Question

A relation f: N —Y N, f(x) = x— 1 is a function. (True / False)

(2) For any non-empty sets A and B (A u B)' = At u B'. (True / False)

(3) If A c B and B c A then A = B. (True / False)

(4) For any matrix A, AA-I = I. (True / False)
(5)he range of a function f : N N, f(x) = x is (true or false)​

Answer 1

Answer:

1) false

Answer

If a function f(x) is differentiable at a point a, then it is continuous at the point a. But the converse is not true.

f(x)=∣x∣ is contineous but not differentiable at x=0.

Therefore, the given statement is false.

2)true

Answer

Let M = (A ∩ B)' and N = A' U B'

Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'

⇒ x ∉ (A ∩ B)

⇒ x ∉ A or x ∉ B

⇒ x ∈ A' or x ∈ B'

⇒ x ∈ A' U B'

⇒ x ∈ N

Therefore, M ⊂ N …………….. (i)

Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'

⇒ y ∈ A' or y ∈ B'

⇒ y ∉ A or y ∉ B

⇒ y ∉ (A ∩ B)

⇒ y ∈ (A ∩ B)'

⇒ y ∈ M

Therefore, N ⊂ M …………….. (ii)

Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)' = A' U B'

3)false

Answer

a<b

Then,

ac<bc if c is positive

ac>bc if c is negative

So, Given statement is false.

4)true

Answer

It is true, 

Because,

[AA′]′=(A′)′A′=[AA′]

So, AA′ is a symmetric matrix for any matrix A

5)false

Answer

The function is injective (one-to-one) if every element of the codomain is mapped to by at most one element of the domain

The function is surjective (onto) if every element of the codomain is mapped to by at least one element of the domain

f(x)=2x 

Domain of f is N 

Codomain of f is N

Every element of N(codomain) is mapped to only one element in N(domain)

i.e. f(x)=f(y)

⟹2x=2y⟹x=y

Hence f is one one function

Every even element of N(codomain) has corresponding value in N(domain).

But, for any odd number in N(codomain) has no corresponding value in Ndomain.

Hence, f is not onto function.

Step-by-step explanation:

hope this will help

Answer 2

Answer:

1. false

2. true

3.false

4.true

5.false

hope it goes well