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.