MATHEMATICS REASONING QUIZ-5
Dear Readers,
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. When does the value of the statement p(∧r)⇔(r∧q) become false?
Q2.For the circuit shown below, the Boolean polynomial is
:
Solution
For the given circuit, Boolean polynomial is (~p∧q)∨(p∨~q).
For the given circuit, Boolean polynomial is (~p∧q)∨(p∨~q).
Q3. Consider the proposition : “ If the pressure increases, the volume decreases:. The negation of this proposition is
Solution
We know that p→q≅∼q→∼p So, the given statement is equivalent to: If the volume does not decrease, the pressure does not increase
We know that p→q≅∼q→∼p So, the given statement is equivalent to: If the volume does not decrease, the pressure does not increase
Q4. Which of the following propositions is a tautology?
Solution
We have, p↔q≅∼p∨q ∴(p↔q)∨(p∧∼q) =(∼p∨q)∨(p∧∼q) ={(∼p∨q)∨p}∧{(∼p∨q)∨∼q)} ={(∼p∨p)∨q}∧{∼p∨(q∨∼q)} =(t∨q)∧(∼p∨t) =t∧t=t
We have, p↔q≅∼p∨q ∴(p↔q)∨(p∧∼q) =(∼p∨q)∨(p∧∼q) ={(∼p∨q)∨p}∧{(∼p∨q)∨∼q)} ={(∼p∨p)∨q}∧{∼p∨(q∨∼q)} =(t∨q)∧(∼p∨t) =t∧t=t
Q5.Which of the following is a contradiction?
Solution
(p∧q)∧(∼(p∨q)) ≅(p∧q)∧(∼p∧∼q) ≅q∧(p∧∼p)∧∼q ≅q∧c∧∼≅c So, statement in option (a) is a contradiction
(p∧q)∧(∼(p∨q)) ≅(p∧q)∧(∼p∧∼q) ≅q∧(p∧∼p)∧∼q ≅q∧c∧∼≅c So, statement in option (a) is a contradiction
Q6. Let p and q be two properties. Then, the contrapositive of the implication p→q is
Q7.The statement p∨∼p is
Q8.The contrapositive of 2x+3=9⇒x≠4 is
Solution
The contrapositive of p→q is ∼q→∼p So, the contrapositive of 2x+3=2→x≠4 is x=4→2x+3≠9
The contrapositive of p→q is ∼q→∼p So, the contrapositive of 2x+3=2→x≠4 is x=4→2x+3≠9
Q9.If p⇒(∼p∨q) is false, the truth value of p and q are respectively
Solution
p⇒(∼p∨q) is false means p is true and ∼p∨q is false ⇒p is true and both ∼p and q are false ⇒p is true and q is false
p⇒(∼p∨q) is false means p is true and ∼p∨q is false ⇒p is true and both ∼p and q are false ⇒p is true and q is false
Q10. Negation of "Pairs is in France and Londan is in England" is
Solution
Let p:Pairs is in France and q: London is in England Given, p∧q Its negation is ~(p∧q)≡∼p∨∼q Hence, paris is not in France or London is not in England.
Let p:Pairs is in France and q: London is in England Given, p∧q Its negation is ~(p∧q)≡∼p∨∼q Hence, paris is not in France or London is not in England.
