Relations and Functions Test - 86

Given that:

\(f:R \rightarrow R\), given by \(f(x)=e^{x}\).

One − one:

Let \(x_{1}\) and \(x_{2}\) be any two elements in the domain (R), such that \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(f\left(x_{1}\right)\):

\(\Rightarrow f\left(x_{1}\right)=e^{x_{1}}\)

Now, \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow {e}^{{x}_{1}}={e}^{{x}_{2}}\)

\(\Rightarrow {x}_{1}={x}_{2}\)

\(\therefore {f}\) is one-one function.

Onto:

We know that:

Range of \(e^{x}\) is \((0, \infty)=R^{+}\)

\(\Rightarrow\) Co-domain \(=R\)

Both are not same.

\(\therefore {f}\) is not onto function.

If the co-domain is replaced by \({R}^{+}\), then the co-domain and range become the same and in that case, \({f}\) is onto and therefore, it is a bijection.

Given here:

1. \(f(x)=x^{3}\)

\(f\left(x_{1}\right)=x_{1}^{3}\) and

\(f\left(x_{2}\right)=x_{2}^{3}\)

\(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow x_{1}^{3}=x_{2}^{3}\)

This is one-one function.

This statement is correct.

2. \(f(a)=f(b)\)

\(\Rightarrow a=b\)

Which is a definition of one-one function.

Let's check the given relation for its type one by one.

Reflexive: Every line is parallel to itself. It means that \(\mathrm{lRl}\) for all \(\mathrm{l \in L}\). Therefore, \(\mathrm{R}\) is reflexive.

Symmetric: If a line \(\mathrm{l}\) is parallel to \(\mathrm{m}\), then \(\mathrm{m}\) is parallel to l, i.e., if \(\mathrm{lRm} \Rightarrow \mathrm{mRl}, \forall \mathrm{~l}, \mathrm{m} \in \mathrm{L}\). Therefore, \(\mathrm{R}\) is symmetric.

Transitive: If a line \(\mathrm{l}\) is parallel to \(\mathrm{m}\) and \(\mathrm{m}\) is parallel to \(\mathrm{k}\), then \(\mathrm{l}\) is also parallel to \(\mathrm{k}\), i.e., if \(\mathrm{lRm}\) and \(\mathrm{m R k \Rightarrow l R k, \forall ~l, m}\), \(\mathrm{k} \in \mathrm{L}\). Therefore, \(\mathrm{R}\) is transitive.

Since, the relation \(\mathrm{R}\) is reflexive, symmetric and transitive as well, it is an equivalence relation.

Given:

\({ }^{*}\) is a binary operation on \(Z\) such that:

\(a^{*} b=a+b+1 \forall a, b \in Z\)

Statement 1:\({ }^{*}\) is associative on \(\mathrm{Z}\).

Let \(a, b, c \in Z\)

First lets calculate \(\left(a^{*} b\right)^{*} c\)

Now according to the definition of *, we have:

\(\Rightarrow\left(a^{*} b\right)^{*} c=(a+b+1)^{*} c\)

\(\Rightarrow(a+b+1)^{*} c=(a+b+1)+c+1=a+b+c+2\)

\(\Rightarrow\left(a^{*} b\right)^{*} c=a+b+c+2\)\(\quad\).....(1)

Similarly, lets find out the value of \(a^{*}\left(b^{*} c\right)\)

\(\Rightarrow a^{*}\left(b^{*} c\right)=a^{*}(b+c+1)\)

\(\Rightarrow a^{*}(b+c+1)=a+(b+c+1)+1=a+b+c+2\)

\(\Rightarrow a^{*}\left(b^{*} c\right)=a+b+c+2\)\(\quad\).....(2)

Now from (1) and (2), we get:

\(\left(a^{*} b\right)^{*} c=a^{*}\left(b^{*} c\right) \forall a, b, c \in Z\).

Therefore, statement 1 is true.

Statement 2 :\({ }^{*}\) is commutative on \(Z\)

Let \(a, b \in Z\).

First lets find out \(\mathrm{a}^{*} \mathrm{~b}\)

According to the definition of * we have

\(\Rightarrow a^{*} b=a+b+1\)\(\quad\).....(3)

Similarly lets find out \(b^{*} a\)

Again by the definition of * we have,

\(\Rightarrow b^{*} a=b+a+1\)\(\quad\)......(4)

Now, from (3) and * (4), we have:

\(a^{*} b=b^{\star} a \forall a, b \in Z\).

Therefore, statement 2 is also true.

Given that: 

\(R\) is a relation on \(Z\) and is defined as: 

\(R=\{(a, b): a-b\) is divisible by 7 where \(a, b \in Z\}\)

We know that:

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Therefore,

\(R\) is reflexive as: 

\(a - a=0\) is divisible by 7 for all \(a \in Z\).

Suppose, if \((a, b) \in R\)

\( \Rightarrow 7\) divides \(a-b\) i.e., 

\(\Rightarrow a-b=7 m\), where \(m\in Z \)

\(\Rightarrow b-a=7 n\), where \(n=-m\)

\(\Rightarrow 7\) divides \(b-a\) too, which implies that: 

\((b, a) \in R .\)

Therefore, \(R\) is symmetric.

Suppose, if \((a, b) \in R\) and \((b, c) \in R\), then: 

\(a-b\) and \(b-c\) are divisible by 7.

\(\Rightarrow a-b=7 m\) and \(b-c=7 n\), where \(m, n \in Z\)

\(\Rightarrow a-c=7 q\), where \(q=m+n\)

\(\Rightarrow(a, c) \in R\)

Therefore, \(R\) is transitive.

Thus relation \(R\) is an equivalence relation.

Given function:

\(f:\mathrm{R} \rightarrow\{0,1\}\) such that \(f(x)=\left\{\begin{array}{c}1 \text { if } x \text { is rational } \\ 0 \text { if } x \text { is irrational }\end{array}\right.\)

Let f(x) be any function.

f (x) is onto if range of f (x) = Codomain

Codomain = {0, 1}

Since, on taking a straight line parallel to the x-axis, the group of given function intersect it at many points.

\(\Rightarrow f(x)\) is many-one.

Range of function is \(\{0,1\}\)

As range of \(f(x)=\) Codomain

\(\Rightarrow f(x)\) is onto.

Therefore, \(f(x)\) is many-one onto.

Let \(f(x)\) be any function.

\(f(x)\) is onto if range of \(f(x)=\) Co-domain

The function \(f\) is said to be many-one functions if there exist two or more than two different elements in \(X\) having the same image in \(Y\).

Given function is:

\(f: R \rightarrow\{0,1\}\), such that:

\(f(x)=\left\{\begin{array}{c}1 \text { if } x \text { is rational } \\0 \text { if } x \text { is irrational }\end{array}\right.\)

Co-domain \(=\{0,1\}\)

Since, on taking a straight line parallel to the \(x\)-axis, the group of given function intersect it at many points.

\(\Rightarrow f(x)\) is many-one.

Range of function is \(\{0,1\}\)

As range of \(f(x)=\) Co-domain

\(\Rightarrow f(x)\) is onto.

Therefore, \(f(x)\) is many-one onto.

We know that:

One - One Function/Injective Function:

A function f: A → B is said to be an one - one function, if different elements in A have different f - images in B.

i.e., if \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow x_{1}=x_{2}, \forall x_{1}, x_{2} \in A\),

Onto Function/Surjective Function:

A function f: A → B is said to be an onto function if each element in B has at least one pre-image in A.

Bijective Function:

A function f: A → B is said to be a bijective function if it is both one -one and onto function.

Given here:

\(f: N \rightarrow N\) and is defined as \(f(x)=x+1\)

One - One:

Let \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow x_{1}+1=x_{2}+1\)

\(\Rightarrow x_{1}=x_{2}\)

Therefore, the given function f is one - one.

Onto:

Let \(y=f(x)=x+1\)

\(\Rightarrow x=y-1\)

But if \(\mathrm{y}=1 \in \mathrm{N}\) then \(\mathrm{x}=0 \notin \mathrm{N}\).

Therefore, all the elements of co - domain i.e., N does not has a pre - image in the domain i.e., N.

Therefore, the given function is not onto.

Given that:

\(f(x)=x^{2}+4 x+4\)

Replace \(x\) by \(-x\).

We know that:

If f(x) is even function then f(-x) = f(x)

If f(x) is odd function then f(-x) = -f(x)

\(\Rightarrow \mathrm{f}(-x)=(-x)^{2}+4(-x)+4\)

\(=x^{2}-4 x+4 \quad\left(\because(-x)^{2}=x^{2}\right)\)

\(\Rightarrow f(-x) \neq \pm f(x)\)

Therefore, function is neither odd nor even.

Given,

\(f(x)=8 x^{3}\) and \(g(x)=x^{\frac{1}{3}}\)

\(\operatorname{gof}(x)=g[f(x)]=\left[8 x^{3}\right]^{\frac{1}{3}}\)

We know that:

\(\operatorname{fog}(x)=f[g(x)]\)

\(\Rightarrow\) \(\operatorname{gof}(x)=\left[(2 x)^{3}\right]^{\frac{1}{3}}\)

\(\Rightarrow\) \(\operatorname{gof}(x)=2 x\)

Now, put \(x=2\)

So, \(\operatorname{gof}(2)=4\).