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.