## MATHEMATICS REASONING QUIZ-2

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. If a compound statement r is contradiction, then the truth value of (p⇒q)∧r∧p[p⇒∼r] is
•  TM
•  F
•  T or F
•  None of these

Q2.12. Which of the following is equivalent to p⇒q? :
•  p⇒q
•  q⇒p
•  (p⇒q)∧(q⇒p)
•  None of these

Q3.  When does the inverse of the statement ∼p⇒q results in T?
•   p and q both are true
•  p is true and q is false
•  p is false and q is false
•  Both (b) and (c)

Q4. H: Set of holidays S: Set of Sundays U:Set of day's Then, the Venn diagram of statement, "Every Sunday implies holiday" is
•
•
•
•
Solution
The required venn diagram of given statement is
Q5.Dual of (x'∨y' )'=x∧y is
•  (x'∨y' )=x∨y
•  (x'∧y' )'=x∨y
•  (x'∧y' )'=x∧y
•  None of the above
Solution
Dual of (x'∨y' )'=x∧y is (x'∧y' )=x∨y

Q6. The switching function for the following network is
•  (p∧q∨r)∧t
•  (p∧q∨r)∨t
•  p∨r∧q∨t
•  None of these
Solution
The switching function for the given network is (p∧q∨r)∨t

Q7.The contrapositive of the statement ∼p⇒(p∧∼q) is
•  p⇒(∼p∨q)
•  p⇒(p∧q)
•  p⇒(∼p∧q)
•  ∼p∨q⇒p
Solution
We have, ∼{p∨(∼p∨q)} ≅∼{(p∨∼p)∨q}≅∼(t∨q)≅∼t≅c Also, (p∧∼q)∧∼p≅(p∧∼p∧∼q)≅c∧∼q≅c So, option (a) is correct

Q8.Let p and q be two statements. Then, (∼p∨q)∧(∼p∧∼q) is a
•  Tautology
•  Neither tautology nor contradiction
•  Both tautology and contradiction

Q9.Which of the following propositions is a tautology?
•  (~p∨∼q)∨(p∨∼q)
•  (∼p∨∼q)∧(p∨∼q)
•  ∼p∧(∼p∨∼q)
•  ∼q∧(∼p∨∼q)
Solution
We have, (∼p∨∼q)∨(p∨∼q)=∼p∨(∼q∨(p ∨∼q)) =∼p∨(p∨∼q)=(∼p∨p)∨∼q=t∨∼q=t

Q10. Which of the following is a contradiction?
•  p∨q
•  p∧q
•  p∨∼p
•  p∧∼p

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