Q1(a).Make truth table for the following:(i) P (~q V ~r) Λ p Λ ~q = A(ii) P r V ~q Λ p V r = B

Ans:- (i) P (~q V ~r) Λ p Λ ~q

p Q R ~p ~r (~q V ~r) (~q V ~r) Λ p Λ ~q A

T T T F F F F FT T F F T T F FT F T T F T T TT F F T T T T TF T T F F F F TF T F F T T F TF F T T F T F TF F F T T T F T

(ii) P r V ~q Λ p V r

p Q r ~p ~q Λ p p r B

T T T F F T T

T T F F F F F

T F T T T T T

T F F T T F T

F T T F F T T

F T F F F T TF F T T F T T

F F F T F T T

Q1(b). What are conditional connectives? Give an example of Biconditional Statement.

Ans. CONDITIONAL CONNECTIVE: - If p and q are two propositions, the compound proposition ,it is

denoted by (“ if p then q’”) p q. it is also read as “p implies q” or “p is sufficient for q” or “p only if q”. Here p is called the hypothesis and q the conclusion. Thus any statement of the form p q is called or conditional statement or conditional proposition. The two connectives “” and “” are called conditional connectives.BICONDITIONAL STATEMENT: -

If p and q are two propositions, the the compound statement (pq) Λ (q p) is the biconditional of p and q. It is denoted by (p q), and is read as “p if and only q”.It can also be defined as “p is necessary and sufficient condition for q” or “p implies q and q implies p”.The example of biconditional statement are:● Subham will win the match if and only if he practice regularly.● He drinks if and only if milk is warm.

Q1(c). Write down suitable mathematical statement that can be represented by the following symbolic properties.

(i) ( . x) ( . y) p(ii) . (x) ( . y)( . z) p

Ans. (i) ( . x Є N) ( . y Є z ) ( . z Є H) (yz/x . Q)

(ii) ( . x Є N) ( . y Є z)( . z Є H) (yz/x . Q)

Q2(a). Explain different methods of proof with the help of one example each.Ans. PROOF: - A proof of a proposition p is a mathematical argument consisting of a sequence of statements p1,p2,p3,………,pn from which p logically follows. Thus p is the conclusion of that argument.The statement that is provided to be true is called Theorem.

METHOD OF PROOF : -The different method of proof can be categories as Direct proof , indirect proof and Counter example.

DIRECT PROOF: -A direct proof of p q is a logically valid

argument that begins with the assumptions that p is true and in one or more applications of the law of detachment concludes that q must be true.Therefore to construct a direct proof of p q, start by assuming that p is true. In one or more steps of the form pq1, q1q2,qnq, it is concluded that q is true.

For example: - “the square of an even integer is an even integer”.Symbolically the statement is written as: -( . x Є Z ) (p(x) q(x))

Where p(x) : x is evenAnd q(x) : x² is even.

Now the direct proof is as follows:(Assuming that p(x) is true )

Let x be an even number.Then x = 2n for some integer n.

(Applying the definition of an even number) x² = (2n)² = 4n² = 2(2n²) x² is even i.e. q(x) is true

INDIRECT PROOF: -In indirect proof there are two method for

proving pq . They are proof by contra positive and proof by contradiction. (i) Proof by contra positive: - In this method we can use the fact that the propositions pq is logically equivalent toits contra positive( –q –p).i.e. pq ≡ (~q ~p).

For example: - “If shiva does not agree with the communist then he is not orthodox”.

To prove If x , y Є Z such that xy is odd, then both x and y are odd.

Let p : xy is oddq : both x and y are odd.

So ~p : xy is even ~q : both xand y are even

We have to prove pq , by proving ~q ~p. Assume ~q is true, and suppose x is even.Then x = 2n for some n Є NTherefore xy = 2nyBut xy is even by definition Hence ~p is true.

(ii) Proof by contradiction: - This method is also called reduction ad absurdum(a Latin phrase) because it relies on reducing a given assumption to an absurdity. In this method to prove, q is true let assume that q is false (i.e. ~q is true). By a logical argument, we reach to a contradiction r Λ ~r for some statement that is always false also.Therefore, q must be true.

For example: - Show that √5 is rational.

Proof by Contradiction: -Suppose that √5 is rational. Thus there exist positive integer a and b.Such that √5 = a/b

a = √5b a² = 5b² --------------------- (1) 5 │a² = 5 │a.

Therefore by definition, a = 5c for some c Є Z . a² = 25c²

But from eq (1) a² = 5b² 25c² = 5b² 5c² = b² 5 │b² = 5 │b

Hence it is found that 5 divides both a and b. Which contradicts that a have no common factor.Hence √5 is rational.

COUNTER EXAMPLE: -

A counter example to a statement is an example that shows that the statement is false. The name itself suggests that it is an example to counter a given statement.

For example: - To disprove the statement “if n is an odd integer, then n is prime.” Let look for an odd integer which is not prime number. 15 is onesuch integer.So n = 15 is a counter example to the given statement.

Q2(b). Show that √17 is rational.

SOL: -Suppose that √17 is rational. Thus there exist positive integer a and b.Such that √17 = a/b

a = √17b a² = 17b² --------------------- (1) 17 │a² = 17 │a.

Therefore by definition, a = 17c for some c Є Z . a² = 289c²

But from eq (1) a² = 17b² 289c² = 17b² 17c² = b² 17 │b² = 17 │b

Hence it is found that 17 divides both a and b. Which contradicts that a have no common factor.Hence √17 is rational.

Q3(a). What is Logic Circuit? Explain how Boolean algebra methods are used in Logic Circuit Design.Ans: - Logic Circuits are structures which are built from certain elementary circuits called logic gates. They are simply electronic circuits which operates on one or more input signals to perform a particular logical function and producing only one output.

A Boolean expression can be transformed from an algebraic expression into a logic diagram composed of AND, OR and NOT gates. A logic diagram include an inverter circuit for each

variable present in the complement form. For example if any logic expression is defined by F = A+B.C then it is clear that equivalent logic circuit will require one AND gate and one OR gate.Similarly the expression F = A + B’ requires one OR gate and one NOT gate.

For example: - Consider F = A.B +C’

Q3(b). If p and q are statement show whether the statement [(~p~q) Λ (q)] (p V q) = A is a tautology.Ans: - The truth table of given statement is: -

p q ~p ~q (~p~q) (~p~q) Λ(q) (p V q) AT T F F T T T TT F F T T F T TF T T F F F T TF F T T T F F T

From the above table it is clear that given statement is a Tautology.Q4(a). Make logic circuit for the following Boolean Expressions:

(i) (x’. y + z) + (x + z)’

(ii) x’.y’ + y.z’ + z’.x’ + x’ + yAns: - (x’. y + z) + (x + z)’

x’.y’ + y.z’ + z’.x’ + x’ + y

Q4(b). Find Dual of Boolean expression for the output of following logic circuit?

Ans.

Hence by solving the Boolean expression for the output of the above logic circuit is : [{( A’.B ). B’ }’ + C]’

Q4(c). Set A, B and C areA = {1,2,3,4,9}, B = {1,2}, C = {2,5,11}. Find A∩BUC and

AUB∩C.Ans: - (i) A ∩ B = {1,2}

A∩BUC = {1,2,5,11}

(ii) B ∩ C = {2}AUB∩C = {1,2,3,4,9}

Q5(a). Draw Venn Diagram to represent the following:(i) (A ∆ B) U (C ~A)(ii) (A U B) ∩ (B~C)

Ans: - (i (A ∆ B) U (C ~A)

(iii) (A U B) ∩ (B~C)

(A ∆ B)(C~A)

(A ∆ B)UC~A

(A U B)(B~C)

(AUB)∩(B~C)

Q5(b). Give geometric representation for following:(i) {3} x R(ii) (1, 2) x (2, -3)

Ans: - The geometric diagram fir {3} X R is as follows:

The geometric diagram fir (1,2) X(2,-3) is as follows:(1, 2) X (2, -3) = {(1,2),(2,2), (1,-3),(2,-3)

Q5(c). Explain the use of Counter example.Ans: - As the name “counter example” itself suggests that it is an example to counter a given statement. A counter example to a statement is an example that shows that the statement is false.

For example: - The statement “All men speak English”. it is quite likely that someone does not know English.(i.e., for whom the statement is not true). A counter example to a statement proves that p is false i.e., ~p is true.Example: - (. a Є R)( . b Є R)

[(a² = b² )] (a = b)

For disapproving above statement a pair of real numbers a and b is needed for which a² = b² but a ≠ b i.e. a counter example. One such counter example is a =1 and b = -1.

Q6(a). What is inclusion-exclusion principle? Also explain one application of inclusion exclusion princliple.Ans: - Inclusion-Exclusion principle:-

Let A and B be any finite sets. Then n(A U B) = n(A) + n(B) – n(A ∩ B)

in other words to find the number n(A U B) of element in the union A U B add n(A) and n(B) and then subtract n(A ∩ B) i.e. “include” n(A) and n(B) and “exclude” n(A ∩ B). This principle hods for any number of sets.Application of Inclusion-Exclusion :-

Inclusion exclusion principle is used for counting the number of surjective functions, finding probability and finding the number of derangements.

For example: - Find the probability of the student in a college studying Japanese given that: All student have to study one language out of Hindi, Spanish, and Japanese. 65 study Hindi,45 study Spanish and 42 study Japanese. Further 20 study Hindi and Spanish, 25 study Hindi and Japanese, 15 study Spanish and Japanese and 8 study all three language.

Solution: - The total no. of student is │H U S U J │ where H, S and J are no. of student of Hindi, Spanish and Japanese respectively. By Inclusion Exclusion Principle, we have:│HUSUJ│ = │H│+│S│+│J│- │H∩S│-│H∩J│-│S∩J│+│H∩S∩J │

= 65 + 45 + 42 – 20 – 25 – 15 – 8 = 100

Therefore the required probability is [ │J│/ 100 = 0.42 ]

Q6(b). Find the inverse of the following(i) f(x) = (x³ + 5)/(x - 3), x ≠ 3

Solution: Interchanging x and y, we get x = (y³ + 5)/(y - 3)

Solving for y, we getY = √(-2x - 5)

Replacing y by f ¹־ (x), we getf ¹־ (x) = √(-2x - 5)

(ii) f(x) = (x³ - 8)/(x² - 4), x ≠ ± 2

Solution: Interchanging x and y, we getx = (y³ - 8)/(y² - 4)

Solving for y, we getY = (-3x + 8)

Replacing y by f ¹־ (x), we getf ¹־ (x) = (-3x + 8)

Q7(a). How many 4 digit numbers are even? How many 4 digit numbers are composed of Odd Digits?

Ans: For the number to be even the last digit should be even. So the last fourth position of the number can be filled up in 5 ways. If the Third position is filled by 0 then the Second position can be filled in 8 ways and the First position can be filled in 7 ways.

Thus the number of even number with 0 in third position and all digits distinct are 7.8.1.5 = 280 by thee multiplication principle.

If 0 is placed in second position then number of even numbers are 7.1.8.5 = 280.

If the third position is filled by digit other than 0,the number of even number are 6.7.8.5 = 1680.

Similarly it the second position is filled by digit other than 0, then the number of even number are 6.7.8.5 = 1680.

Therefore the total number of 4 digit number= 280+280+1680+1680= 3920

By multiplication principle the 4 digit number composed of odd digit.= 5.5.5.5 = 625

Q7(b). How many different 15 persons committees can be formed each containing at least 8 women and at least one man from a set of 10 Women and 12 men.Solution: Total no of persons = 15

Total no of women = 10Total no of men = 12

The committee of 15 persons can be possible only in 3 ways, either (10 women and 5 men) or (9 women and 6 men) or (7 women and 7 men). 10C10 + 12C5 + 10C9 . 12C6 + 10C8 . 12C7

! 10 / !10!(10-10) . !12 / !5!(12 - 5) + !10 / !9!(10-9) . !12 / !6!(12-6) + !10/!8!(10-8) + !12 / !7 1(12-7)

792 + 9240 + 35640 45672

Q7(c). Explain Bijective mapping with an example.Ans: BIJECTIVE MAPPING: -A mapping f : A b is said to be bijective if and only there is both one to one and onto mapping between both A and B.

For example let f be the function from A to B where A = (1,2,3,4) and B = (a,b,c,d) with f(1) = d, f(2) = b, f(3) = c, f(4) = a, then f is bijective function, f is one-one since the function takes distinct values. It is also onto since every element of B is the image of some element in A. Hence f is a bijective function.

Q8(a). What are Demorgan’s Law?

F