Important Questions for CBSE Class 11 Maths Chapter 14 - Mathematical Reasoning

4 Marks Questions

1. Give three examples of sentences that are not statements. Give reasons for the statements.

Ans: 

(i) “Rani is a beautiful girl” is not a statement. This sentence depends on perspective. Rani may look beautiful to some people whereas she may not look beautiful to other people. Thus, we cannot logically say that the sentence is true.

(ii) “Shut the door” is not a statement. It is just a sentence giving direction to someone. There is no question that it is true or false.

(iii) “Yesterday was Friday” sentence is not a statement. This sentence is true only if it is spoken on Saturday and false if it is spoken on other days of the week. Here, truth or false depends upon the time at which it is spoken and not o mathematical reasoning.

2. Write the negation of the following statements.

(i) Chennai is the capital of Tamil Nadu.

Ans: Chennai is not the capital of Tamil Nadu.

(ii) Every natural number is an integer.

Ans: Every natural number is not an integer.

3. Find the component statements of the following compound statements and check whether they are true or false.

(i) The number 3 is prime or it is odd.

Ans:

The given compound statements can be written as component statements as:

p: “number 3 is prime”

q: “number 3 is odd”.

The given statement is connected by connective “or”

Since, both the statements are true, hence, the given statement is true.

4. Check whether the following pair of statements are negative of each other. Give reasons for your answer.

(i) x + y = y + x is true for every real numbers x and y.

(ii) There exists real number x and y for which x + y = y + x.

Ans: The statements are:

p: “x + y = y + x is true for every real number x and y”

q: “There exists real number x and y for which x + y = y + x”

The negation of p is 

$ \sim {\text{p}}\,$: “There are real numbers x and y for which ${\text{x}}\,{\text{ + }}\,{\text{y}}\, \ne \,{\text{y}}\,{\text{ + }}\,{\text{x}}$”

Therefore, the given statements are not negations of each other and they can be true at the same time.

Note: p is always true whatever x and y value may be whereas $ \sim {\text{p}}$ is always false.

5. Write the contra-positive and converse of the following statements.

(i) If x is a prime number, then x is odd.

Ans: We know that, if a statement ${\text{p}}\, \Rightarrow \,{\text{q}}$, then its contra-positive will be $ \sim {\text{q }} \Rightarrow {\text{ }} \sim {\text{p}}$ and its converse will be ${\text{q }} \Rightarrow {\text{ p}}$.

(i) Contra-positive: “If x is not odd, then it is not a prime number.”

Converse: “If x is odd, then it is a prime number.”

(ii) If the two lines are parallel, then they do not intersect in the same plane.

Ans:  Contra-positive: “If two lines intersect in the same plane, then they are not parallel.”

Converse: “If two lines do not intersect in the same plane, then they are parallel.”

6. Show that the statement

p: “If x is a real number such that ${{\text{x}}^{{\text{3}}\,}}{\text{ + }}\,{\text{4x}}\,{\text{ = }}\,{\text{0}}$, then x is 0” is true by

(i) Direct Method

Ans: Given: p: “If x is a real number such that ${{\text{x}}^{{\text{3}}\,}}{\text{ + }}\,{\text{4x}}\,{\text{ = }}\,{\text{0}}$, then x is 0”

(i) Direct method: Consider,

x² + 4x + = 0, x 𝜖 R

⇒ x ( x² + 4x ) = 0,  x 𝜖 R ⇒ x=0

(∴ if x 𝜖 R then x² + 4x ≥ 4  )

Note: At least one of the numbers is surely zero, if the product of two numbers is zero.

Therefore, p is a true statement.

 (ii) Method of Contradiction

Ans:  Consider, x is a nonzero real number

⇒ x²  > 0 (∴ square of non zero real number is always positive)

⇒ x² + 4x > 4 ⇒ x² + 4x ≠ 0

⇒ x ( x² + 4x )  ≠ 0  (∴  x ≠ 0 and x² + 4x   ≠ 0 )

⇒ x³ + 4x ≠ 0, which is a contradiction

hence,x=0

(iii) Method of Contra-Positive:

Ans: Consider, q: “${\text{x }} \in {\text{ R and }}{{\text{x}}^{\text{3}}}{\text{  +  4x  =  0}}$”

And r: “x = 0”

Thus, the given statement p is ${\text{q }} \Rightarrow {\text{ r}}$

Its contra-positive is $ \sim {\text{r }} \Rightarrow {\text{ }} \sim {\text{q}}$

i.e. “if x is a nonzero real number then ${{\text{x}}^{\text{3}}}{\text{  +  4x  =  0}}$ is also non zero.”

Now,

x ≠ 0 , x 𝜖 R ⇒ x² > 0 ⇒  x² + 4 > 4  ⇒ x² + 4 ≠ 0

 ⇒  x( x² +  4 ) ≠ 0  ⇒ x³ + 4x ≠ 0 i,e ∼ r  ⇒ ∼ q

Therefore, the statement $ \sim {\text{r }} \Rightarrow {\text{ }} \sim {\text{q}}$ is always true.

Hence, ${\text{q }} \Rightarrow {\text{ r}}$ is also true.

Note: ‘Method of contradiction’ is another form of ‘contra-positive method’ while proving an implication.

7. Given the below two statements

p: 25 is a multiple of 5.

q: 25 is a multiple of 8.

Write the compound statement connecting these two statements with “and” and “or”. In both the cases check the validity of the compound statement.

Ans: Case 1: By using “and”, we get the compound statement “p and q”

i.e., “25 is multiple of 5 and 8.”

It is not a true statement as q is always false. ${\text{(}}\because {\text{ 25 is not a multiple of 8)}}$

Case 2: By using “or”, we get the compound statement “p or q”

i.e., “25 is multiple of 5 or 8.”

It is a true statement as p is always true. ${\text{(}}\because {\text{ 25 is a multiple of 5)}}$

8. Write the following statement in five different ways, conveying the same meaning.

p: If a triangle is equiangular, then it is an obtuse angled triangle.

Ans: Given: “If a triangle is equiangular, then it is an obtuse angled triangle”.

The given statement can be written as follows:

1. “A triangle is equiangular only if it is an obtuse angled triangle”.

2. “If a triangle is not obtuse angled triangle then it is not an equiangular triangle”.

3. “Equiangularity is a sufficient condition for the triangle to be obtuse angled”.

4. “A triangle being obtuse angled, is a necessary condition for it to be equiangular”.

5. “A triangle is obtuse angled if it is equiangular”.

Important Related Links for CBSE Class 11 

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