OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Chapter Test

 

Question 1. for each of the following compound statements, first identify the connective words and then break it into component statements.
(i) All rational numbers are real and all real numbers are not complex.
(ii) Square of an integer is positive or negative.
(iii) x = 2 and x = 3 are the roots of the equation \( 3x^2 - x - 10 = 0 \).
Answer:
(i) The connective word is 'AND'.
Component statement p: All rational numbers are real.
Component statement q: All real numbers are not complex.
(ii) The connective word is 'OR'.
Component statement p: Square of an integer is positive.
Component statement q: Square of an integer is negative.
(iii) The connective word is 'AND'.
Component statement p: \( x = 2 \) is a root of the equation \( 3x^2 - x - 10 = 0 \).
Component statement q: \( x = 3 \) is a root of the equation \( 3x^2 - x - 10 = 0 \).
Understanding connective words helps us to break down complex statements into simpler parts, making them easier to analyze logically.
In simple words: For each sentence, we find the word that joins two ideas (like "and" or "or"). Then, we write down each of those two ideas as separate, smaller sentences.

🎯 Exam Tip: Pay close attention to words like "and", "or", "if...then", and "if and only if" as these are the main connectives to identify. Each component statement should be a complete sentence on its own.

 

Question 2. Identify the quantifier in the following statements and write the negation of the statements.
(i) There exists a number which is equal to its square.
(ii) For every real numbers x, x is less than x + 1.
(iii) There exists a capital for every state in India.
Answer:
(i) Quantifier: 'there exists'.
Negation of the given statement: There does not exist a number which is equal to its square. (Alternatively, For every number, it is not equal to its square.)
(ii) Quantifier: 'For every'.
Negation of the given statement: It is false that for every real number x, x is less than x + 1. (Alternatively, There exists a real number x such that x is not less than x + 1.)
(iii) Quantifier: 'There exists'.
Negation of the given statement: There exists a state in India which does not have a capital. (Alternatively, For every state in India, it does not have a capital.)
Quantifiers are crucial in mathematics as they specify the number of elements for which a statement is true.
In simple words: First, we find the word that tells us "how many" (like "there exists" or "for every"). Then, we write the opposite of the original statement by changing that quantifier and the main idea.

🎯 Exam Tip: To negate a statement with "there exists", use "for every" and negate the predicate. To negate a statement with "for every", use "there exists" and negate the predicate. Remember, "not equal" is the negation of "equal".

 

Question 3. For any statement 'p' prove that \( \sim (\sim p) = p \).
Answer:
We can prove this using a truth table:

From the truth table, we can see that the column for \( p \) and the column for \( \sim (\sim p) \) have the exact same truth values. This shows that they are logically equivalent.
In simple words: If you say "not not true," it means the same as "true." So, the opposite of the opposite of something is just the original thing.

🎯 Exam Tip: This is the law of double negation. In a truth table, two statements are logically equivalent if their truth value columns are identical for all possible combinations of truth values of their component statements.

 

Question 4. Write the converse, contradiction and contrapositive of the statement 'If x + 3 = 9, then x = 6.'
Answer:
Let p be the statement: \( x + 3 = 9 \).
Let q be the statement: \( x = 6 \).
The given statement is \( p \implies q \).

Its converse is \( q \implies p \).
i.e., If \( x = 6 \), then \( x + 3 = 9 \).

Its contradiction (as presented in the source) is \( q \implies \sim p \).
i.e., If \( x = 6 \), then \( x + 3 \neq 9 \).

Its contrapositive is \( \sim q \implies \sim p \).
i.e., If \( x \neq 6 \), then \( x + 3 \neq 9 \).
The contrapositive of a statement is logically equivalent to the original statement, meaning they always have the same truth value.
In simple words: We have "If P, then Q." The converse is "If Q, then P." The contrapositive is "If not Q, then not P." The term "contradiction" here means a statement that is true when the converse is false.

🎯 Exam Tip: Remember these transformations: Converse swaps p and q (\( q \implies p \)). Contrapositive swaps and negates both (\( \sim q \implies \sim p \)). The negation of \( p \implies q \) is \( p \wedge \sim q \).

 

Question 5. For any statements, p and q, prove that \( p \implies q = (\sim p \vee q) \).
Answer:
We can prove this using a truth table for \( p \implies q \) and \( \sim p \vee q \):

From column III (\( p \implies q \)) and column V (\( \sim p \vee q \)), we can observe that they have identical truth values for all possible combinations of \( p \) and \( q \). This means that \( p \implies q \) is logically equivalent to \( \sim p \vee q \). This equivalence is often used to simplify logical expressions.
In simple words: The statement "If P, then Q" means the same thing as "Not P, or Q." This truth table shows that both statements are always true or false at the exact same time.

🎯 Exam Tip: This logical equivalence \( (p \implies q) \equiv (\sim p \vee q) \) is fundamental in mathematical logic and is frequently used to simplify or prove other logical statements. Memorize this identity.

 

Question 6. Write the following implications \( (p \implies q) \) in the form \( (\sim p \vee q) \) and write its negation. 'If △ABC is isosceles then the base angles A and B are equal.'
Answer:
Let p be the statement: △ABC is isosceles.
Let q be the statement: The base angles A and B are equal.

First, we write \( (p \implies q) \) in the form \( (\sim p \vee q) \):
\( \sim p \vee q \): Either △ABC is not an isosceles triangle or the base angles A and B are equal.

Now, we write its negation. The negation of \( (p \implies q) \) is \( \sim (p \implies q) \equiv p \wedge \sim q \).
So, the negation is: △ABC is an isosceles triangle AND the base angles A and B are not equal.

Here is the truth table for \( p \implies q \) and \( \sim p \vee q \) (used to demonstrate equivalence):

From column III and V, we observe that \( p \implies q = (\sim p \vee q) \).
Therefore, the negation of \( (\sim p \vee q) \) is \( \sim (\sim p \vee q) \equiv \sim (\sim p) \wedge \sim q \equiv p \wedge \sim q \).
In simple words: We take the "if-then" statement and change it to an "either-or" statement using "not" for the first part. To find the opposite (negation) of the original statement, we say the "if" part is true AND the "then" part is false.

🎯 Exam Tip: Remember De Morgan's Laws: \( \sim(A \vee B) \equiv (\sim A \wedge \sim B) \) and \( \sim(A \wedge B) \equiv (\sim A \vee \sim B) \). These are vital when negating compound statements like implications.