+40=0, and 3+41=0
Hence (Z4, +4) is a group.
(2 MARK)
Further since a+b and b+a leave the same remainder when divided by 4,
a+4b=b+4a
(Z4, +4) is an abelian group.
(1 MARK)