Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.
12th Maths Part 1 Mathematical Logic Miscellaneous Exercise 1 Questions And Answers Maharashtra Board
Question 1.
Select and write the correct answer from the given alternatives in each of the following questions:
i) If p β§ q is false and p β¨ q is true, the ________ is not true.
(A) p β¨ q
(B) p β q
(C) ~p β¨ ~q
(D) q β¨ ~p
Solution:
(b) p β q.
(ii) (p β§ q) β r is logically equivalent to ________.
(A) p β (q β r)
(B) (p β§ q) β ~r
(C) (~p β¨ ~q) β ~r
(D) (p β¨ q) β r
Solution:
(a) p β (q β r) [Hint: Use truth table.]
(iii) Inverse of statement pattern (p β¨ q) β (p β§ q) is ________.
(A) (p β§ q) β (p β¨ q)
(B) ~(p β¨ q) β (p β§ q)
(C) (~p β§ ~q) β (~p β¨ ~q)
(D) (~p β¨ ~q) β (~p β§ ~q)
Solution:
(c) (~p β§ ~q) β (~p β¨ ~ q)
(iv) If p β§ q is F, p β q is F then the truth values of p and q are ________.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
Solution:
(b) T, F
(v) The negation of inverse of ~p β q is ________.
(A) q β§ p
(B) ~p β§ ~q
(C) p β§ q
(D) ~q β ~p
Solution:
(a) q β§ p
(vi) The negation of p β§ (q β r) is ________.
(A) ~p β§ (~q β ~r)
(B) p β¨ (~q β¨ r)
(C) ~p β§ (~q β ~r)
(D) ~p β¨ (~q β§ ~r)
Solution:
(d) ~p β¨ (q β§ ~r)
(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ζ x β A such that x + 3 = 8
(B) Ζ x β A such that x + 2 < 9
(C) β±― x β A, x + 6 β₯ 9
(D) Ζ x β A such that x + 6 < 10
Solution:
(c) Ζ x β A, x + 6 β₯ 9.
Question 2.
Which of the following sentences are statements in logic? Justify. Write down the truth
value of the statements :
(i) 4! = 24.
Solution:
It is a statement which is true, hence its truth value is βTβ.
(ii) Ο is an irrational number.
Solution:
It is a statement which is true, hence its truth value is βTβ.
(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is βFβ. β¦.[T β§ F β‘ F]
(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.
(v) cos
2
ΞΈ β sin
2
ΞΈ = cos2ΞΈ for all ΞΈ β R.
Solution:
It is a statement which is true, hence its truth value is βTβ.
(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is βFβ.
Question 3.
Write the truth values of the following statements :
(i) \(\sqrt {5}\) is an irrational but \(3\sqrt {5}\) is a complex number.
Solution:
Let p : \(\sqrt {5}\) is an irrational.
q : \(3\sqrt {5}\) is a complex number.
Then the symbolic form of the given statement is p β§ q.
The truth values of p and q are T and F respectively.
β΄ the truth value of p β§ q is F. β¦ [T β§ F β‘ F]
(ii) β±― n β N, n
2
+ n is even number while n
2
β n is an odd number.
Solution:
Let p : β±― n β N, n
2
+ n is an even number.
q : β±― n β N, n
2
β n is an odd number.
Then the symbolic form of the given statement is p β§ q.
The truth values of p and q are T and F respectively.
β΄ the truth value of p β§ q is F. β¦ [T β§ F β‘ F].
(iii) Ζ n β N such that n + 5 > 10.
Solution:
Ζ n β N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n β₯ 6, where n β N, satisfy n + 5 > 10).
(iv) The square of any even number is odd or the cube of any odd number is odd.
Solution:
Let p : The square of any even number is odd.
q : The cube of any odd number is odd.
Then the symbolic form of the given statement is p β¨ q.
The truth values of p and q are F and T respectively.
β΄ the truth value of p β¨ q is T. β¦ [F β¨ T β‘ T].
(v) In β ABC if all sides are equal then its all angles are equal.
Solution:
Let p : ABC is a triangle and all its sides are equal.
q : Its all angles are equal.
Then the symbolic form of the given statement is p β q
If the truth value of p is T, then the truth value of q is T.
β΄ the truth value of p β q is T. β¦ [T β T β‘ T].
(vi) β±― n β N, n + 6 > 8.
Solution:
β±― n β N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 β N, n = 2 β N do not satisfy n + 6 > 8).
Question 4.
If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of each of the following statement :
(i) Ζ x β A such that x + 8 = 15.
Solution:
True
(ii) β±― x β A, x + 5 < 12.
Solution:
False
(iii) Ζ x β A, such that x + 7 β₯ 11.
Solution:
True
(iv) β±― x β A, 3x β€ 25.
Solution:
False
Question 5.
Write the negations of the following :
(i) β±― n β A, n + 7 > 6.
Solution:
The negation of the given statements are :
Ζ n β A, such that n + 7 β€ 6.
OR Ζ n β A, such that n + 7 β― 6.
(ii) Ζ x β A, such that x + 9 β€ 15.
Solution:
β±― x β A, x + 9 > 15.
(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.
Question 6.
Construct the truth table for each of the following :
(i) p β (q β p)
Solution:
(ii) (~p β¨ ~q) β [~(p β§ q)]
Solution:
(iii) ~(~p β§ ~q) β¨ q
Solution:
(iv) [(p β§ q) β¨ r] β§ [~r β¨ (p β§ q)]
Solution:
(v) [(~p β¨ q) β§ (q β r)] β (p β r)
Solution:
Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(p β q) β§ ~q)] β ~p
Solution:
All the entries in the last column of the above truth table are T.
β΄ [(p β q) β§ ~q)] β ~p is a tautology.
(ii) [(p β¨ q) β§ ~p] β§ ~q
Solution:
All the entries in the last column of the above truth table are F.
β΄ [(p β¨ q) β§ ~p] β§ ~q is a contradiction.
(iii) (p β q) β§ (p β§ ~q)
Solution:
All the entries in the last column of the above truth table are F.
β΄ (p β q) β§ (p β§ ~q) is a contradiction.
(iv) [p β (q β r)] β [(p β§ q) β r]
Solution:
All the entries in the last column of the above truth table are T.
β΄ [p β (q β r)] β [(p β§ q) β r] is a tautology.
(v) [(p β§ (p β q)] β q
Solution:
All the entries in the last column of the above truth table are T.
β΄ [(p β§ (p β q)] β q is a tautology.
(vi) (p β§ q) β¨ (~p β§ q) β¨ (p β¨ ~q) β¨ (~p β§ ~q)
Solution:
All the entries in the last column of the above truth table are T.
β΄ (p β§ q) β¨ (~p β§ q) β¨ (p β¨ ~q) β¨ (~p β§ ~q) is a tautology.
(vii) [(p β¨ ~q) β¨ (~p β§ q)] β§ r
Solution:
The entries in the last column are neither T nor all F.
β΄ [(p β¨ ~q) β¨ (~p β§ q)] β§ r is a contingency.
(viii) (p β q) β¨ (q β p)
Solution:
All the entries in the last column of the above truth table are T.
β΄ (p β q) β¨ (q β p) is a tautology.
Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p β¨ q) is T and (p β§ q) is T
Solution:
Since p β¨ q and p β§ q both are T, from the table the truth values of both p and q are T.
(ii) (p β¨ q) is T and (p β¨ q) β q is F
Solution:
Since the truth values of (p β¨ q) is T and (p β¨ q) β q is F, from the table, the truth values of p and q are T and F respectively.
(iii) (p β§ q) is F and (p β§ q) β q is T
Solution:
Since the truth values of (p β§ q) is F and (p β§ q) β q is T, from the table, the truth values of p and q are either T and F respectively or F and T respectively or both F.
Question 9.
Using truth tables prove the following logical equivalences :
(i) p β q β‘ (p β§ q) β¨ (~p β§ ~q)
Solution:
The entries in the columns 3 and 8 are identical.
β΄ p β q β‘ (p β§ q) β¨ (~p β§ ~q).
(ii) (p β§ q) β r β‘ p β (q β r)
Solution:
The entries in the columns 5 and 7 are identical.
β΄ (p β§ q) β r β‘ p β (q β r).
Question 10.
Using rules in logic, prove the following :
(i) p β q β‘ ~ (p β§ ~q) β§ ~(q β§ ~p)
Solution:
By the rules of negation of biconditional,
~(p β q) β‘ (p β§ ~q) β¨ (q β§ ~p)
β΄ ~ [(p β§ ~ q) β¨ (q β§ ~p)] β‘ p β q
β΄ ~(p β§ ~q) β§ ~(q β§ ~p) β‘ p β q β¦ (Negation of disjunction)
β‘ p β q β‘ ~(p β§ ~ q) β§ ~ (q β§ ~p).
(ii) ~p β§ q β‘ (p β¨ q) β§ ~p
Solution:
(p β¨ q) β§ ~ p
β‘ (p β§ ~p) β¨ (q β§ ~p) β¦ (Distributive Law)
β‘ F β¨ (q β§ ~p) β¦ (Complement Law)
β‘ q β§ ~ p β¦ (Identity Law)
β‘ ~p β§ q β¦(Commutative Law)
β΄ ~p β§ q β‘ (p β¨ q) β§ ~p.
(iii) ~(p β¨ q) β¨ (~p β§ q) β‘ ~p
Solution:
~ (p β¨ q) β¨ (~p β§ q)
β‘ (~p β§ ~q) β¨ (~p β§ q) β¦ (Negation of disjunction)
β‘ ~p β§ (~q β¨ q) β¦ (Distributive Law)
β‘ ~ p β§ T β¦ (Complement Law)
β‘ ~ p β¦ (Identity Law)
β΄ ~(p β¨ q) β¨ (~p β§ q) β‘ ~p.
Question 11.
Using the rules in logic, write the negations of the following :
(i) (p β¨ q) β§ (q β¨ ~r)
Solution:
The negation of (p β¨ q) β§ (q β¨ ~ r) is
~ [(p β¨ q) β§ (q β¨ ~r)]
β‘ ~ (p β¨ q) β¨ ~ (q β¨ ~r) β¦ (Negation of conjunction)
β‘ (~p β§ ~q) β¨ [~q β§ ~(~r)] β¦ (Negation of disjunction)
β‘ {~ p β§ ~q) β¨ (~q β§ r) β¦ (Negation of negation)
β‘ (~q β§ ~p) β¨ (~q β§ r) β¦ (Commutative law)
β‘ (~ q) β§ (~ p β¨ r) β¦ (Distributive Law)
(ii) p β§ (q β¨ r)
Solution:
The negation of p β§ (q β¨ r) is
~ [p β§ (q β¨ r)]
β‘ ~ p β¨ ~(q β¨ r) β¦ (Negation of conjunction)
β‘ ~p β¨ (~q β§ ~r) β¦ (Negation of disjunction)
(iii) (p β q) β§ r
Solution:
The negation of (p β q) β§ r is
~ [(p β q) β§ r]
β‘ ~ (p β q) β¨ (~ r) β¦ (Negation of conjunction)
β‘ (p β§ ~q) β¨ (~ r) β¦ (Negation of implication)
(iv) (~p β§ q) β¨ (p β§ ~q)
Solution:
The negation of (~ p β§ q) β¨ (p β§ ~ q) is
~ [(~p β§ q) β¨ (p β§ ~q)]
β‘ ~(~p β§ q) β§ ~ (p β§ ~q) β¦ (Negation of disjunction)
β‘ [~(~p) β¨ ~q] β§ [~p β¨ ~(q)] β¦ (Negation of conjunction)
β‘ (p β¨ ~ q) β§ (~ p β¨ q) β¦ (Negation of negation)
Question 12.
Express the following circuits in the symbolic form. Prepare the switching table :
(i)
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
~ p : the switch S
1
β is closed or the switch S
1
is open
~ q: the switch S
2
β is closed or the switch S
2
is open.
Then the symbolic form of the given circuit is :
(p β§ q) β¨ (~p) β¨ (p β§ ~q).
(ii)
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
r : the switch S
3
is closed.
Then the symbolic form of the given statement is : (p β¨ q) β§ (p β¨ r).
Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
~ p: the switch S
1
β is closed or the switch S
1
is open
~ q: the switch S
2
β is closed or the switch S
2
is open.
Then the given circuit in symbolic form is :
(p β§ ~q) β¨ (~p β§ q) β¨ (~p β§ ~q)
Using the laws of logic, we have,
(p β§ ~q) β¨ (~p β§ q) β¨ (~p β§ ~ q)
= (p β§ ~q) β¨ [(~p β§ q) β¨ (~p β§ ~q) β¦(By Complement Law)
= (p β§ ~q) β¨ [~p β§ (q β¨ ~q)} (By Distributive Law)
= (p β§ ~q) β¨ (~p β§ T) β¦(By Complement Law)
= (p β§ ~q) β¨ ~ p β¦(By Identity Law)
= (p β¨ ~p) β§ (~q β¨ ~p) β¦(By Distributive Law)
= ~q β¨ ~p β¦(By Identity Law)
= ~p β¨ ~p β¦(By Commutative Law)
Hence, the simplified circuit for the given circuit is :
(ii)
Solution:
(ii) Let p : the switch S
1
is closed
q : the switch S
2
is closed
r : the switch S
3
is closed
s : the switch S
4
is closed
t : the switch S
5
is closed
~ p : the switch S
1
β is closed or the switch S
1
is open
~ q : the switch S
2
β is closed or the switch S
2
is open
~ r : the switch S
3
β is closed or the switch S
3
is open
~ s : the switch S
4
β is closed or the switch S
4
is open
~ t : the switch S
5
β is closed or the switch S
5
is open.
Then the given circuit in symbolic form is
[(p β§ q) β¨ ~r β¨ ~s β¨ ~t] β§ [(p β§ q) β¨ (r β§ s β§ t)]
Using the laws of logic, we have,
[(p β§ q) β¨ ~r β¨ ~s β¨ ~ t] β§ [(p A q) β¨ (r β§ s β§ t)]
= [(pβ§ q) β¨ ~(r β§ s β§ t)] β§ [(p β§ q) β¨ (r β§ s β§ t)] β¦ (By De Morganβs Law)
= (p β§ q) β¨ [ ~(r β§ s β§ t) β§ (r β§ s β§ t)] β¦ (By Distributive Law)
= (p β§ q) β¨ F β¦ (By Complement Law)
= p β§ q β¦ (By Identity Law)
Hence, the alternative simplified circuit is :
Question 14.
Check whether the following switching circuits are logically equivalent β Justify.
(A)
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
r : the switch S
3
is closed
(A) The symbolic form of the given switching circuits are
p β§ (q β¨ r) and (p β§ q) β¨ (p β§ r) respectively.
By Distributive Law, p β§ (q β¨ r) β‘ (p β§ q) β¨ (p β§ r)
Hence, the given switching circuits are logically equivalent.
(B)
Solution:
The symbolic form of the given switching circuits are
(p β¨ q) β§ (p β¨ r) and p β¨ (q β§ r)
By Distributive Law,
p β¨ (q β§ r) β‘ (p β¨ q) β§ (p β¨ r)
Hence, the given switching circuits are logically equivalent.
Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
r : the switch S
3
is closed
~p : the switch S
1
β is closed, or the switch S
1
is open
~q : the switch S
2
β is closed or the switch S
2
is open.
Then the symbolic form Of the given circuit is :
(p β§ q β§ ~p) β¨ (~p β§ q β§ r) β¨ (p β§ q β§ r) β¨ (p β§ ~q β§ r)
Using the laws of logic, we have,
(p β§ q β§ ~p) β¨ (~p β§ q β§ r) β¨ (p β§ q β§ r) β¨ (p β§ ~q β§ r)
β‘ (p β§ ~p β§ q) β¨ (~p β§ q β§ r) β¨ (p β§ q β§ r) y (p β§ ~q β§ r) β¦(By Commutative Law)
β‘ (F β§ q) β¨ (~p β§ q β§ r) β¨ (p β§ q β§ r) β¨ (p β§ ~q β§ r) β¦ (By Complement Law)
β‘ F β¨ (~p β§ q β§ r) β¨ (p β§ q β§ r) β¨ (p β§ ~q β§ r) β¦ (By Identity Law)
β‘ (~p β§ q β§ r) β¨ (p β§ q β§ r) β¨ (p β§ ~q β§ r) β¦ (By Identity Law)
β‘ [(~p β¨ p) β§ (q β§ r)] β¨ (p β§ ~q β§ r) β¦ (By Distributive Law)
β‘ [T β§ (q β§ r)] β¨ (p β§ ~q β§ r) = (q β§ r) β¨ (p β§ ~q β§ r) β¦(By Complement Law)
β‘ (q β§ r) β¨ (p β§ ~q β§ r) β¦ (By Identity Law)
β‘ [q β¨ (p β§ ~q)] β§ r β¦ (By Distributive Law)
β‘ [q β¨ p) β§ ((q β¨ ~q)] β§ r β¦ (By Distributive Law)
β‘ [(q β¨ p) β§ T] β§ r β¦(By Complement Law)
β‘ (q β¨ p) β§ r β¦ (By Identity Law)
β‘ (p β¨ q) β§ r β¦(By Commutative Law)
β΄ the alternative arrangement of the new circuit with minimum switches is :
Question 16.
Simplify the following so that the new circuit circuit.
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
~ p : the switch S
1
β is closed or the switch S
1
is open
~ q : the switch S
2
β is closed or the switch S
2
is open.
Then the symbolic form of the given switching circuit is :
(~p β¨ q) β¨ (p β¨ ~q) β¨ (p β¨ q)
Using the laws of logic, we have,
(~p β¨ q) β¨ (p β¨ ~q) β¨ (p β¨ q)
β‘ (~p β¨ q β¨ p β¨ ~q) β¨ (p β¨ q)
β‘ [(~p β¨ p) β¨ (q β¨ ~q)] β¨ (p β¨ q) β¦ (By Commutative Law)
β‘ (T β¨ T) β¨ (p β¨ q) β¦ (By Complement Law)
β‘ T β¨ (p β¨ q) β¦ (By Identity Law)
β‘ T β¦ (By Identity Law)
β΄ the current always flows whether the switches are open or closed. So, it is not necessary to use any switch in the circuit.
β΄ the simplified form of given circuit is :
Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.
Solution:
Let p : the switch S
1
is closed
q : the switch S
2
is closed
r : the switch S
3
is closed
~ q : the switch S
2
β is closed or the switch S
2
is open
~ r : the switch S
3
β is closed or the switch S
3
is open.
Then, the symbolic form of the given switching circuit is : [p β¨ (~ q) β¨ (~ r)] β§ [p β¨ (q β§ r)]
From the table, theβ final columnβ and the column of p are identical. Hence, the given circuit is equivalent to the simple circuit with only one switch S
1
.
the simplified form of the given circuit is :