Class 11 Maths Ncert Solutions Ex 14.5

Class 11 Maths Ncert Solutions Chapter 14 Ex 14.5

Q.1: Prove that p: “If a is real such that a3+ 4a = 0, then a is  0″ is true 

(i). by direct method

(ii). by method of contradiction

(iii). by method of contra positive

 

Sol:

p: “if a is real such that a3+ 4a = 0, then a is 0”

Let q: a is real such that a3+ 4a = 0 r, a is 0.

(i).  To show that “p” is true, we take that “q” must be true and then prove that “r” is true.

Therefore, assume statement “q” be true.

a3+ 4a= 0 a (a2 + 4) = 0

=> a = 0 or a2 + 4 = 0

However, since a is real, so it is O.

Hence, “r” is true.

Thus, the statement is true.

(ii).  To show “p” is true using contradiction, we take that “p” isn’t true.

Let x be a real number such that a3+ 4a = 0 and let x is not 0.

Therefore, a3+ 4a = 0 x (a2 + 4) = 0 a = 0 or a2 + 4 = 0 a = 0 or a2= — 4

However, ‘a’ is real. Thus, a = 0, which is contradiction as we assume that a is not 0.

Hence, the statement “p” is true.

(iii).  To prove “p” to be true by using contrapositive method, let r is false and show that q is false.

Here, “r” is false states that its requirement of the negation for statement r.

This obtains the following statement.

~ r. x is not 0.

It is seen (a2 + 4) cannot be negative so it will be positive.

a ≠ 0 states that product of a positive number with “a” is not zero.

Let us assume the product of a with (a2 + 4).

a (a2+ 4) ≠ 0

a3 + 4a ≠ 0

This proves that “q” isn’t true.

Hence, it is proved that ~ r => ~ q

Therefore, the statement “p” is true.

 

 

Q.2: Prove the statement “For real numbers b and a, b2 = a2 implies that b = a” isn’t true. Give a counter example.

 

Sol:

Using “if- then” the given statement may be written as follows.

If b and a are two real numbers and b2 = a2, then b = a

Let p: b and a are two real numbers and b2 = a2

q: b = a

To prove: given statement is false. For this we need to prove that p. then ~q. To prove this, we need two numbers b and a with b2 = a2 such that b ≠ a (the numbers must be real numbers)

Let b = (-1) and a = 1 b2 = (-1)2 = 1 and a2 = (1)2  = 1

Therefore b2 = a2

However, b ≠ a

Thus, it is proved that given statement is false.

 

 

Q3: By using contrapositive method prove the following statement is true

p: if a is an integer and a2 is even, then a is even.

 

Sol:

p: if a is an integer and a2 is even, then a is even.

Assume q: a is an integer with “a2” even. r. a is even

To show: By using contrapositive method p is true, q is false.

r is false [we assume]

let a is odd number

To show that ‘q’ is false, we have to prove that ‘a’ isn’t an integer or a2 isn’t even

‘a’ isn’t even implies that a2 is not even

Therefore, ‘q’ is false. Thus, given statement i.e p is true.