Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 14 Mathematical Reasoning.

## Kerala Plus One Maths Chapter Wise Previous Questions Chapter 14 Mathematical Reasoning

### Plus One Maths Mathematical Reasoning 3 Marks Important Questions

Question 1.

i) Write the converse of the statement. (IMP-2010)

p: If a divides b then b is a multiple of a.

ii) Consider the compound statement,

p: 2 + 2 is equal to 4 or 6

1) Write the component statements.

2) Is the compound statement true? Why?

Answer:

i) Converse statement is “If a is a multiple of b then a divides b.”

ii) 1) q: 2+2 is equal to 4

r. 2+2 is equal to 6.

2) q is true and r is false, so p is true.

Question 2.

Verify by method of contradiction p. √2 is irrational. (IMP-2012)

Answer:

Assume that √2 is rational. Then can be written in the form √2 = \(\frac { p }{ q }\) , where p and q are integers without common factors.

Squaring; 2 = \(\frac { p² }{ q² }\)

=> 2q² = p²

=> 2 divides p² => 2 divides p

Therefore, p = 2k for some integer k.

=> p² = 4k²

=> 2q² = 4k²

=> q² = 2k²

=> 2 divides q² => 2 divides q

Hence p and g have common factor 2, which

contradicts our assumption. Therefore, √2 is irrational.

Question 3.

i) Write the negation of the following . statement, ‘Every natural number is an integer’. (MARCH-2013)

ii) Write the contrapositive and converse of the following statement, ‘If x is a prime number, then x is odd ’.

Answer:

Negation of the statement is ‘Every natural ’ number is not an integer’,

ii) The contrapositive statement, ‘If x is not odd, then x is not a prime number.’

The converse of the statement, ‘If x is odd, then x is a prime number’.

Question 4.

i) Write the component statement of the following statement: “All rational

numbers are real and all real numbers are complex. (IMP-2014)

ii) Write the contrapositive and converse

of the following statement: ‘If a number is divisible by 9, then it is divisible by 3.’

(Imp (Commerce) – 2014)

Answer:

i) p: All rational numbers are real,

q: All real numbers are complex,

ii) Contrapositive:

If a number is not divisible by 3, it is not divisible by 9.

Converse:

If a number is divisible by 3 then it is divisible by 9.

Question 5.

i) Write the negation of the statement: the sum of 3 and 4 is 9. (IMP-2014)

ii) Write the component statements of ‘Chandigarh is the capital of Haryana and Uttar Pradesh.’

iii) Write the converse of the statement: ‘If a number n is even, then n² is even.’

Answer:

i) Negation: ‘The sum of 3 and 4 is not equal to 9.’

ii) p: Chandigarh is the capital of Haryana.

q: Chandigarh is the capital of Uttar Pradesh.

iii) Converse: If a number n² is even then n is even.

### Plus One Maths Mathematical Reasoning 4 Marks Important Questions

Question 1.

i) Write the negation of the statement.“Both the diagonals of a rectangle have the same length.” (MARCH-2013)

ii) Prove the statement, “Product of two odd integers is odd,” by proving its contrapositive.

Answer:

i) “Both the diagonals of a rectangle do not have the same length.”

ii) Let us name the statements as below p: ab is odd. q: a, b is odd.

We have to check p => q is true or not, that is by checking its contrapositive statement

~ q =>~ p

~ q: ab is even.

Let a and b be two even numbers. Then, a = 2n and b = 2m, where m and n are any integer.

a x b = 2n(2m) = 4 nm

Then product of a and b is even. That is ~ p is true. Hence by the contrapositive principle we say that “Product of two odd integers is odd,”

Question 2.

Consider the compound statement “ √5 is a rational number or irrational number”. (IMP-2011)

i) Write the component statements of

above and check whether these component statements are true or false.

ii) Check whether the compound statement is true or false.

(Imp (Commerce) – 2011)

Answer:

i) The component statements are

p: √5 is a rational number

q: √5 irrational number.

Here p is false and q is true,

ii) In this compound statement “or” is exclusive, p is false and q is true and therefore compound statement is true.

Question 3.

i) Write the converse of the statement: “If a number n is even, then n² is even” (MARCH-2011)

ii) Verify by method of contradiction: “ √2 is irrational”.

Answer:

i) “If n² is even, then n is even”

ii) Assume that √2 is rational. Then √2 can

be written in the form √2=\(\frac { p }{ q }\), where p and q are integers without common factors.

Squaring; 2 = \(\frac { p² }{ q² }\)

=> 2q²= p²

=> 2 divides p² => 2 divides p

Therefore, p = 2k tot some integer k.

=> p² = 4k²

=> 2q² = 4k²

=> q² = 2k²

=> 2 divides

q² => 2 divides q

Hence p and g have common factor 2, which contradicts our assumption.

Therefore, √2 is irrational.

Question 4.

Which of the following sentences are statements? Give reason for your answer. (IMP-2012)

a) The cube of a natural number is an odd number.

b) The product of (- 4) and (- 5) is 20.

Write the negation of the following statements and check whether the resulting statements are true.

a) √2 is rational.

b) Every natural number is greater than zero.

Answer:

a) This sentence is a statement. Since for a particular natural number it is true

and for other it is false. (1)³ = 1 and 2³ = 8

b) This sentence is a statement. Since the product is always 20 and true.

a) √2 is not rational. The negation statement is true.

b) Every natural number is not greater than zero. The negation statement is false.

Question 5.

Consider the statement, “If x is an integer and x² is even, then x is also even.” (MARCH-2012)

i) Write the converse of the statement.

ii) Prove the statement by the contra-positive method.

Answer:

i) Converse of the statement is “If x is an even number, then x is an integer and x² is even.”

ii) The contrapositive of a statement p => q is the statement ~ q => ~ p .

The contrapositive statement is “If x is an odd integer, then x² is odd.”

Let x is an odd number.

Then x= 2n+1

x² =(2n+1)² =4n² +4n+1

= 4(n² +n)+1

Which is odd.

ie; if q is not true then p is not true.

Question 6.

i) Write the negation of the following statement: “All triangles are not equilateral triangles”. (MARCH-2013)

ii) Verify by the method of contradiction. p:√7 is irrational.

Answer:

i) All triangles are equilateral triangles

ii) Assume that is rational. Then √7 can be written in the form√7 =\(\frac { p }{ q }\), where p and q are integers without common factors.

Squaring; 7 = \(\frac { p² }{ q² }\)

=> 7q² = p²

=> 7 divides p² => 7 divides p

Therefore, p = 7k tot some integer k.

=>p² = 49k²

7q² = 49k²

=>q² = 7k²

=>7 divides q² =>7 dividesq

Hence p and q have common factor 7, which contradicts our assumption.

Therefore, √7 is irrational.

Question 7.

i) Write the contrapositive of the statement. “If x is a prime number, then x is odd.” (IMP-2013)

ii) Verify by the method of contradiction p : √5 is irrational.

Answer:

i) Contrapositive statement is “If x is not odd, then x is not prime number.”

ii) Assume that √5 is rational. Then √5 can be written in the form √5 = \(\frac { p }{ q }\), where p and q are integers without common factors.

=> 5q² = p²

=> 5 divides p²=> 5 divides p

Therefore, p = 5k for some integer k.

=>p² = 25k²

=> 5q² = 25k²

=> q² = 5k²

=> 5 divide q² => 5 divides q

Hence p and g have common factor 5, which contradicts our assumption.

Therefore, √5 is irrational.

Question 8.

i) Write the negation of the following statement : “ √5 is not a complex number.” (MARCH-2014)

ii) Verify using the method of contradiction:

“p: √2 is irrational number.”

Answer:

i) Negation statement:” √5 is a complex number.”

ii) Assume that √2 is rational. Then √2 can

be written in the form √2=\(\frac { p }{ q }\), where p and q are integers without common factors.

Squaring; 2 = \(\frac { p² }{ q² }\)

=> 2q²= p²

=> 2 divides p² => 2 divides p

Therefore, p = 2k tot some integer k.

=> p² = 4k²

=> 2q² = 4k²

=> q² = 2k²

=> 2 divides

q² => 2 divides q

Hence p and g have common factor 2, which contradicts our assumption.

Therefore, √2 is irrational.

Question 9.

i) Write the negation of the statement: “√7 is rational.” (MARCH-2015)

ii) Prove that“√7 is rational.” by the method of contradiction.

(March – 2015)

Answer:

i) Negation is : “ √7 is not rational.”

ii) Assume that is rational. Then √7 can be written in the form√7 =\(\frac { p }{ q }\), where p and q are integers without common factors.

Squaring; 7 = \(\frac { p² }{ q² }\)

=> 7q² = p²

=> 7 divides p² => 7 divides p

Therefore, p = 7k tot some integer k.

=>p² = 49k²

7q² = 49k²

=>q² = 7k²

=>7 divides q² =>7 dividesq

Hence p and q have common factor 7, which contradicts our assumption.

Therefore, √7 is irrational.

Question 10.

i) Which of the following is the contrapositive of the statement (IMP-2015)

a) q => p

b) ~ p =>~ q

c) ~ q =>~ p

d) p =>~ q

ii) Prove by contrapositive method. “If x is an integer and x² is even then x is also even.”

(Imp-2015)

Answer:

i) c) ~ q =>~ p

ii) The contrapositive of a statement p => q is the statement ~ q => ~ p .

The contrapositive statement is “If x is an odd integer, then x² is odd.”

Let x is an odd number.

Then x= 2n+1

x² =(2n+1)² =4n² +4n+1

= 4(n² +n)+1

Which is odd.

ie; if q is not true then p is not true.

Question 11.

i) Write the negation of the statement: “Every natural number is greater than zero.” (MARCH-2016)

ii) Verify by the method of contradiction: “P: √13 is irrational.”

Answer:

i) Negation of the statement: “It is false that every natural number is greater than zero.”

ii) Assume that √13 is rational. Then √13 can be written in the form √13 = \(\frac { p }{ q }\) ,

where p and q are integers without common factors.

Squaring; 13 = \(\frac { p² }{ q² }\)

=>13q² = p²

=>13 divides p² => 13 divides p

Therefore, p = 13k for some integer k.

=> p² = 169k²

=> 13q² = 169k²

q²= 13k²

=>13 divides q²

=> 13 divides q

Hence p and q have common factor 13, which contradicts our assumption.

Therefore, √13 is irrational.

Question 12.

i) Write the negation of the statement

“ √2 is not a complex number.”

ii) Prove by the method of contradiction,

P: √11 is irrational.”

Answer:

i) “√2 is a complex number.”

ii) Assume that √11 is rational. Then √11 can be written in the form√11 = \(\frac { p }{ q }\),

where p and q are integers without common factors.

Squaring; 11 = \(\frac { p² }{ q² }\)

=> 11q² =p²

=> 11 divides p² => 11 divides p

Therefore, p =11k for some integer k.

=>p² =121 k²

=>11q² = 121k²

=> q² = 11k²

=> 11 divides q² => 11 divides q

Hence p and q have common factor 11, which contradicts our assumption. Therefore,

√11 is irrational.

Question 13.

i) Write the contrapositive of the statement “If a number is divisible by 9, then it is divisible by 3.”

ii) Prove by the method of contradiction.

“P: √5 is irrational.”

Answer:

i) “If a number is not divisible by 9, then it is not divisible by 3.”

ii) Assume that √5 is rational. Then √5 can be written in the form √5 = \(\frac { p }{ q }\), where p and q are integers without common factors.

=> 5q² = p²

=> 5 divides p²

=> 5 divides p

Therefore, p = 5k for some integer k.

=>p² = 25k²

=> 5q² = 25k²

=> q² = 5k²

=> 5 divide q² => 5 divides q

Hence p and g have common factor 5, which contradicts our assumption.

Therefore, √5 is irrational.