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Published on: 13/05/2022
QB365 provides detailed and simple solution for every book back questions in class 12 Maths subject.It will helps to get more idea about question pattern in every book back questions with solution.
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Take MCQ Maths Test1.
Let \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) ,C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)be any three boolean matrices of the same type.
Find (A∧B)∨C
2.
Let \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) ,C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)be any three boolean matrices of the same type.
Find AΛB
3.
Establish the equivalence property p ➝ q ≡ ㄱp ν q
4.
Show that ¬(p↔️q) ≡ p↔️¬q
5.
Show that p ➝ q and q ➝ p are not equivalent
6.
Verify whether the following compound propositions are tautologies or contradictions or contingency
( p ⟶ q) ↔️ (~ p ⟶ q)
7.
Verify whether the following compound propositions are tautologies or contradictions or contingency
(( p V q)∧ ¬ p) ➝ q
8.
Verify whether the following compound propositions are tautologies or contradictions or contingency
(p ∧ q) ∧ ¬ (p ∨ q)
9.
Prove that q ➝ p ≡ ¬p ➝ ¬q
10.
Show that ¬( p ∧ q) ≡ ¬p V ¬q
1.
Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\((A\wedge B)\vee C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
2.
Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 1\wedge 1 & 0\wedge 0 \end{matrix}\begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 0\wedge 0 & 1\wedge 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \) [∵ a៱b=max(a,b)]
3.
| p | q | ㄱp | p ➝ q | ㄱp ν q |
| T | T | F | T | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
The entries in the columns corresponding to p → q and ㄱp ν q are identical and hence they are equivalent.
4.
| p | q | p↔️q | ~(p↔️q) | ~q | p↔️~q |
| T | T | T | F | F | F |
| T | F | F | T | T | T |
| F | T | F | T | F | T |
| F | F | T | F | T | F |
The entries in column (4) and (6) are identical ~(p↔️q) ≡ p↔️~q
5.
| p | q | p ➝ q | q ➝ p |
| T | T | T | T |
| T | F | F | T |
| F | T | T | F |
| F | F | T | T |
The entries in column (3) and column (4) are not identical.
6.
| p | q | p ⟶ q | ~p | ~p ⟶ q | ( p ⟶ q) ↔️ (~p ⟶ q) |
| T | T | T | F | T | T |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | F | F |
Since this is neither a tautology not a contradiction
( p ⟶ q) ↔️ (~p ⟶ q) is a contingency.
7.
(( p V q)∧ ~p)) ➝ q
| p | q | p V q | ~p | ( p V q) ∧ ~q | ( p V q) ∧ ~q |
| T | T | T | F | F | T |
| T | F | T | F | F | T |
| F | T | T | T | T | T |
| F | F | F | T | F | T |
The statement (( p V q)∧ ~p) ➝ q is a tautology.
8.
| p | q | p ∧ q | p ∨ q | ~p ∨ q | (p ∧ q) ∧ ~(p ∨ q) |
| T | T | T | T | F | F |
| T | F | F | T | F | F |
| F | T | F | T | F | F |
| F | F | F | F | T | F |
The statement (p ∧ q) ∧ ~(p ∨ q)
9.
| p | q | q ➝ p | ~p | ~q | ~q ➝ ~p |
| T | T | T | F | F | T |
| T | F | T | F | T | T |
| F | T | F | T | F | F |
| F | F | T | T | T | T |
The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved
10.
~( p ∧ q) ≡ ~p V ~q
| p | q | p ∧ q | ~(p ∧ q) | ~p | ~p | ~p V ~q |
| T | T | T | F | F | F | F |
| T | F | F | T | F | T | T |
| F | T | F | T | T | F | T |
| F | F | F | T | T | T | T |
From column (4) and column (7), the entries are Identical.
ஃ ~( p ∧ q) ≡ ~p V ~q are identical and hence they are equivalent.
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Tamilnadu Stateboard 12th Standard Subjects
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