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A relation R:A→AR:A→A is said to be reflexive if, for every element aa in the set AA (where AA is a non-empty set), the ordered pair (a,a)(a,a) belongs to the relation RR. In simpler terms, reflexive relations include every element paired with itself in the set.

To prove that the function f:R→Rf:R→R given by f(x)=2xf(x)=2x is one-to-one (injective), we need to show that if f(x1)=f(x2)f(x1)=f(x2), then x1=x2x1=x2 for all x1,x2x1,x2 in the domain.

Let's assume f(x1)=f(x2)f(x1)=f(x2): 2x1=2x22x1=2x2

Now, we'll solve for x1x1 and x2x2: x1=x2x1=x2

Since x1=x2x1=x2, it means that for any two inputs x1x1 and x2x2 that produce the same output under the function f(x)=2xf(x)=2x, those inputs must be the same. This proves that the function f(x)=2xf(x)=2x is one-to-one.

To prove that (f+g)∘h=f∘h+g∘h(f+g)∘h=f∘h+g∘h, let's start by understanding what (f+g)∘h(f+g)∘h means:

(f+g)∘h(x)=(f+g)(h(x))=f(h(x))+g(h(x))(f+g)∘h(x)=(f+g)(h(x))=f(h(x))+g(h(x))

Now, let's find (f∘h+g∘h)(x)(f∘h+g∘h)(x):

f∘h(x)+g∘h(x)=f(h(x))+g(h(x))f∘h(x)+g∘h(x)=f(h(x))+g(h(x))

This expression is identical to what we found for (f+g)∘h(x)(f+g)∘h(x). Hence, we can conclude that (f+g)∘h=f∘h+g∘h(f+g)∘h=f∘h+g∘h.

To find (g∘f)(x)(g∘f)(x), which is the composition of g(x)g(x) with f(x)f(x), we substitute f(x)f(x) into g(x)g(x) wherever we see xx.

Given:

f(x)=∣x∣f(x)=∣x∣ g(x)=∣5x+1∣g(x)=∣5x+1∣

We first find f(x)f(x):

f(x)=∣x∣f(x)=∣x∣

And then substitute it into g(x)g(x):

g(f(x))=∣5(∣x∣)+1∣g(f(x))=∣5(∣x∣)+1∣

Now, ∣x∣∣x∣ can be either xx if x≥0x≥0 or −x−x if x<0x<0.

So, ∣5(∣x∣)+1∣∣5(∣x∣)+1∣ will be:

If x≥0x≥0: g(f(x))=∣5x+1∣g(f(x))=∣5x+1∣

If x<0x<0: g(f(x))=∣−5x+1∣g(f(x))=∣−5x+1∣

To determine the principal value of

The identity matrix of order n is denoted by I∩

It is a square matrix with dimensions n×nn×n where all the elements on the main diagonal (from the top left to the bottom right) are 1, and all other elements are 0.