Miscellaneous Exercise Relations and Function Solution - NCERT Class 12


Chapter 1 Ex.1.ME Question 1

Let \(f:R \to R\) be defined as \(f(x)=\text{1}0x~+\text{7}\). Find the function \(g:R \to R\) such that \(gof = fog = {{\rm{I}}_R}\).

 

Solution

Video Solution

 

\(f:R \to R\) is defined as \(f(x)=\text{1}0x~+\text{7}\)

For one-one:

\(f\left( x \right) = f\left( y \right)\) where \(x,y \in R\)

\(\begin{align}& \Rightarrow 10x + 7 = 10y + 7\\& \Rightarrow x = y\end{align}\)

\(\therefore f\)is one-one.

For onto:

\(y \in R,\,\,{\text{Let }}y = 10x + 7\)

\[ \Rightarrow x = \frac{{y - 7}}{{10}} \in R\]

For any \(y \in R,\) there exists \(x = \frac{{y - 7}}{{10}} \in R\) such that

\(f\left( x \right) = f\left( {\frac{{y - 7}}{{10}}} \right) = 10\left( {\frac{{y - 7}}{{10}}} \right) + 7 = y - 7 + 7 = y\)

\(\therefore f\)is onto.

Thus, \(f\) is an invertible function.

Let us define \(g:R \to R\)as\(g\left( y \right) = \frac{{y - 7}}{{10}}\).

Now,

\(gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {10x + 7} \right) = \frac{{\left( {10x + 7} \right) - 7}}{{10}} = \frac{{10x}}{{10}} = 10\)

And,

\(fog\left( y \right) = f\left( {g\left( y \right)} \right) = f\left( {\frac{{y - 7}}{{10}}} \right) = 10\left( {\frac{{y - 7}}{{10}}} \right) + 7 = y - 7 + 7 = y\)

\(\therefore gof = {{\text{I}}_R}\) and \(fog = {{\text{I}}_R}\)

Hence, the required function \(g:R \to R\) as\(g\left( y \right) = \frac{{y - 7}}{{10}}\).

Chapter 1 Ex.1.ME Question 2

Let \(f:W \to W\) be defined as\(f(n) = n - 1\), if is odd and \(f\left( n \right) = n + 1\), if \(n\) is even. Show that \(f\) is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

 

Solution

Video Solution

 

\(f:W \to W\)is defined as \(f\left( n \right) = \left\{ \begin{gathered} n - 1\,,\,\,{\text{If }}n{\text{ is odd}} \hfill \\ n + 1\,,\,\,{\text{If }}n{\text{ is even}} \hfill \\ \end{gathered} \right\}\)

For one-one:

\(f\left( n \right) = f\left( m \right)\)

If \(n\)is odd and\(m\) is even, then we will have\(n - 1 = m + 1\).

\( \Rightarrow n - m = 2\)

Similarly, the possibility of \(n\) being even and \(m\) being odd can also be ignored under a similar argument.

∴Both \(n\) and \(m\) must be either odd or even.

Now, if both \(n\) and \(m\) are odd, then we have:

\[\begin{align} & f\left( n \right)=~f\left( m \right) \\ & \Rightarrow ~n~-\text{ 1 }=~m~-\text{1 } \\ & \Rightarrow ~n~=~m \\ \end{align}\]

Again, if both \(n\) and \(m\) are even, then we have:

\[\begin{align} & f\left( n \right)=~f\left( m \right) \\ & \Rightarrow ~n~+\text{ 1 }=~m~+\text{1} \\ & \Rightarrow ~n~=~m \\ \end{align}\]

\(\therefore f\) is one-one.

For onto:

Any odd number \(2r + 1\)in co-domain N is the image of \(2r\) in domain N and any even number \(2r\) in co-domain N is the image of \(2r + 1\)in domain N.

\(\therefore f\) is onto.

\(f\)is an invertible function.

Let us define \(g:W \to W\) as \(f\left( m \right) = \left\{ \begin{array}{l}m - 1\,,\,\,{\text{If }}m{\text{ is odd}}\\m + 1,\,\,\,{\text{If }}m{\text{ is even}}\end{array} \right\}\)

When \(r\) is odd

\(gof\left( n \right) = g\left( {f\left( n \right)} \right) = g\left( {n - 1} \right) = n - 1 + 1 = n\)

When \(r\) is even

\(gof\left( n \right) = g\left( {f\left( n \right)} \right) = g\left( {n + 1} \right) = n + 1 - 1 = n\)

When \(m\) is odd

\(fog\left( n \right) = f\left( {g\left( m \right)} \right) = f\left( {m - 1} \right) = m - 1 + 1 = m\)

When \(m\) is even

\(fog\left( m \right) = f\left( {g\left( m \right)} \right) = f\left( {m + 1} \right) = m + 1 - 1 = m\)

\(\therefore gof = {{\text{I}}_W}\) and \(fog = {{\text{I}}_W}\)

\(f\) is invertible and the inverse of \(f\) is given by \({f^{ - 1}} = g\), which is the same as \(f\).

inverse of \(f\) is \(f\) itself.

Chapter 1 Ex.1.ME Question 3

If \(f:R \to R\) be defined as \(f(x) = {x^2} - 3x + 2\), find \(f\left( {f\left( x \right)} \right)\).

 

Solution

Video Solution

 

\(f:R \to R\) is defined as \(f(x) = {x^2} - 3x + 2\).

\[\begin{align}f\left( {f\left( x \right)} \right) &= f\left( {{x^2} - 3x + 2} \right)\\& = {\left( {{x^2} - 3x + 2} \right)^2} - 3\left( {{x^2} - 3x + 2} \right) + 2\\& = \left( {{x^4} + 9{x^2} + 4 - 6{x^3} - 12x + 4{x^2}} \right) + \left( { - 3{x^2} + 9x - 6} \right) + 2\\ &= {x^4} - 6{x^3} + 10{x^2} - 3x\end{align}\]

Chapter 1 Ex.1.ME Question 4

Show that function \(f:R \to \left\{ {x \in R: - 1 < x < 1} \right\}\) be defined by \(f(x) = \frac{x}{{1 + \left| x \right|}}\), \(x \in R\) is one-one and onto function.

 

Solution

Video Solution

 

\(f:R\to \left\{ x\in R:-1<x<1 \right\}\) is defined by \(f(x) = \frac{x}{{1 + \left| x \right|}}\), \(x \in R\).

For one-one:

\(f\left( x \right) = f\left( y \right)\)  where \(x,y \in R\)

\( \Rightarrow \frac{x}{{1 + \left| x \right|}} = \frac{y}{{1 + \left| y \right|}}\)

If \(x\) is positive and \(y\) is negative,

\(\begin{align}\frac{x}{{1 + \left| x \right|}} = \frac{y}{{1 + \left| y \right|}}\\ \Rightarrow 2xy = x - y\end{align}\)

Since, \(x\) is positive and \(y\) is negative,

\(x > y \Rightarrow x - y > 0\)

\(2xy\) is negative.

\(2xy \ne x - y\)

Case of \(x\) being positive and \(y\) being negative, can be ruled out.

\(\therefore\; x\) and \(y\) have to be either positive or negative.

If \(x\) and \(y\) are positive,

\[\begin{align}&f\left( x \right) = f\left( y \right)\\ &\Rightarrow \frac{x}{{1 + x}} = \frac{y}{{1 + y}}\\& \Rightarrow x - xy = y - xy\\ &\Rightarrow x = y\end{align}\]

\(\therefore f\) is one-one.

For onto:

Let \(y \in R\)such that \( - 1 < y < 1\).

If \(x\) is negative, then there exists \(x = \frac{y}{{1 + y}} \in R\) such that

\[f\left( x \right) = f\left( {\frac{y}{{1 + y}}} \right) = \frac{{\left( {\frac{y}{{1 + y}}} \right)}}{{1 + \left| {\frac{y}{{1 + y}}} \right|}} = \frac{{\frac{y}{{1 + y}}}}{{1 + \left( {\frac{{ - y}}{{1 + y}}} \right)}} = \frac{y}{{1 + y - y}} = y\]

If \(x\) is positive, then there exists \(x = \frac{y}{{1 - y}} \in R\)such that

\[f\left( x \right) = f\left( {\frac{y}{{1 - y}}} \right) = \frac{{\left( {\frac{y}{{1 - y}}} \right)}}{{1 + \left| {\frac{y}{{1 - y}}} \right|}} = \frac{{\frac{y}{{1 - y}}}}{{1 + \left( {\frac{y}{{1 - y}}} \right)}} = \frac{y}{{1 - y + y}} = y\]

\(\therefore f\)is onto.

Hence, \(f\) is one-one and onto.

Chapter 1 Ex.1.ME Question 5

Show that function \(f:R \to R\) be defined by\(f(x) = {x^3}\) is injective.

 

Solution

Video Solution

 

\(f:R \to R\) is defined by\(f(x)={{x}^{3}}\)

For one-one:

\(f\left( x \right) = f\left( y \right)\) where\(x,y \in R\)

\[{x^3} = {y^3}........................\left( 1 \right)\]

We need to show that \(x = y\)

Suppose \(x \ne y\), their cubes will also not be equal.

\[ \Rightarrow {x^3} \ne {y^3}\]

This will be a contradiction to \(\left( 1 \right)\).

\(\therefore x = y\). Hence, \(f\) is injective.

Chapter 1 Ex.1.ME Question 6

Give examples of two functions \(f:N \to Z\) and \(g:Z \to Z\) such that \(gof\) is injective but \(g\) is not injective.

(Hint: Consider \(f\left( x \right) = x\)and \(g\left( x \right) = \left| x \right|\))

 

Solution

Video Solution

 

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

Let us first show that \(g\)is not injective.

\(\begin{align}\left( { - 1} \right) = \left| { - 1} \right| = 1\\\left( 1 \right) = \left| 1 \right| = 1\\\therefore \left( { - 1} \right) = g\left( 1 \right),\,\,\,\,\,{\text{but }} - 1 \ne 1\end{align}\)

\(\therefore g\)is not injective.

\(gof:N \to Z\)is defined as \(gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( x \right) = \left| x \right|\)

\(x,y \in N\)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.

\(\begin{align}\therefore \left| x \right| = \left| y \right|\\ \Rightarrow x = y\end{align}\)

\(\therefore \) \(gof\) is injective.

Chapter 1 Ex.1.ME Question 7

Given examples of two functions \(f:N \to N\) and \(g:N \to N\) such that \(gof\) is onto but \(f\) is not onto.

(Hint: Consider \(f\left( x \right) = x + 1\)and \(g\left( x \right) = \left\{ \begin{array}{l}x - 1,\,\,\,{\text{if }}x > 1\\1,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}x = 1\end{array} \right\}\))

 

Solution

Video Solution

 

Define \(f:N \to Z\)as \(f\left( x \right) = x + 1\)and \(g:Z \to Z\)as \(g\left( x \right) = \left\{ \begin{array}{l}x - 1,\,\,\,{\text{if }}x > 1\\1,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}x = 1\end{array} \right\}\)

Let us first show that \(g\) is not onto.

Consider element 1 in co-domain \(N\). This element is not an image of any of the elements in domain \(N\).

\(\therefore f\) is not onto.

\(g:N \to N\) is defined by

\(gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {x + 1} \right) = x + 1 - 1 = x \qquad \left[ {x \in N \Rightarrow x + 1 > 1} \right]\)

For \(y \in N\), there exists \(x = y \in N\) such that \(gof\left( x \right) = y\).

\(\therefore gof\) is onto.

Chapter 1 Ex.1.ME Question 8

Given a non-empty set \(X\), consider \(P\left( X \right)\) which is the set of all subsets of \(X\).

Define the relation \(R\) in \(P\left( X \right)\) as follows:

For subsets \(A,B\) in \(P\left( X \right)\), \(ARB\) if and only if \(A \subset B\). Is \(R\) an equivalence relation on \(P\left( X \right)\)? Justify you answer.

 

Solution

Video Solution

 

Since every set is a subset of itself, \(ARA\) for all\(A \in P\left( X \right)\).

\(\therefore R\)is reflexive.

Let \(ARB \Rightarrow A \subset B\)

This cannot be implied to \(B \subset A\).

If \(A = \left\{ {{\text{1}},{\text{2}}} \right\}\) and \(B = \left\{ {{\text{1}},{\text{ 2}},{\text{3}}} \right\}\), then it cannot be implied that \(B\) is related to\(A\).

\(\therefore \) \(R\) is not symmetric.

If \(ARB\) and\(BRC\), then \(A \subset B\)and\(B \subset C\).

\(\begin{align} &\Rightarrow A \subset C \hfill \\& \Rightarrow ARC\end{align} \)

\(\therefore\;R\) is transitive.

\(R\) is not an equivalence relation as it is not symmetric.

Chapter 1 Ex.1.ME Question 9

Given a non-empty set \(X\), consider the binary operation \(^*\): \({\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right){\text{ }}\)given by \(A*B = A \cap B\,\,\,\forall \,A,B\) in \(P\left( X \right)\) is the power set of \(X\). Show that \(X\) is the identity element for this operation and \(X\) is the only invertible element in \(P\left( X \right)\)with respect to the operation \(^*\).

 

Solution

Video Solution

 

\({\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right){\text{ }}\)given by \(A*B = A \cap B\,\,\,\forall \,A,B\) in \(P\left( X \right)\)

\(A \cap X = A = X \cap A\) for all \(A \in P\left( X \right)\)

\( \Rightarrow A*X = A = X*A\) for all \(A \in P\left( X \right)\)

\(X\) is the identity element for the given binary operation \(^*\).

An element \(A \in P\left( X \right)\)is invertible if there exists \(B \in P\left( X \right)\)such that

\(A*B = X = B*A\) [As \(X\)is the identity element]

Or

\(A \cap B = X = B \cap A\)

This case is possible only when \(A = X = B\).

\(X\) is the only invertible element in \(P\left( X \right)\)with respect to the given operation \(^*\).

Chapter 1 Ex.1.ME Question 10

Find the number of all onto functions from the set \(\left\{ {1,2,3, \ldots ,n} \right\}\) to itself.

 

Solution

Video Solution

 

Onto functions from the set \(\left\{ {1,2,3, \ldots ,n} \right\}\) to itself is simply a permutation on \(n\)symbols

\(1,2,3, \ldots ,n\).

Thus, the total number of onto maps from \(\left\{ {1,2,3, \ldots ,n} \right\}\) to itself is the same as the total number of permutations on \(n\) symbols \(1,2,3, \ldots ,n\), which is\(n!\).

Chapter 1 Ex.1.ME Question 11

Let \({\text{S}} = \left\{ {a,b,c} \right\}\)and \(T = \left\{ {{\text{1}},{\text{ 2}},{\text{3}}} \right\}\). Find \({{F}^{-1}}\) of the following functions \(F\) from \(S\) to \(T\), if it exists.

(i) \(F = \left\{ {\left( {a,3} \right),\left( {b,2} \right),\left( {c,1} \right)} \right\}\)

(ii) \(F = \left\{ {\left( {a,2} \right),\left( {b,1} \right),\left( {c,1} \right)} \right\}\)

 

Solution

Video Solution

 

\({\text{S}} = \left\{ {a,b,c} \right\},T = \left\{ {{\text{1}},{\text{ 2}},{\text{3}}} \right\}\)

(i) \(F:S \to T\)is defined by \(F = \left\{ {\left( {a,3} \right),\left( {b,2} \right),\left( {c,1} \right)} \right\}\)

\( \Rightarrow F\left( a \right) = 3,\,\,F\left( b \right) = 2,\,\,F\left( c \right) = 1\)

Therefore, \({F^{ - 1}}:T \to S\)is given by \({F^{ - 1}} = \left\{ {\left( {3,a} \right),\left( {2,b} \right),\left( {1,c} \right)} \right\}\)

(ii) \(F:S \to T\)is defined by \(F = \left\{ {\left( {a,2} \right),\left( {b,1} \right),\left( {c,1} \right)} \right\}\)

Since, \(F\left( b \right) = F\left( c \right) = 1\), \(F\) is not one-one.

Hence, \(F\) is not invertible i.e., \({F^{ - 1}}\) does not exists.

Chapter 1 Ex.1.ME Question 12

Consider the binary operations\(^*\): \(R \times R \to R\)and \(o:R \times R \to R\) defined as \(a*b = \left| {a - b} \right|\) and \(aob = a,\,\,\forall a,b \in R\). Show that \(^*\) is commutative but not associative \(o\)is associative but not commutative. Further, show that\(\forall a,b,c \in R\), \(a*\left( {boc} \right) = \left( {a*b} \right)o\left( {a*c} \right)\). [ If it is so, we say that the operation \(^*\) distributes over the operation\(o\)]. Does \(o\) distribute over\(^*\)? Justify your answer.

 

Solution

Video Solution

 

It is given that \(^*\): \(R \times R \to R\)and \(o:R \times R \to R\) defined as \(a*b = \left| {a - b} \right|\) and \(aob = a,\,\,\forall a,b \in R\).

For \(a,b \in R\), we have \(a*b = \left| {a - b} \right|\) and \(b*a = \left| {b - a} \right| = \left| { - \left( {a - b} \right)} \right| = \left| {a - b} \right|\)

\(\therefore a*b = b*a\)

\(\therefore \) The operation \(^*\)is commutative.

\(\begin{align}\left( {1*2} \right)*3 = \left( {\left| {1 - 2} \right|} \right)*3 = 1*3 = \left| {1 - 3} \right| = 2\\1*\left( {2*3} \right) = 1*\left( {\left| {2 - 3} \right|} \right) = 1*1 = \left| {1 - 1} \right| = 0\end{align}\)

\(\therefore \left( {1*2} \right)*3 \ne 1*\left( {2*3} \right)\) where \(1,2,3 \in R\)

\(\therefore \)The operation * is not associative.

Now, consider the operation\(o\):

It can be observed that \(1o2 = 1\)and \(2o1 = 2\).

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

\(\therefore \)The operation \(o\)is not commutative.

Let \(a,~b,~c~\in R\). Then, we have:

\(\begin{align}\left( {aob} \right)oc = aoc = a\\ao\left( {boc} \right) = aob = a\\ \Rightarrow \left( {aob} \right)oc = ao\left( {boc} \right)\end{align}\)

\(\therefore \)The operation \(o\) is associative.

Now, let \(a,~b,~c~\in ~R\), then we have:

\(\begin{align}&a*\left( {boc} \right) = a*b = \left| {a - b} \right|\\&\left( {a*b} \right)o\left( {a*c} \right) = \left( {\left| {a - b} \right|} \right)o\left( {\left| {a - c} \right|} \right) = \left| {a - b} \right|\end{align}\)

Hence, \(a*\left( {boc} \right) = \left( {a*b} \right)o\left( {a*c} \right)\)

Now,

\(\begin{gathered} 1o\left( {2*3} \right) = 1o\left( {\left| {2 - 3} \right|} \right) = 1o1 = 1 \hfill \\ \left( {1o2} \right)*\left( {1o3} \right) = 1*1 = \left| {1 - 1} \right| = 0 \hfill \\ \end{gathered} \)

\(\therefore 1o\left( {2*3} \right) \ne \left( {1o2} \right)*\left( {1o3} \right)\) where \(1,2,3 \in R\)

\(\therefore \)The operation \(o\)does not distribute over\(^*\).

Chapter 1 Ex.1.ME Question 13

Given a non - empty set \(X\), let \(^*\): \({\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right)\)be defined as \(A*B = \left( {A - B} \right) \cup \left( {B - A} \right)\), \(\forall A,B \in P\left( X \right)\). Show that the empty set \(\Phi \) is the identity for the operation \(^*\) and all the elements \(A\) of \(P\left( X \right)\) are invertible with \({A^{ - 1}} = A\).

(Hint: \(\left( {A - \Phi } \right) \cup \left( {\Phi - A} \right) = A\) and \(\left( {A - A} \right) \cup \left( {A - A} \right) = A*A = \Phi \)).

 

Solution

Video Solution

 

It is given that \(^*\): \({\text{P}}\left( X \right) \times {\text{P}}\left( X \right) \to {\text{P}}\left( X \right)\)is defined as \(A*B = \left( {A - B} \right) \cup \left( {B - A} \right)\),\(\forall A,B \in P\left( X \right)\)

\(A \in P\left( X \right)\) then,

\(\begin{align}A*\Phi = \left( {A - \Phi } \right) \cup \left( {\Phi - A} \right) = A \cup \Phi = A\\\Phi *A = \left( {\Phi - A} \right) \cup \left( {A - \Phi } \right) = \Phi \cup A = A\\\therefore A*\Phi = A = \Phi *A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{for all }}A \in P\left( X \right)\end{align}\)

\(\Phi \) is the identity for the operation \(^*\).

Element \(A\in P\left( X \right)\) will be invertible if there exists \(B \in P\left( X \right)\)such that

\(A*B = \Phi = B*A\) [As \(\Phi \)is the identity element]

\(A*A = \left( {A - A} \right) \cup \left( {A - A} \right) = \Phi \cup \Phi = \Phi \) for all \(A \in P\left( X \right)\).

All the elements \(A\)of \(P\left( X \right)\)are invertible with \({A^{ - 1}} = A\).

Chapter 1 Ex.1.ME Question 14

Define a binary operation \(^*\) on the set \(\left\{ {0,{\text{1}},{\text{2}},{\text{3}},{\text{4}},{\textabc}} \right\}\) as

\(a + b = \left\{ \begin{array}{l}a + b,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}a + b < 6\\a + b - 6\,\,\,\,\,\,{\text{if }}a + b \ge 6\end{array} \right\}\)

Show that zero is the identity for this operation and each element \(a \ne 0\) of the set is invertible with \(6 - a\) being the inverse of \(a\).

 

Solution

Video Solution

 

Let \(X = \left\{ {0,{\text{1}},{\text{2}},{\text{3}},{\text{4}},{\text{5}}} \right\}\)

The operation \(^*\) is defined as \(a + b = \left\{ \begin{gathered} a + b,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}a + b < 6 \hfill \\ a + b - 6\,,\,\,\,\,{\text{if }}a + b \geqslant 6 \hfill \\ \end{gathered} \right\}\)

An element \(e \in X\) is the identity element for the operation \(^*\), if\(a*e = a = e*a\,\,\,\,\,\,\forall a \in X\)

For \(a \in X\),

\(\begin{align}a*0 = a + 0 = a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {a \in X \Rightarrow a + 0 < 6} \right]\\0*a = 0 + a = a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {a \in X \Rightarrow 0 + a < 6} \right]\end{align}\)

\(\therefore a*0 = a = 0*a\,\,\,\forall a \in X\)

Thus, 0 is the identity element for the given operation \(^*\).

An element \(a \in X\) is invertible if there exists \(b \in X\) such that \(a*b = 0 = b*a\).

i.e., \(\left\{ \begin{array}{l}a + b = 0 = b + a,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if }}a + b < 6\\a + b - 6 = 0 = b + a - 6\,\,\,\,\,\,{\text{if }}a + b \ge 6\end{array} \right\}\)

\( \Rightarrow a = - b\,\,{\text{or }}b = 6 - a\)

\(X = \left\{ {0,{\text{1}},{\text{2}},{\text{3}},{\text{4}},{\text{5}}} \right\}\) and \(a,b \in X\). Then \(a \ne - b\).

\(\therefore b = 6 - a\) is the inverse of \(a\) for all \(a \in X\).

Inverse of an element \(a \in X\), \(a \ne 0\) is \(6 - a\) i.e., \(a - 1 = 6 - a\).

Chapter 1 Ex.1.ME Question 15

Let\(A = \left\{ { - {\text{1}},0,{\text{1}},{\text{2}}} \right\}\), \(B = \left\{ { - {\text{4}}, - {\text{2}},0,{\text{2}}} \right\}\)and \(f,g:A \to B\) be functions defined by \({x^2} - x\), \(x \in A\)and \(g\left( x \right) = 2\left| {x - \frac{1}{2}} \right| - 1,\,\,x \in A\). Are \(f\) and \(g\) equal?

 

Solution

Video Solution

 

It is given that \(A = \left\{ { - {\text{1}},0,{\text{1}},{\text{2}}} \right\}\), \(B = \left\{ { - {\text{4}}, - {\text{2}},0,{\text{2}}} \right\}\)

Also, \(f,g:A \to B\) is defined by \({x^2} - x\), \(x \in A\) and \(g\left( x \right) = 2\left| {x - \frac{1}{2}} \right| - 1,\,\,x \in A\).

\(\begin{align}&f\left( { - 1} \right) = {\left( { - 1} \right)^2} - \left( { - 1} \right) = 1 + 1 = 2\\&g\left( { - 1} \right) = 2\left| {\left( { - 1} \right) - \frac{1}{2}} \right| - 1 = 2\left( {\frac{3}{2}} \right) - 1 = 3 - 1 = 2\\& \Rightarrow f\left( { - 1} \right) = g\left( { - 1} \right)\end{align}\)

\(\begin{align}&f\left( 0 \right) = {\left( 0 \right)^2} - 0 = 0\\&g\left( 0 \right) = 2\left| {0 - \frac{1}{2}} \right| - 1 = 2\left( {\frac{1}{2}} \right) - 1 = 1 - 1 = 0\\& \Rightarrow f\left( 0 \right) = g\left( 0 \right)\end{align}\)

\(\begin{align}&f\left( 1 \right) = {\left( 1 \right)^2} - 1 = 0\\&g\left( 1 \right) = 2\left| {1 - \frac{1}{2}} \right| - 1 = 2\left( {\frac{1}{2}} \right) - 1 = 1 - 1 = 0\\& \Rightarrow f\left( 1 \right) = g\left( 1 \right)\end{align}\)

\(\begin{align}&f\left( 2 \right) = {\left( 2 \right)^2} - 2 = 2\\&g\left( 2 \right) = 2\left| {2 - \frac{1}{2}} \right| - 1 = 2\left( {\frac{3}{2}} \right) - 1 = 3 - 1 = 2\\& \Rightarrow f\left( 2 \right) = g\left( 2 \right)\end{align}\)

\(\therefore f\left( a \right) = g\left( a \right)\,\,\,\,\,\,\,\forall a \in A\)

Hence, the functions \(f\) and \(g\) are equal.

Chapter 1 Ex.1.ME Question 16

Let\(A = \left\{ {1,2,3} \right\}\). Then number of relations containing \(\left( {{\text{1}},{\text{2}}} \right)\)and \(\left( {{\text{1}},{\text{3}}} \right)\)which are reflexive and symmetric but not transitive is,

A. \(1\)

B. \(2\)

C. \(3\)

D. \(4\)

 

Solution

Video Solution

 

The given set is \(A = \left\{ {1,2,3} \right\}\).

The smallest relation containing \(\left( {{\text{1}},{\text{2}}} \right)\) and \(\left( {{\text{1}},{\text{3}}} \right)\) which are reflexive and symmetric but not transitive is given by,

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

This is because relation \(R\) is reflexive as \(\left\{ {\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right)} \right\} \in R\).

Relation \(R\) is symmetric as \(\left\{ {\left( {1,2} \right),\left( {2,1} \right)} \right\} \in R\,\,\,{\text{and}}\,\,\,\left\{ {\left( {1,3} \right)\left( {3,1} \right)} \right\} \in R\).

Relation \(R\) is transitive as \(\left\{ {\left( {3,1} \right),\left( {1,2} \right)} \right\} \in R\) but \(\left( {3,2} \right) \notin R\).

Now, if we add any two pairs \(\left( {{\text{3}},{\text{2}}} \right)\) and \(\left( {{\text{2}},{\text{3}}} \right)\) (or both) to relation \(R\), then relation \(R\) will become transitive.

Hence, the total number of desired relations is one.

The correct answer is A.

Chapter 1 Ex.1.ME Question 17

Let \(A = \left\{ {1,2,3} \right\}\). Then number of equivalence relations containing \(\left( {{\text{1}},{\text{2}}} \right)\) is,

A. \(1\)

B. \(2\)

C. \(3\)

D. \(4\)

 

Solution

Video Solution

 

The given set is \(A = \left\{ {1,2,3} \right\}\).

The smallest equivalence relation containing \(\left( {{\text{1}},{\text{2}}} \right)\) is given by;

\({R_1} = \left\{ {\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),\left( {1,2} \right),\left( {2,1} \right)} \right\}\)

Now, we are left with only four pairs i.e., \(\left( {{\text{2}},{\text{3}}} \right),\left( {{\text{3}},{\text{2}}} \right),\left( {{\text{1}},{\text{3}}} \right){\text{and }}\left( {{\text{3}},{\text{1}}} \right).\)

If we odd any one pair [say\(\left( {{\text{2}},{\text{3}}} \right)\)] to\({R_{\text{1}}}\), then for symmetry we must add\(\left( {{\text{3}},{\text{2}}} \right)\). Also, for transitivity we are required to add \(\left( {{\text{1}},{\text{3}}} \right)\) and \(\left( {{\text{3}},{\text{1}}} \right)\).

Hence, the only equivalence relation (bigger than \({R_1}\)) is the universal relation.

This shows that the total number of equivalence relations containing \(\left( {{\text{1}},{\text{2}}} \right)\) is two.

The correct answer is B.

Chapter 1 Ex.1.ME Question 18

Let \(f:R \to R\) be the Signum Function defined as\(f\left( x \right) = \left\{ \begin{array}{l}1,\,\,\,x > 0\\0,\,\,\,x = 0\\ - 1,\,\,\,x < 0\end{array} \right\}\) and \(g:R \to R\) be the greatest integer function given by \(g\left( x \right) = \left[ x \right]\), where \(\left[ x \right]\) is greatest integer less than or equal to \(x\). Then does \(fog\) and \(gof\) coincide in \(\left( {0,{\text{1}}} \right]\)?

 

Solution

Video Solution

 

It is given that \(f:R \to R\)be the Signum Function defined as\(f\left( x \right) = \left\{ \begin{array}{l}1,\,\,\,x > 0\\0,\,\,\,x = 0\\ - 1,\,\,\,x < 0\end{array} \right\}\)

Also\(g:R \to R\)is defined as \(g\left( x \right) = \left[ x \right]\), where \(\left[ x \right]\)is greatest integer less than or equal to \(x\).

Now let \(x \in (0,1]\),

\(\left[ x \right] = 1\)if \(x = 1\) and \(\left[ x \right] = 0\) if \(0 < x < 1\).

\(\begin{align} \therefore fog\left( x \right) &= f\left( {g\left( x \right)} \right) = f\left( {\left[ x \right]} \right) = \left\{ \begin{gathered} f\left( 1 \right),\,\,\,\,{\text{if }}x = 1 \hfill \\ f\left( 0 \right),\,\,\,\,{\text{if }}x \in \left( {0,1} \right) \hfill \\ \end{gathered} \right\} = \left\{ \begin{gathered} 1,\,\,\,{\text{if }}x = 1 \hfill \\ 0,\,\,\,{\text{if }}x \in \left( {0,1} \right) \hfill \\ \end{gathered} \right\} \hfill \\ gof\left( x \right)& = g\left( {f\left( x \right)} \right) \hfill \\ & = g\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {x > 0} \right] \hfill \\& = \left[ 1 \right] = 1 \hfill \\ \end{align} \)

Thus, when \(x \in \left( {0,1} \right)\), we have \(fog\left( x \right) = 0\)and \(gof\left( x \right) = 1\).

Hence, \(fog\) and \(gof\) does not coincide in \(\left( {0,{\text{1}}} \right]\).

Chapter 1 Ex.1.ME Question 19

Number of binary operations on the set \(\{a,b\}\)are

A. \(10\)

B. \(16\)

C. \(20\)

D. \(8\)

 

Solution

Video Solution

 

A binary operation \(^*\) on \(\{a,~b\}\)is a function from \(\left\{ {a,b} \right\} \times \left\{ {a,b} \right\} \to \left\{ {a,b} \right\}\)

i.e., \(^*\) is a function from \(\left\{ {\left( {a,a} \right),\left( {a,b} \right),\left( {b,a} \right),\left( {b,b} \right)} \right\} \to \left\{ {a,b} \right\}\)

Hence, the total number of binary operations on the set \(\{a,~b\}\) is \({2^4} = 16\).

The correct answer is B.

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