Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1

Question 1.
Select and write the correct answer from the given alternatives in each of the following questions :
i) If p ∧ q is false and p ∨ q is true, the ________ is not true.
(A) p ∨ q
(B) p ↔ q
(C) ~p ∨ ~q
(D) q ∨ ~p
Solution:
(b) p ↔ q.

(ii) (p ∧ q) → r is logically equivalent to ________.
(A) p → (q → r)
(B) (p ∧ q) → ~r
(C) (~p ∨ ~q) → ~r
(D) (p ∨ q) → r
Solution:
(a) p → (q → r) [Hint: Use truth table.]

(iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~p ∧ ~q) → (~p ∨ ~q)
(D) (~p ∨ ~q) → (~p ∧ ~q)
Solution:
(c) (~p ∧ ~q) → (~p ∨ ~ q)

(iv) If p ∧ q is F, p → 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

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) The negation of inverse of ~p → q is ________.
(A) q ∧ p
(B) ~p ∧ ~q
(C) p ∧ q
(D) ~q → ~p
Solution:
(a) q ∧ p

(vi) The negation of p ∧ (q → r) is ________.
(A) ~p ∧ (~q → ~r)
(B) p ∨ (~q ∨ r)
(C) ~p ∧ (~q → ~r)
(D) ~p ∨ (~q ∧ ~r)
Solution:
(d) ~p ∨ (q ∧ ~r)

(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ǝ x ∈ A such that x + 3 = 8
(B) Ǝ x ∈ A such that x + 2 < 9
(C) Ɐ x ∈ A, x + 6 ≥ 9
(D) Ǝ x ∈ A such that x + 6 < 10
Solution:
(c) Ǝ x ∈ A, x + 6 ≥ 9.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

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 ‘T’.

(ii) π is an irrational number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is ‘F’. ….[T ∧ F ≡ F]

(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.

(v) cos2θ – sin2θ = cos2θ for all θ ∈ R.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

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 ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(ii) Ɐ n ∈ N, n2 + n is even number while n2 – n is an odd number.
Solution:
Let p : Ɐ n ∈ N, n2 + n is an even number.
q : Ɐ n ∈ N, n2 – n is an odd number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F].

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) Ǝ n ∈ N such that n + 5 > 10.
Solution:
Ǝ n ∈ N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n ≥ 6, where n ∈ 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 ∨ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ∨ q is T. … [F ∨ T ≡ T].

(v) In ∆ 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 → q
If the truth value of p is T, then the truth value of q is T.
∴ the truth value of p → q is T. … [T → T ≡ T].

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) Ɐ n ∈ N, n + 6 > 8.
Solution:
Ɐ n ∈ N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 ∈ N, n = 2 ∈ 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) Ǝ x ∈ A such that x + 8 = 15.
Solution:
True

(ii) Ɐ x ∈ A, x + 5 < 12.
Solution:
False

(iii) Ǝ x ∈ A, such that x + 7 ≥ 11.
Solution:
True

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) Ɐ x ∈ A, 3x ≤ 25.
Solution:
False

Question 5.
Write the negations of the following :
(i) Ɐ n ∈ A, n + 7 > 6.
Solution:
The negation of the given statements are :
Ǝ n ∈ A, such that n + 7 ≤ 6.
OR Ǝ n ∈ A, such that n + 7 ≯ 6.

(ii) Ǝ x ∈ A, such that x + 9 ≤ 15.
Solution:
Ɐ x ∈ A, x + 9 > 15.

(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 6.
Construct the truth table for each of the following :
(i) p → (q → p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 1

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 2

(iii) ~(~p ∧ ~q) ∨ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 3

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 4

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 5

Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(p → q) ∧ ~q)] → ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 6
All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q)] → ~p is a tautology.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) [(p ∨ q) ∧ ~p] ∧ ~q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 7
All the entries in the last column of the above truth table are F.
∴ [(p ∨ q) ∧ ~p] ∧ ~q is a contradiction.

(iii) (p → q) ∧ (p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 8
All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

(iv) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 9
All the entries in the last column of the above truth table are T.
∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) [(p ∧ (p → q)] → q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 10
All the entries in the last column of the above truth table are T.
∴ [(p ∧ (p → q)] → q is a tautology.

(vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 11
All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) is a tautology.

(vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 12
The entries in the last column are neither T nor all F.
∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(viii) (p → q) ∨ (q → p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 13
All the entries in the last column of the above truth table are T.
∴ (p → q) ∨ (q → p) is a tautology.

Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p ∨ q) is T and (p ∧ q) is T
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 14
Since p ∨ q and p ∧ q both are T, from the table the truth values of both p and q are T.

(ii) (p ∨ q) is T and (p ∨ q) → q is F
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 15
Since the truth values of (p ∨ q) is T and (p ∨ q) → q is F, from the table, the truth values of p and q are T and F respectively.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 16
Since the truth values of (p ∧ q) is F and (p ∧ q) → 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 ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 17
The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 18
The entries in the columns 5 and 7 are identical.
∴ (p ∧ q) → r ≡ p → (q → r).

Question 10.
Using rules in logic, prove the following :
(i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)
Solution:
By the rules of negation of biconditional,
~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q
∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q … (Negation of disjunction)
≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
(p ∨ q) ∧ ~ p
≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)
≡ F ∨ (q ∧ ~p) … (Complement Law)
≡ q ∧ ~ p … (Identity Law)
≡ ~p ∧ q …(Commutative Law)
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
~ (p ∨ q) ∨ (~p ∧ q)
≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)
≡ ~p ∧ (~q ∨ q) … (Distributive Law)
≡ ~ p ∧ T … (Complement Law)
≡ ~ p … (Identity Law)
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 11.
Using the rules in logic, write the negations of the following :
(i) (p ∨ q) ∧ (q ∨ ~r)
Solution:
The negation of (p ∨ q) ∧ (q ∨ ~ r) is
~ [(p ∨ q) ∧ (q ∨ ~r)]
≡ ~ (p ∨ q) ∨ ~ (q ∨ ~r) … (Negation of conjunction)
≡ (~p ∧ ~q) ∨ [~q ∧ ~(~r)] … (Negation of disjunction)
≡ {~ p ∧ ~q) ∨ (~q ∧ r) … (Negation of negation)
≡ (~q ∧ ~p) ∨ (~q ∧ r) … (Commutative law)
≡ (~ q) ∧ (~ p ∨ r) … (Distributive Law)

(ii) p ∧ (q ∨ r)
Solution:
The negation of p ∧ (q ∨ r) is
~ [p ∧ (q ∨ r)]
≡ ~ p ∨ ~(q ∨ r) … (Negation of conjunction)
≡ ~p ∨ (~q ∧ ~r) … (Negation of disjunction)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~ [(p → q) ∧ r]
≡ ~ (p → q) ∨ (~ r) … (Negation of conjunction)
≡ (p ∧ ~q) ∨ (~ r) … (Negation of implication)

(iv) (~p ∧ q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∧ q) ∨ (p ∧ ~ q) is
~ [(~p ∧ q) ∨ (p ∧ ~q)]
≡ ~(~p ∧ q) ∧ ~ (p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∨ ~q] ∧ [~p ∨ ~(q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~ p ∨ q) … (Negation of negation)

Question 12.
Express the following circuits in the symbolic form. Prepare the switching table :
(i)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 19
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q: the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given circuit is :
(p ∧ q) ∨ (~p) ∨ (p ∧ ~q).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 21

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 20
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed.
Then the symbolic form of the given statement is : (p ∨ q) ∧ (p ∨ r).
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 22

Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 23
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p: the switch S1‘ is closed or the switch S1 is open
~ q: the switch S2‘ is closed or the switch S2 is open.
Then the given circuit in symbolic form is :
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~q)
Using the laws of logic, we have,
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~ q)
= (p ∧ ~q) ∨ [(~p ∧ q) ∨ (~p ∧ ~q) …(By Complement Law)
= (p ∧ ~q) ∨ [~p ∧ (q ∨ ~q)} (By Distributive Law)
= (p ∧ ~q) ∨ (~p ∧ T) …(By Complement Law)
= (p ∧ ~q) ∨ ~ p …(By Identity Law)
= (p ∨ ~p) ∧ (~q ∨ ~p) …(By Distributive Law)
= ~q ∨ ~p …(By Identity Law)
= ~p ∨ ~p …(By Commutative Law)
Hence, the simplified circuit for the given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 24

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 25
Solution:
(ii) Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
s : the switch S4 is closed
t : the switch S5 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open
~ s : the switch S4‘ is closed or the switch S4 is open
~ t : the switch S5‘ is closed or the switch S5 is open.
Then the given circuit in symbolic form is
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]
Using the laws of logic, we have,
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~ t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]
= [(p∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … (By De Morgan’s Law)
= (p ∧ q) ∨ [ ~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … (By Distributive Law)
= (p ∧ q) ∨ F … (By Complement Law)
= p ∧ q … (By Identity Law)
Hence, the alternative simplified circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 26

Question 14.
Check whether the following switching circuits are logically equivalent – Justify.
(A)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 27
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
(A) The symbolic form of the given switching circuits are
p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) respectively.
By Distributive Law, p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Hence, the given switching circuits are logically equivalent.

(B)
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 28
Solution:
The symbolic form of the given switching circuits are
(p ∨ q) ∧ (p ∨ r) and p ∨ (q ∧ r)
By Distributive Law,
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Hence, the given switching circuits are logically equivalent.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 29
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~p : the switch S1‘ is closed, or the switch S1 is open
~q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form Of the given circuit is :
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
≡ (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) y (p ∧ ~q ∧ r) …(By Commutative Law)
≡ (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Complement Law)
≡ F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [(~p ∨ p) ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) … (By Distributive Law)
≡ [T ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) = (q ∧ r) ∨ (p ∧ ~q ∧ r) …(By Complement Law)
≡ (q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [q ∨ (p ∧ ~q)] ∧ r … (By Distributive Law)
≡ [q ∨ p) ∧ ((q ∨ ~q)] ∧ r … (By Distributive Law)
≡ [(q ∨ p) ∧ T] ∧ r …(By Complement Law)
≡ (q ∨ p) ∧ r … (By Identity Law)
≡ (p ∨ q) ∧ r …(By Commutative Law)
∴ the alternative arrangement of the new circuit with minimum switches is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 30

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 16.
Simplify the following so that the new circuit circuit.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 31
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
~ p : the switch S1‘ is closed or the switch S1 is open
~ q : the switch S2‘ is closed or the switch S2 is open.
Then the symbolic form of the given switching circuit is :
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
Using the laws of logic, we have,
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
≡ (~p ∨ q ∨ p ∨ ~q) ∨ (p ∨ q)
≡ [(~p ∨ p) ∨ (q ∨ ~q)] ∨ (p ∨ q) … (By Commutative Law)
≡ (T ∨ T) ∨ (p ∨ q) … (By Complement Law)
≡ T ∨ (p ∨ q) … (By Identity Law)
≡ T … (By Identity Law)
∴ the current always flows whether the switches are open or closed. So, it is not necessary to use any switch in the circuit.
∴ the simplified form of given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 32

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 33
Solution:
Let p : the switch S1 is closed
q : the switch S2 is closed
r : the switch S3 is closed
~ q : the switch S2‘ is closed or the switch S2 is open
~ r : the switch S3‘ is closed or the switch S3 is open.
Then, the symbolic form of the given switching circuit is : [p ∨ (~ q) ∨ (~ r)] ∧ [p ∨ (q ∧ r)]
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 34
From the table, the’ final column’ and the column of p are identical. Hence, the given circuit is equivalent to the simple circuit with only one switch S1.
the simplified form of the given circuit is :
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1 35

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