# Show that t

Given that R={(a ,b ):a ≤ b}

For Reflexive:

Since a=a a ,

therefore a ≤ a always.

Hence (a, a) always belongs to R
a . Therefore, R is reflexive.

For Symmetric:

If a ≤b

then b≥ a
b ≤a .

Example:

(2,4) ∈ ℝ as 2 ≤ 4.

But (4,2) ∉ ℝ as 4  is greater than 2.

Hence if (a, b) belongs to R, then (b, a) does not always belong to R.

Hence R is not symmetric.

For Transitive:

If a ≤b  -----(1) and

b ≤c  -----(2)

Add (1) and (2) to get,

a + b ≤ b +c

Hence a ≤c.

Hence if (a, b)∈R and (b, c)∈R, then (a, c)∈R
a, b, c .

Hence, R is transitive.

OR

f(x)=x2+x+1, f:

A function is one-one if f(a)=f(b)

a=b

f(a)=f(b)

a2+a+1=b2+b+1

a2- b2+a -b=0 or (a-b)(a+b+1)=0

Hence, a=b or a + b=-1

Since a, b ∈ℕ, therefore a + b=-1 is not possible.

Hence a=b.

Since f(a)=f(b)

a=b

Therefore, f(x) is one-one.

Let y= x2+x+1

differentiating with respect to x, we get,

y’=2x+1>0 x ∈ℕ, hence f is an increasing function.

The range of y={3,7,13,21…} which is not equal to .

Since the range is not equal to codomain, therefore, f is not onto.

Let S be the range of f.

Then f(x)=x2+x+1,

f:S

y= x2+x+1

x2+x+1-y=0

Using the quadratic formula, we get,

There are two possibilities  for f-1(x).

As f(1) = 3

So, f-1(3) = 1

Hence, f-1:S→ℕ

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