Relations and Functions Test - 27

Given here, 

\(f(x)=x^{4}-\frac{1}{x^{4}}\)\(\quad\)......(1)

Replace \(\mathrm{x}\) by \(\frac{1}{\mathrm{x}}\), we get,

\(f\left(\frac{1}{x}\right)=\frac{1}{x^{4}}-x^{4}\)\(\quad\)......(2)

So, from (1) and (2), 

\(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}^{4}-\frac{1}{\mathrm{x}^{4}}+\frac{1}{x^{4}}-\mathrm{x}^{4}\)

\(=0\)

Hence, the correct option is (D).

Given: 

\(\mathrm{f}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}}\) and \(\mathrm{g}(\mathrm{x})=\ln \mathrm{x}\)

\(\operatorname{gof}(x)=g(f(x))\)

\(\operatorname{gof}(x)=g\left(e^{2 x}\right)\)

\(\operatorname{gof}(x)=\ln e^{2 x}\)

\(\operatorname{gof}(x)=2 x \ln e\)

\(\operatorname{gof}(x)=2 x \quad \{\because \ln e=1\}\)

Hence, the correct option is (B).

We know that:

In BODMAS, 

\(B \rightarrow\) Brackets, \(O \rightarrow\) Of, \(D \rightarrow\) Division, \(M \rightarrow\) Multiplication, \(A \rightarrow\) Addition, \(S \rightarrow\) Subtraction.

Given:

\(a * b=\sqrt{a^{2}+b^{2}}\)

\(\therefore(3 * 4) * 5\)

\(=\sqrt{(3 * 4)^{2}+5^{2}}\)

\(=\sqrt{\left(\sqrt{3^{2}+4^{2}}\right)^{2}+5^{2}}\)

\(=\sqrt{3^{2}+4^{2}+5^{2}}\)

\(=\sqrt{9+16+25}\)

\(=\sqrt{50}\)

\(=5 \sqrt{2}\)

Hence, the correct option is (B).

Given here, 

\(R=\{(a, b): a, b \in Z\) and \((a+b)\) is even \(\}\)

We know that:

R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Therefore, 

1. Since, \(a+a=2 a\), which is even, so \(R\) is reflexive.

2. If \(a+b\) is even, then \(b+a\) will also be even. So, \(R\) is symmetric.

3. Let, \(a=3, b=5\), and \(c=7\), then:

\(a+b=3+5=8\) (which is even), 

b \(+c=5+7=12\) (which is again even) 

and, \(a+c=3+7=10\) (which is also even)

So, R is transitive.

Therefore, \(R\) is an equivalence relation on \(Z\).

Hence, the correct option is (B).

Given here, 

\(f(x)=\frac{x+1}{x-1}\)

We know that:

\(f \circ g(x)=f[g(x)]\)

\(f\{f(x)\}=\frac{f(x)+1}{f(x)-1}\)

\(=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}\)

\(=\frac{2 x}{2}\)

\(=x\)

Hence, the correct option is (C).

Given that: 

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

\(a^{*} b=a+b+1 \forall a, b \in Z\) 

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

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 ∈ Z and because e is the identity element of Z with respect to given operation *.

i.e., a * e = a =  e * a ∀ a ∈ Z.

According to the definition of *, we have:

⇒ a * e = a + e + 1 = a ∀ a ∈ Z

⇒ e = - 1

Therefore, - 1 is the identity element of Z with respect to given operation *.

Hence, the correct option is (A).

If the function is onto and one-to-one, then it is called as Bijection.

If a function is said to be one-to-one, then it is called as Injective.

If a function is said to be onto, then it is called as Surjective.

If f is a bijection from set A to set B, then the function g is called as Inverse function.

Hence, the correct option is (B).

Given, 

\(f(x)=3 x, g(x)=3 x^{2}+9\) 

Then, 

\((f+g)(2)=f(2)+g(2)\)

\(=(3 \times 2)+\left(3 \times 2^{2}+9\right)\)

\(=6+21\)

\(=27\)

\(\therefore\) The value of \((f+g)(2)\) is 27.

Hence, the correct option is (C).

Given that: 

\(R=\left\{(a, b): a^{2} \geq b\right\}\)

As we know:

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

We know that: 

\(\mathrm{a}^{2} \geq \mathrm{a}\)

Therefore, \((\mathrm{a}, \mathrm{a}) \in \mathrm{R}\), for all \(\mathrm{a} \in \mathrm{Z}\).

Thus, relation \(\mathrm{R}\) is reflexive.

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

\(\Rightarrow a^{2} \geq b\) but \(b^{2} \ngeq a\) for all \(a, b \in Z\).

So, if \((a, b) \in R\), then it does not implies that \((b, a)\) also belongs to \(\mathrm{R}\).

Thus, relation \(\mathrm{R}\) is not symmetric.

Now, let \((a, b) \in R\) and \((b, c) \in R\).

\(\Rightarrow a^{2} \geq b\) and \(b^{2} \geq c\)

This does not implies that \(\mathrm{a}^{2} \geq \mathrm{c}\), therefore \((\mathrm{a}, \mathrm{c})\) does not belong to \(\mathrm{R}\) for all \(\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{Z}\).

Thus, relation \(\mathrm{R}\) is not transitive.

Hence, the correct option is (D).

Given: 

\(y=5^{\log x}\)

Here, we have to find the inverse of the given function.

By applying log to base 5 on both sides of \(y=5^{\log x}\), we get:

Let \(f: A \rightarrow B\) be a bijective function.

Step - I: Put \(f(x)=y\)

Step - II: Solve the equation \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) to obtain \(\mathrm{x}\) in terms of \(\mathrm{y}\). 

Interchange \(x\) and \(y\) to obtain the inverse of the given function \(f\)

\(\Rightarrow \log _{5} y=\log _{5}\left(5^{\log x}\right)\)

\(\Rightarrow \log _{5} y=\log x\)

\(\Rightarrow \frac{\log y}{\log 5}=\log x\)

\(\Rightarrow \log \left[y^{\frac{1}{\log 5} 5}\right]=\log x\)

\(\Rightarrow x=y^{\frac{1}{\log 5}}\)

Hence, the correct option is (B).

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

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

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

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

Onto:

We know that: 

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

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

Both are not same.

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

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

Hence, the correct option is (A).

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.

Hence, the correct option is (A).

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.

Hence, the correct option is (D).

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.

Hence, the correct option is (C).

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.

Hence, the correct option is (D).

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.

Hence, the correct option is (D).

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.

Hence, the correct option is (D).

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.

Hence, the correct option is (B).

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.

Hence, the correct option is (C).

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:

\(f \circ g(x)=f[g(x)]\)

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

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

Now, put \(x=2\)

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

Hence, the correct option is (B).

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

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

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

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

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

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

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

Hence, the correct option is (D).

Let A = QXQ 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 = QXQ 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).

Hence, the correct option is (D).

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.

Hence, the correct option is (B).

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: 

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

Hence, the correct option is (C).

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.

Hence, the correct option is (A).

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.

Hence, the correct option is (C).

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 *.

Hence, the correct option is (D).

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.

Hence, the correct option is (D).

Given, 

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

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

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

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

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

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

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

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

Hence, the correct option is (B).

Given condition: 

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

Now,

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

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

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

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

Hence, the correct option is (B).