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The negation of the statement “A circle is an ellipse” is

(a) An ellipse is a circle.

(b) An ellipse is not a circle.

(c) A circle is not an ellipse.

(d) A circle is an ellipse.

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is an Objective Type Questions as classified in NCERT Exemplar

    L e t     p : A     c i r c l e     i s     a n     e l l i p s e .           ∼ p : A     c i r c l e     i s     n o t     a n     e l l i p s e . H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( c ) .

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Let Δ , ∇ ∈ { ∧ , ∨ } be such that p ∇ q ⇒ ( ( p Δ q ) ∇ r ) is a tautology. Then ( p ∇ q ) Δ   r  is logically equivalent to:
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Raj Pandey

Case – I Δ ≡ ∇ ≡ ∧  

              ( p ∧ q ) → ( ( p ∧ q ) ∧ r )  

              it can be false if r is false,

              so not a tautology

              Case – II If Δ ≡ ∇ ≡ ∨  

              ( p ∨ q ) ?     → ( ( p ∨ q ) ∨ r ) ≡ tautology

              then ( p ∨ q ) ∨ r ≡ ( p Δ r ) ∨ q   

              Case – I

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Negation of the Boolean statement ( p ∨ q ) ⇒ ( ( ∼ r ) ∨ p )  is equivalent to
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Vishal Baghel

pva ( ∼ r   v   p )

∼ ( p ∨ q ) ∨ ( ∼ r v p )

its negation as asked in question

( p ∨ q ) ∧ ( ∼ p ∧ r )

= ( p ∧ ∼ p ∧ r ) ∨ ( q ∧ r ∧ ∼ p )

= ( p ∧ r ∧ ∼ p )                   [ a s     p ∧ ∼ p     i s     f a l s e ]

The remainder when (2021)2023 is divided by 7 is:
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Raj Pandey

2 0 2 1 ≡ − 2 mod (7)

⇒ ( 2 0 2 1 ) 2 0 2 3 ≡ ( − 2 ) 2 0 2 3 m o d ( 7 )   …… (i)

Now,   ( − 2 ) 3 ≡ − 1 m o d ( 7 )  

⇒ ( − 2 ) 2 0 2 3 ≡ ( − 2 ) m o d ( 7 ) ≡ 5 m o d ( 7 )       ……. (ii)

(i) & (ii)

⇒ ( 2 0 2 1 ) 2 0 2 3 ≡ 5 m o d ( 7 )  

∴  Remainder = 5

Let p, q, r be three logical statements. Consider the compound statements

  S 1 : ( ( ∼ p ) ∨ q ) ∨ ( ( ∼ p ) ∨ r )     a n d  

  S 2 : p → ( q ∨ r )  

 Then, which of the following is NOT true?
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Raj Pandey

kindly consider the following Image 

 

The statement ( ∼ ( p ⇔ ∼ q ) ) ∧ q     i s :  
R
Raj Pandey

∼ ( p ⇔ ∼ q ) ∧ q = ( p ⇔ ∼ q ) ∧ q     i s :

∴ ( ∼ ( P ⇔ ∼ Q ) ) ∧ q is equivalent to ( p ⇒ q ) ∧

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