OP Malhotra Class 11 Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(g)

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S Chand Class 11 ICSE Maths Solutions Chapter 27 Mathematical Reasoning Ex 27(g)

Question 1.
Show that the statement p : ‘If x is a real number such that x³ + 4x = 0, then x = 0’ is true by
(i) Direct method,
(ii) Method of contradiction,
(iii) Method of contrapositive.
Solution:
Let q : x ∈ R s.t x³ + 4x = 0
r : x = 0
Then p :q → r
(i) Then q be true ⇒ x³ + 4x = 0
⇒ x (x² + 4x) = 0 ⇒ x = 0 (∵ x ∈ R)
⇒ r be true
Thus, q ⇒ r is true ⇒ p is true statement.

(ii) Let x ≠ 0 and let x = k, where k e R and k be a root of x² + 4x = 0
⇒ k³ + 4k = 0 ⇒ k (k² + 4) = 0
⇒ k = 0 (∵ k ∈ R)
since k² + 4 ≠ 0
∴ which contradict the our assumption that k ≠ 0
∴ our supposition is wrong.
∴ x = 0
Thus q ⇒ r is true.

(iii) Let r be not true ∴ x ≠ 0
Let x = p ≠ 0 ∈ R
∴ x³ + 4x = p³ + 4p – p (p² + 4) ≠ 0
[∵ p ≠ 0, p² + 4
⇒ q be not true
∴ ~ r ⇒ ~ q is true
Hence q ⇒ r is true.

Question 2.
Show that the statement ‘For any real numbers a and b, a² = b² implies that a = b’ is not true by giving counter examples.
Solution:
Let us take a = 2 and b = – 2
Here a² = b² = 4 but a ≠ b

Question 3.
Show that the following statement is true by the method of contrapositive.
p : If x is an integer and x² is even, then x is also even.
Solution:
Let q : x is an integer and x² is even r : x is also even
Then p : q ⇒ r be the given statement.
We want to check, whether the statement q ⇒ r is true by contrapositive method.
i.e. we want to check the validity of ~ r ⇒ ~ q
Let ~ r be true ⇒ r be not true
⇒ x is not even
⇒ x = 2 k + 1, where k ∈ I
⇒ x² = (2k + 1)² = 4k² + 4k + 1
=4k(k + 1) + 1 = 8k’ + 1,
where k’ ∈ I [∵ product of n consecutive integers is divisible by n!]
⇒ x² is not even integer
Thus ~ r ⇒ ~ q be true
Hence q ⇒ r be a true statement.

Question 4.
Given below are two statements :
p : 30 is a multiple of 5.
q : 30 is a multiple of 9.
Write the compound statement, connecting these two statements with ‘and’ and ‘or’. In both cases, check the validity of the compound statement.
Solution:
Given p : 30 is a multiple of 5 (True)
q : 30 is a multiple of 9 (False)
Then compound statement p ∧ q : 30 is a multiple of 5 and 9
Its truth value be false since one of the component statement is false and hence p ∧ q is not valid.
The compound statement p ∨ q : 30 is a multiple of 5 or 9
Its truth value be true and it is a valid statement.

Question 5.
Verify by the method of contradiction that √7 is irrational.
Solution:
Let p be the given statement
i.e. p : √7 is irrational
Let if possible p be not true i.e. √7 is a rational number ⇒ √7 = \(\frac { a }{ b }\)
where a, b ∈ l and (a, b) = 1 i.e. have no common factor.
⇒ 7 = \(\frac{a^2}{b^2}\)
⇒ a² = 7b²
⇒ 7 divides a² ⇒ 7 divides a
⇒ a = 7k, where k ∈ I
⇒ (7k)² = 7b² ⇒ 7k² = b²
⇒ 7 divides b² ⇒ 7 divides b
⇒ 7 is a common factor of both a and b. which contradicts the fact that a and b have no common factor and hence our supposition is wrong. Thus p is true. Hence, √7 be an irrational number.