Relations and Functions Test - 87

Let \({f}({x})={y}\)

\(\therefore {f}^{-1}({y})={x}\)

Now, \({y}=2 {x}+6\)

\(2 {x}={y}-6\)

\({x}=\frac{{y}-6}{2}\)

\(\therefore {f}^{-1}({y})=\frac{{y}}{2}-3\)

\({f}^{-1}({x})=\frac{{x}}{2}-3\)

Let A = Q × Q and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for all (a, b), (c, d) belongs to A.

Let e = (a', b') be the identity element of A.

\(\Rightarrow\) p * e = p = e * p

Consider, p * e = p

(a, b) * (a', b') = (a, b)

\(\Rightarrow\) (aa', b + ab') = (a, b)

\(\Rightarrow\) aa' = a, b + ab' = b

\(\Rightarrow\) a' = 1, b' = 0

\(\Rightarrow\) e = (1, 0) be the identity element of A.

Therefore, A = Q × Q and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for all (a, b), (c, d) belongs to A, then the identity element in A is (1, 0).

Let two lines \(\ell_{1}, \ell_{2} \in \mathrm{L}\)

Now \(\left(\ell_{1}, \ell_{2}\right) \in R\)

Only when \(\ell_{1} \perp \ell_{2}\)

1. For reflexivity:

Let \(\ell_{1} \in \mathrm{L}\)

But \(\left(\ell_{1}, \ell_{1}\right) \notin R\)

Since, no line is perpendicular to itself.

Therefore, \(\mathrm{R}\) is not reflexive.

2. For symmetric:

Let, \(\left(\ell_{1}, \ell_{2}\right) \in \mathrm{L}\)

Now, if \(\ell_{1} \perp \ell_{2}\), then this implies that \(\ell_{2}\) is also perpendicular to \(\ell_{1}\).

So, \(\left(\ell_{1}, \ell_{2}\right) \in L\)

Therefore, \(\mathrm{R}\) is symmetric.

3. For Transitive:

Let \(\left(\ell_{1}, \ell_{2}\right) \in L\) and \(\left(\ell_{2}, \ell_{3}\right) \in L\)

\(\ell_{1} \perp \ell_{2}\) and \(\ell_{2} \perp \ell_{3}\)

\(\ell_{1}\) is not perpendicular to \(\ell_{3}\).

\(\left(\ell_{1}, \ell_{3}\right) \notin \mathrm{R}\)

So, R is not transitive.

It is given that:

\(a * b=\frac{a b}{2} ; a, b \in Q-\{0\}\)

Let the identity element for * be \(\mathrm{i}\), i.e.,

\(\mathrm{a}\times \mathrm{i}=\mathrm{a}\)

\(\therefore \mathrm{a} * \mathrm{i}=\frac{\mathrm{ai}}{2}\)

\(\Rightarrow \mathrm{a}=\frac{\mathrm{ai}}{2}\)

\(\Rightarrow \mathrm{i}=2\)

Therefore, the identity element for the operation * is:

\(\mathrm{i}={2}\)

Given set is:

\(\mathrm{A}=\{1,2,3\}\)

and, the relation is:

\(R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1, 3)\}\)

Let \(A\) be a set in which the relation \(R\) defined.

1. \(R\) is said to be a reflexive relation, if:

\((a, a) \in R\)

2. \(R\) is said to be a symmetric relation, if:

\((a, b) \in R \Rightarrow(b, a)\) \(\in \mathrm{R}\)

3. \(R\) is said to be a transitive relation, if:

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

Since, \(1,2,3 \in A\) and \((1,1),(2,2),(3,3) \in R\)

Therefore, \( \mathrm{R}\) is Reflexive.

Now, \(1,2,3 \in \mathrm{R}\)

\((1,2),(2,3) \in \mathrm{R} \Rightarrow(1,3) \in \mathrm{R}\)

So, \( \mathrm{R}\) relates 1 to 2 and 2 to 3, then \(\mathrm{R}\) also relates 1 to 3.

Therefore, \( \mathrm{R}\) is Transitive.

Here, \(\mathrm{R}\) is not symmetric relation, as:

\((\mathrm{a}, \mathrm{b}) \in \mathrm{R} \neq(\mathrm{b}, \mathrm{a}) \in \mathrm{R}\)

Therefore,

The relation, \(\mathrm{R}=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3) \}\) on a set \(\mathrm{A}=\{1,2,3\}\) is reflexive, transitive but not symmetric.

Let us check for each option:

(A)Given:

\(f(x)=|x|, \forall x \in R\)

As we know that,

\(f(x)=|x|\)

\(\Rightarrow f(x)=\left\{\begin{array}{cc}-x, & x<0 \\ x, & x \geq 0\end{array}\right.\)

So, \(f(-1)=-(-1)=1\) and \(f(1)=1\)

\(\Rightarrow f(-1)=f(1)\), but \(-1 \neq 1\)

\(\therefore\) The property, \(f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\), does not hold true \(\forall x_{1}, x_{2} \in R\).

Therefore, the function \(\mathrm{f}(x)=|x|, \forall x \in R\) is not an injective function.

(B) Given:

\(f(x)=x^{2}, \forall x \in R\)

Let \(x_{1}=1\) and \(x_{2}=-1\)

\(\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{x}_{1}^{2}=1\)

and, \(\mathrm{f}\left(\mathrm{x}_{2}\right)=\mathrm{x}_{2}^{2}=1\)

\(\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)\), but \(-1 \neq 1\)

\(\therefore\) The property, \(f\left(x_{1}\right)=f\left(x_{2}\right)\)\( \Rightarrow x_{1}=x_{2}\), does not hold true \(\forall x_{1}, x_{2} \in R\).

Therefore, the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}, \forall \mathrm{x} \in \mathrm{R}\) is not an injective function.

(C) Given:

\(f(x)=-x, \forall x \in R\)

Let \(x_{1}\) and \(x_{2}\) be any two real numbers.

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

If \(\mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)\)

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

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

\(\therefore\) The property, \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow x_{1}=x_{2}\), holds true \(\forall x_{1}, x_{2} \in R\)

Therefore, the function \(f(x)=-x, \forall x \in R\) is an injective function.

Given * is the binary operator defined on \(R\) such that,

\(a^{*} b=a+b-a b \forall a, b \in R\)

Let e be the identity element of \(R\) with respect to *.

Now, as we know that, if e is an identity element of a non-empty set S with respect to a binary operation *.

Then,

a * e = e * a = a ∀ a ∈ S

Let a ∈ R and because e is the identity element of R with respect to given operation *

i.e., a * e = a = e * a ∀ a ∈ R

Now,

According to the definition of *, we have:

a * e = a + e – ae = a ∀ a ∈ R ---(1)

∴ e – ae = 0

e(1 – a) = 0

e = 0 or (1 – a) = 0

Therefore, 0 is the identity of R with respect to give operation *.

For a, b in the set of children, we say that a and b are related if a has the same father as b.

Now, we will check for each property one by one.

The relation is obviously reflexive as a has the same father as a.

Therefore, aRa .... (1)

If a has the same father as b, then that means both a and b have the same father.

Therefore, b also has the same father as a. Thus, the relation is symmetric.

Therefore, aRb implies bRa .... (2)

Now, assume that aRb and bRc.

That means a has the same father as b and b has the same father as c. This implies that a, b and c have the same father.

Therefore, a has the same father as c. Thus, the relation is transitive.

Therefore, aRb, bRc implies aRc .... (3)

Thus, from (1), (2) and (3), we conclude that the given relation is an equivalence relation.

Given,

\(f(x+1)=x^{2}-3 x+2\)

\(=x^{2}-2 x-x+2\)

\(=x(x-2)-(x-2)\)

\(=(x-2)(x-1)\)

\(=(x+1-3)(x+1-2)\)

\(\therefore f(x)=(x-3)(x-2)\)

\(=x^{2}-3 x-2 x+6\)

\(=x^{2}-5 x+6\)

Given condition:

\(\mathrm{a}{*} \mathrm{b}=\frac{2 \mathrm{ab}}{5}\)

Now,

\(\Rightarrow 2{*} \mathrm{x}=3^{-1}\)

\(\Rightarrow \frac{2(2)(\mathrm{x})}{5}=\frac{1}{3}\)

\(\Rightarrow 4 \mathrm{x}=\frac{5}{3}\)

\(\mathrm{x}=\frac{5}{12}\)