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Maths NCERT Exemplar Solutions Class 12th Chapter One Ncert Solutions Maths class 12th Maths Class 12th

The number of choices for $Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}$ , such that $(p Δ q) \Rightarrow ((p Δ ~ q) \lor (~ p) Δ q)$ is a tautology, is :

Option 1 -

Option 2 -

Option 3 -

Option 4 -

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$(p Δ q) \Rightarrow (p Δ \sim q) \lor (\sim p Δ q)$

Case I

When $Δ$ is same as $\lor .$

Then $(p Δ \sim q) \lor (- p Δ q)$ becomes

$(p \lor \sim q) \lor (\sim p \lor q)$ which is always true, so x becomes tautology.

Case II

When $Δ$ is same as $\lor$

Then $(p \land q) \Rightarrow (p \land \sim q) \lor (\sim p \land q)$ becomes $p \land q$ is T, then $(p \land \sim q) \lor (\sim p \land q)$ is false, so x cannot be tautology.

Case III

When $Δ$ is same as $\Rightarrow$

Then $(p \Rightarrow \sim q) \lor (- p \Rightarrow q)$ is same as $(\sim p \lor \sim q) \lor (p \lor q)$ which is true, so x becomes tautology.

Case IV

When $Δ$ is same as $\Leftrightarrow$

Then $(p \Leftrightarrow q) \Rightarrow (p \Leftrightarrow \sim q) \lor (\sim p \Leftrightarrow q)$

$p \Leftrightarrow q$ is true when p and q have same truth values $p \Leftrightarrow \sim q a n d \sim p \Leftrightarrow q$ both are false. Hence x cannot be tautology.

<math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mi>Δ</mi> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mi>Δ</mi> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>∼</mo> <mi>p</mi> <mi>Δ</mi> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> Case IWhen  <math> <mrow> <mi>Δ</mi> </mrow> </math>  is same as <math> <mrow> <mo>∨</mo> <mo>.</mo> </mrow> </math>  Then  <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mi>Δ</mi> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>−</mo> <mi>p</mi> <mi>Δ</mi> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> becomes <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∨</mo> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>∼</mo> <mi>p</mi> <mo>∨</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> which is always true, so x becomes tautology.Case IIWhen  <math> <mrow> <mi>Δ</mi> </mrow> </math>  is same as <math> <mrow> <mo>∨</mo> </mrow> </math>  Then  <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∧</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∧</mo> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>∼</mo> <mi>p</mi> <mo>∧</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math>  becomes <math> <mrow> <mi>p</mi> <mo>∧</mo> <mi>q</mi> </mrow> </math> is T, then <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∧</mo> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>∼</mo> <mi>p</mi> <mo>∧</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> is false, so x cannot be tautology.Case IIIWhen  <math> <mrow> <mi>Δ</mi> </mrow> </math>  is same as <math> <mrow> <mo>⇒</mo> </mrow> </math>  Then  <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>⇒</mo> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>−</mo> <mi>p</mi> <mo>⇒</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> is same as <math> <mrow> <mrow> <mo>(</mo> <mrow> <mo>∼</mo> <mi>p</mi> <mo>∨</mo> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∨</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> which is true, so x becomes tautology.Case IVWhen  <math> <mrow> <mi>Δ</mi> </mrow> </math> is same as <math> <mrow> <mo>⇔</mo> </mrow> </math>  Then   <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>⇔</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>⇔</mo> <mo>∼</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mrow> <mo>(</mo> <mrow> <mo>∼</mo> <mi>p</mi> <mo>⇔</mo> <mi>q</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math> <mrow> <mi>p</mi> <mo>⇔</mo> <mi>q</mi> </mrow> </math> is true when p and q have same truth values <math> <mrow> <mi>p</mi> <mo>⇔</mo> <mo>∼</mo> <mi>q</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mo>∼</mo> <mi>p</mi> <mo>⇔</mo> <mi>q</mi> </mrow> </math>  both are false. Hence x cannot be tautology.

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The number of choices for $Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}$ , such that $(p Δ q) \Rightarrow ((p Δ ~ q) \lor (~ p) Δ q)$ is a tautology, is :

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The number of choices for Δ ∈ { ∧ , ∨ , ⇒ , ⇔ } , such that ( p Δ q ) ⇒ ( ( p Δ ~ q ) ∨ ( ~ p ) Δ q ) is a tautology, is :

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The number of choices for $Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}$ , such that $(p Δ q) \Rightarrow ((p Δ ~ q) \lor (~ p) Δ q)$ is a tautology, is :