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

(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.

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) cos^{2}θ – sin^{2}θ = cos2θ for all θ ∈ R.

Solution:

It is a statement which is true, hence its truth value is ‘T’.

(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, n^{2} + n is even number while n^{2} – n is an odd number.

Solution:

Let p : Ɐ n ∈ N, n^{2} + n is an even number.

q : Ɐ n ∈ N, n^{2} – 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].

(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].

(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

(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.

Question 6.

Construct the truth table for each of the following :

(i) p → (q → p)

Solution:

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]

Solution:

(iii) ~(~p ∧ ~q) ∨ q

Solution:

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]

Solution:

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)

Solution:

Question 7.

Determine whether the following statement patterns are tautologies contradictions or contingencies :

(i) [(p → q) ∧ ~q)] → ~p

Solution:

All the entries in the last column of the above truth table are T.

∴ [(p → q) ∧ ~q)] → ~p is a tautology.

(ii) [(p ∨ q) ∧ ~p] ∧ ~q

Solution:

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:

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:

All the entries in the last column of the above truth table are T.

∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

(v) [(p ∧ (p → q)] → q

Solution:

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:

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:

The entries in the last column are neither T nor all F.

∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

(viii) (p → q) ∨ (q → p)

Solution:

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:

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:

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.

(iii) (p ∧ q) is F and (p ∧ q) → q is T

Solution:

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:

The entries in the columns 3 and 8 are identical.

∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

(ii) (p ∧ q) → r ≡ p → (q → r)

Solution:

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).

(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.

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)

(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)

Solution:

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

~ p : the switch S_{1}‘ is closed or the switch S_{1} is open

~ q: the switch S_{2}‘ is closed or the switch S_{2} is open.

Then the symbolic form of the given circuit is :

(p ∧ q) ∨ (~p) ∨ (p ∧ ~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 ∨ q) ∧ (p ∨ 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}‘ is closed or the switch S_{1} is open

~ q: the switch S_{2}‘ is closed or the switch S_{2} 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 :

(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}‘ is closed or the switch S_{1} is open

~ q : the switch S_{2}‘ is closed or the switch S_{2} is open

~ r : the switch S_{3}‘ is closed or the switch S_{3} is open

~ s : the switch S_{4}‘ is closed or the switch S_{4} is open

~ t : the switch S_{5}‘ is closed or the switch S_{5} 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 :

Question 14.

Check whether the following switching circuits are logically equivalent – 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 ∧ (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)

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.

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}‘ is closed, or the switch S_{1} is open

~q : the switch S_{2}‘ is closed or the switch S_{2} 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 :

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}‘ is closed or the switch S_{1} is open

~ q : the switch S_{2}‘ is closed or the switch S_{2} 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 :

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}‘ is closed or the switch S_{2} is open

~ r : the switch S_{3}‘ is closed or the switch S_{3} is open.

Then, the symbolic form of the given switching circuit is : [p ∨ (~ q) ∨ (~ r)] ∧ [p ∨ (q ∧ r)]

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 S_{1}.

the simplified form of the given circuit is :