Mathematical Reasoning Class 11 OP Malhotra Exe-27G ISC Maths Ch-27 Solutions. In this article you would learn about Validating and Invalidity Statements. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Mathematical Reasoning Class 11 OP Malhotra Exe-27G ISC Maths Solutions Ch-27
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-27 | Mathematical Reasoning |
| Writer | O.P. Malhotra |
| Exe-27(G) | Validating and Invalidity Statements. |
Validating and Invalidity Statements
Mathematical Reasoning Class 11 OP Malhotra Exe-27G ISC Maths Ch-27 Solutions.
Que-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.
Sol: 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.
Que-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.
Sol: Let us take a = 2 and b = – 2
Here a2 = b2 = 4 but a ≠ b
Que-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.
Sol: 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.
Que-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.
Sol: 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.
Que-5: Verify by the method of contradiction that √7 is irrational.
Sol: 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 = a/b
where a, b ∈ l and (a, b) = 1 i.e. have no common factor.
⇒ 7 = a²/b²
⇒ 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.
–: End of Mathematical Reasoning Class 11 OP Malhotra Exe-27G ISC Maths Ch-27 Solutions. :–
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