Access to comprehensive ISC Mathematics Class 11 OP Malhotra Solutions Chapter 27 Mathematical Reasoning Chapter Test encourages independent learning.
S Chand Class 11 ICSE 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 3x2 – x – 10 = 0.
Solution:
(i) The connective word be ‘AND’ and component statements are :
p : all rational numbers are real.
q : all real numbers are not complex.
(ii) The connective word is ‘OR’ and component statements are ;
p : square of an integer is positive
q : square of an integer is negative
(iii) The connective word is ‘AND’ and component statements are ;
p : x = 2 be a root of eqn. 3x2 – x – 10 = 0
q : x = 3 be a root of eqn. 3x2 – x – 10 = 0
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.
Solution:
(i) Here, the quantifier be ‘there exists’ Negation of given statement: There does not exists a number which is equal to its square.
(ii) Here, the quantifier be “For every.” Negation of given statement be given by It is false that for every real number x, x is less than x + 1.
(iii) Here, the quantifier be “There exists.” Negation of given statement be given by There exists a state in India which does not have a capital.
Question 3.
For any statement ‘p’ prove that ~(~p) ≡ p.
Solution:
| p | ~p | ~(~p) |
| T | F | T |
| F | T | F |
From truth tables we observe that, p ≡ ~ (~p)
Question 4.
Write the converse, contradiction and contrapositive of the statement
‘If x + 3 = 9, then x = 6.’
Solution:
The component statements are ;
let p : x + 3 = 9
q : x = 6
Then given statement be p ⇒ q
Its converse be q ⇒ p
i.e. if x = 6 Then x + 3 = 9
Contradiction of given statement is x = 6 ⇒ x + 3 ≠ 9
Contrapositive of given statement is ~q ⇒ ~p i.e. if x ≠ 6 then x + 3 ≠ 9
Question 5.
For any statements, p and q, prove that p ⇒ q = (~p ∨ q).
Solution:
The truth table for p ⇒ q and (~p ∨ q) is given as under:
| I | II | III | IV | V |
| p | q | p ⇒ q | ~p | ~p ∨ q |
| T | T | T | F | T |
| T | F | F | F | F |
| F | T | T | T | T |
From column III and V, we observe that
P ⇒ q ≡ (~p ∨ q)
Question 6.
Write the following implications (p ⇒ q) in the form (~ p ∨ q) and write its negation. ‘If △ABC is isosceles then the base angles A andB are equal.’
Solution:
The truth table for p ⇒ q and (~p ∨ q) is given as under :
| I | II | III | IV | V |
| p | q | p ⇒ q | ~p | ~p ∨ q |
| T | T | T | F | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
From column III and V, we observe that p ⇒ q ≡ (~ p ∨ q)
∴ ~(~p ∨ q) = ~ (~p) ∧ (~q) = p ∧ ~q
Let p : △ABC is isosceles
q : The base angles A and B are equal.
Then ~ p ∨ q : either △ABC is not an isosceles triangle or the base angles A and B are equal.
Thus the negation of given statement be (p ∧ ~ q)
i.e. △ABC be an isosceles triangle and the base angles A and B are not equal.