Relations and Functions

It is given that*: $R\times R\to$ and $o:R\times R\to R$ is defined as

$a*b=\left|a-b\right|and,aob=a,\&mnForE;a,b\in R.$

For $a,b\in R$ , we have:

$\begin{array}{l}a*b=\left|a-b\right|\\ b*a=\left|b-a\right|=\left|-\left(a-b\right)\right|=\left|a-b\right|\\ \therefore a*b=b*a\end{array}$

$\therefore$ The operation* is commutative.

It can be observed that,

$\begin{array}{l}\left(1.2\right).3=\left(\left|1-2\right|\right).3=1.3=\left|1-3\right|=2\\ 1*\left(2*3\right)=1*\left(\left|2-3\right|\right)=1*1=\left|1-1\right|=0\\ \therefore \left(1*2\right)*3=1*\left(2*3\right)\left(where,1,2,3\in R\right)\end{array}$

$\therefore$ The operation* is not associative.

Now, consider the operation o:

It can be observed that $1o2=1,and,2o1=2.$

$\therefore 1o2\ne 2o1\left(where,1,2\in R\right)$

$\therefore$ The operation o is not commutative.

Let, $a,b,c\in R$ .

S = {a, b, c}, T = {1, 2, 3}

(i) F: S → T is defined as:

F = { (a, 3), (b, 2), (c, 1)}

⇒ F (a) = 3, F (b) = 2, F (c) = 1 

Therefore, F⁻¹ : T → S is given by

F⁻¹  = { (3, a), (2, b), (1, c)}.

(ii) F: S → T is defined as:

F = { (a, 2), (b, 1), (c, 1)}

Since F (b) = F (c) = 1, F is not

Onto functions from the set {1, 2, 3, …, n} to itself is simply a permutation on n symbols 1, 2, …, n.

Thus, the total number of onto maps from {1, 2, …, n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n!

28. Given, f (x)=

The given fxⁿ is valid for all x such that x – 1 ≥ 0 ⇒x≥ 1

∴ Domain of f (x)= [1,∞)

As x ≥ 1

⇒ x – 1 ≥ 1 – 1

⇒ x – 1 ≥ 0

⇒ ≥ 0

⇒ f (x) ≥ 0

So, range of f (x)= [0,∞ )

Let S be a non-empty set and P (S) be its power set. Let any two subsets A and B of S.

It is given that: $P\left(X\right)xP\left(X\right)\to P\left(X\right)$ is defined as $A.B=A\cap B\forall A,B\in P\left(X\right)$

We know that $A\cap X=A=X\cap A\forall A\in P\left(X\right)$

$\Rightarrow A.X=A=X.A\forall A\in P\left(X\right)$

Thus, X is the identity element for the given binary operation*.

Now, an element is  $A\in P\left(X\right)$ invertible if there exists $B\in P\left(X\right)$ such that

$A*B=X=B*A$&

Since every set is a subset of itself, ARA for all A ∈ P (X).

∴R is reflexive.

Let ARB ⇒ A ⊂ B.

This cannot be implied to B ⊂ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

∴ R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒

Define $f:{\rm N}\to {\rm N}$ by

$f\left(x\right)=x+1$

And,  $g:{\rm N}\to {\rm N}$ by,

$g\left(x\right)\left\{\begin{array}{cc}\hfill x-1\hfill & \hfill if,x>1\hfill \\ \hfill 1\hfill & \hfill f,x=1\hfill \end{array}\right\}$

We first show that g is not onto.

For this, consider element 1 in co-domain N. it is clear that this element is not an image of any of the elements in domain.

$\therefore f$ is not onto.

Now,  $gof:{\rm N}\to {\rm N}$ is defined by,

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(x+1\right)=\left(x+1\right)-1\left[x,in{\rm N}=>\left(x+1\right)>1\right]\\ =x\end{array}$

Then, it is clear that for $y\in {\rm N}$ , there exists

27. Given, f (x)= $\frac{x^2+2x+1}{x^2-8x+12}$

The given function is valid if denominator is not zero.

So, if x² – 8x+12=0.

⇒ x² – 2x – 6x+12=0

⇒ x (x – 2) –6 (x – 2)=0

⇒ (x – 2) (x – 6)=0

⇒ x=2 and x=6.

So,  f (x) will be valid for all real number x except x=2,6.

∴ Domain of f (x)=R – {2,6}

Define $f:N\to Z$ as $f\left(x\right)$ and $g:Z\to Z$ as $g\left(x\right)=\left|x\right|$

We first show that g is not injective.

It can be observed that:

$\begin{array}{l}g\left(-1\right)=\left|-1\right|=1\\ g\left(1\right)=\left|1\right|=1\\ \therefore g\left(-1\right)=g\left(1\right),but,-1\ne 1\end{array}$

$\therefore g$ is not injective.

Now, $gof:N\to Z$ is defined as

$gof\left(x\right)=y\left(f\left(x\right)\right)=y\left(x\right)=\left|x\right|$

Let $x,y\in N$ such that $gof\left(x\right)-gof\left(y\right)$

$\Rightarrow \left|x\right|=\left|y\right|$

Since $x,y\in N$ , both are positive.

$\therefore \left|x\right|=\left|y\right|\Rightarrow x=y$

Hence, gof is injective

f: R → R is given as f (x) = x³.

Suppose f (x) = f (y), where x, y ∈ R.

⇒ x³ = y³  … (1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.