Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 28

EASY

KCET (UG)

IMPORTANT

Earn 100

The negation of the compound statement $q \lor ~ (p \land r)$ is

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Important Questions on Mathematical Reasoning

EASY

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 29

Negation of the proposition

(p \lor (~ q)) \lor (q)

is equivalent to

MEDIUM

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 30

The proposition

~ ((p \lor q) \land (~ q \land ~ p))

is equivalent to :

MEDIUM

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 31

The statement

(p \to (q \to p)) \to (p \to (p \lor q))

is :

MEDIUM

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 32

The proposition

p \to ~ (p \land ~ q)

is equivalent to :

EASY

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 33

If

a :

Vijay becomes a doctor,

b :

Ajay is an engineer. Then the statement ' Vijay becomes a doctor if and only if Ajay is an engineer' can be written in symbolic form as

EASY

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 35

Negation of

(p \land q) \to (~ p \lor r)

is

EASY

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 36

Negation of

p \leftrightarrow q

is

EASY

KCET (UG)

IMPORTANT

Mathematics Crash Course KCET (UG)>Chapter 5 - Mathematical Reasoning>Exercise 1>Q 37

If $p, q$ and $r$ are simple propositions with truth values $T, F$ and $T$ , then the truth value of $((~ p \lor q) \land ~ p) \Rightarrow r$ is