## Mathematical Reasoning Quiz-14

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced
.

Q1. Let p and q be two statements. Then, p∨q is false, if
•  p is false and q is true
•  Both p and q are false
•  Both p and q are true
•  None of these
Solution
(b)

Q2.For any two statements p and q, ~p∨q∨(~p∧q)is logically equivalent to
•  p
•  ~p
•  q
•  ~q
Solution
(b) p∨q∨(∼p∧q) ∼p∧∼q∨(∼p∧q) ≡∼p∧(∼q∨q) ≡∼p

Q3.  In which of the following cases, p⇒q is true?
•  p is true, q is true
•  p is false, q is true
•  p is true, q is false
•  None of these
Solution
(a)

Q4. ~p∧q is logically equivalent to
•  p→q
•  q→p
•  ~(p→q)
•  ~(q→p)
Solution
(d)

Q5.A proposition is called a tautology, if it is
•  Always T
•  Always F
•  Sometimes T, sometimes F
•  None of the above
Solution
(a)

Q6. Let S be a non-empty subset of R. Consider the following statement P: There is a rational number x∈S such that x>0. Which of the following statements is the negation of the statement P?
•  There is a rational number x∈S such that x≤0
•  There is no rational number x∈S such that x≤0
• Every rational number x∈S satisfies x≤0
•  x∈S and x≤0⇒x is not rational
Solution
(c) P : There is rational number x∈S such that x>0 ~P : Every rational number x ∈S satisfies x≤0

Q7.Let p and q be two propositions. Then the inverse of the implication p→q is
•  q → p
•  ~p → ~q
•  p→q
•  ~q → ~p
Solution
(b) By definition, the inverse of implication p→q is ∼p→∼q

Q8.Dual of x∧yx=x is
•  x∨y∧x=x
•  x∨y∨x=x
•  x∧yx∧x=x
•  None of these
Solution
(a)

Q9.p∧q→p is
•  A tautology
•  Neither a tautology n or a contradiction
•  None of these
Solution
(a) Clearly, p∧q→p is always true. So, it is a tautology ALITER We know that p→q≅∼p∨q ∴p∧q→p≅∼(p∧q)∨p≅(∼p∨∼q)∨p ≅(∼p∨p)∨∼q≅t∨q≅t

Q10. If statements p and r are false and q is true, then truth value of ∼p⇒(q∧r)∨r is
•  T
•  F
•  Either T or F
• Neither T or F
Solution
(b)

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