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CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
APR2016
ASSESSMENT_CODE BC0039_APR2016
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
5592
QUESTION_TEXT
Define
i.Union sets
ii.Intersection sets
iii.Difference sets
iv.Complement of a set
SCHEME OF
EVALUATION
i.Union Sets: The union of two sets A and B denoted by AUB is the set
of elements which belongs to A or B or both.That is, AUB={x:x∈A or
x∈B}
ii.Intersection of sets: The intersection of two sets A and B denoted by
A∩B is the set of elements which belongs to both A and B.That is,
A∩B={x:x∈A and x∈B}
iv.Difference of sets: The difference of two sets A and B, denoted by
A–B is the set of elements of A which are not the elements of B. That
is, A–B={x:x∈A, x∈B}
v.Complement of a set: The complement of a set A with respect to the
universal set U is defined as U–Aand is denoted by A` or A c. That is
A`={x:x∈U, x∉A}
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
5593
QUESTION_TEXT
Let R be the set of real numbers. Define f:R→R by
i.f(x)=2x+3
ii.f(x)=x 3 for every x in R, prove that f is one–one?
i.f(x)=2x+3
Solution: Let f(x 1)=f(x 2) for some x 1, x 2∈R
2x 1+3=2x 2+3
X 1=x 2
SCHEME OF EVALUATION Thus for every x 1, x 2∈R, f(x 1)=f(x 2) implies x 1=x 2.
Therefore f is one–one.
ii.f(x)=x 3
Solution: Let f(x 1)=f(x 2) for some x 1,x 2∈R
=
X 1=x 2
Therefore f is one–one
QUESTION_T
DESCRIPTIVE_QUESTION
YPE
QUESTION_ID 73263
QUESTION_T a.
EXT
b.
Show that [p  (p  q)] 
is a tautology.
Verify that [p  (p  q)] 
is a contradiction.
a.) Now we write down the truth table
Observing the table we can conclude that [p  (p  q)] 
is always true. Hence [p
SCHEME OF
 (p  q)]  is a tautology.
EVALUATION b. Solution: Now we write down the truth table
Observing the table, we can conclude that [p  (p  q)] 
false. Hence [p  (p  q)] 
is always
is a contradiction.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
105218
QUESTION_TEXT
Prove that if H is a non empty finite subset of a group G and H is
closed under multiplication, then H is a subgroup of G.
Proof: Suppose H is a non empty finite subset of a group G and H is
closed under multiplication
SCHEME OF
EVALUATION
Now, we have to show that H is a subgroup of G.
It is enough to show that a ∈H ⟹ a–1 ∈H.
Since, H is a non empty, there exists a ∈ H.
Now, a∈ H ⟹ a2 ∈H.
Similarly a3 ∈H, …… am ∈H, ……..
Therefore H≥ {a, a2, …….}. Since H is finite, we name that there must
be repetitions in a, a2….. Therefore there exists integers r, s with r>s >
0
Such that ar = as
⟹ ar. a–s = a0
⟹ ar–s = e ⟹ e ∈H (since r–s >0, a∈ H ⟹ar–s ∈H)
Since r–s–1 ≥0, we have ar–s–1 ∈H and
a. ar–s–1 = ar–s = e ∈H.
Therefore ar–s–1 acts as the inverse of a in H.
Hence H is a subgroup of G
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
105224
Prove that if V be a vector space over F and W is a nonempty subset of
V. Then the following conditions are equivalent.
QUESTION_TEXT
i.
W is a subspace of V
ii.
α, β ∈ F and W1, W2 ∈ W implies αW1 + βW2 ∈ W.
(i) ⇒ (ii): α ∈ F, w1 ∈ W ⇒ αw1 ∈ W; β ∈ F, w2 ∈ W ⇒ βw2 ∈ W.
SCHEME OF
EVALUATION
Since αw1, βw2 ∈ W and W is subspace we have αw1 + βw2 ∈ W.
(ii) ⇒ (i): Let w1, w2 ∈ W. Since 1 ∈ F we have that w1 + w2 = 1.w1
+ 1. w2 = αw1 + βw2 ∈ W (here α = 1 and β = 1). Therefore + is a
closure operation on W.
Since W ⊆ V, the associative law holds.
x ∈ W ⇒ x, x ∈ W ⇒ 0 = 1.x + (1) = αx + βx ∈ W (here α = 1 and β =
-1).
Therefore the additive identity 0 is in W.
Let x ∈ W. Then x, x ∈ W ⇒ -x = 0. x + (-1)x = αx + βx (here α = 1
and β = -1).
Therefore the additive inverse –x ∈ W for any element x ∈ W.
Hence (W, +) is a group. Since W ⊆ V and V is Abelian, we have that
(W, +) is also Abelian. Also it is easy to verify that the other conditions
of vector space.
Therefore W is a subspace of V.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
105226
QUESTION_TEXT
Define the following with one example.
Set, Subset, cardinality of a set, Universal set
Set: A well defined collection of objects is called a set.
Ex: B= {1, 3, 5, 7, 9}
Subset: If every element of a set A is also an element of a Set B, then A is
said to be a sub set of B.
Ex: N is the Set of Natural no’s:
SCHEME OF
EVALUATION
Z is the Set of integers.
Then NCZ (N is a Subset of Z)
Cardinality of a set: If A is a finite set, then the Cardinality of A is the total
number of elements that comprise the set and denoted by N(A).
Ex: A = {1, 2, 3, 4, 5}
Then n(A) = 5
Universal Set: If all the Sets are subsets of a fixed Set, then this Set is called
the universal set & denoted by U
Ex: In the Study of theory of numbers the set Z of integers is considered as
the universal set.