Miscellaneous Exercise Class 12 Chapter 1

It is given that \( f: R \rightarrow R \) be defined as \( f(x)=10 x+7 \)
Let \( f(x)=f(y) \), where \( x, y \in R \).
\(
\Rightarrow 10 x+7=10 y+7\)
\(\Rightarrow x=y
\)
\( \Rightarrow \mathrm{f} \) is a one - one function.
For \( \mathrm{y} \in\mathrm{R} \), let \( \mathrm{y}=10 x+7 \).
\(
\Rightarrow x=\frac{y-7}{10} \in \mathrm{R}
\)
Therefore, for any \( \mathrm{y} \in \mathrm{R} \), there exists \( x=\frac{y-7}{10} \in \mathrm{R} \) such that
\(
f(x)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y
\)
\( \Rightarrow \mathrm{f} \) is onto.
\( \Rightarrow \mathrm{f} \) is an invertible function.
Let us define \( \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} \) as \( \mathrm{g}(\mathrm{y})=\frac{y-7}{10} \)
Now, we get:
\(
\operatorname{g\circ f}(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}(10 x+7)\)
\(=\frac{(10 x+7)-7}{10}=\frac{10 x}{10}=10
\)
And,
\( f\circ g (\mathrm{y})=\mathrm{f}(\mathrm{g}(\mathrm{y}))=\mathrm{f}\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=\mathrm{y}-7+7=y \)
\( \Rightarrow g\circ f =\mathrm{IR} \) and \( \mathrm{g\circ f}=\mathrm{IR} \)
Therefore, the required function \( \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} \) is defined as \( \mathrm{g}(\mathrm{y})=\frac{y-7}{10} \).

It is given that \( \mathrm{f}: \mathrm{W} \rightarrow \mathrm{W} \) be defined as
\( \mathrm{f}(\mathrm{n})=\left\{\begin{array}{l}n-1,\text { if n is odd } \\ n+1, \text { if n is even}\end{array}\right. \)
Let \( \mathrm{f}(\mathrm{n})=\mathrm{f}(\mathrm{m}) \)
We can see that if n is odd and m is even, then we will have \( \mathrm{n}-1=\mathrm{m}+1 \).
\(
\Rightarrow \mathrm{n}-\mathrm{m}=2
\)
\( \Rightarrow \) this is impossible
Similarly, the possibility of \( n \) being even and \( m \) being odd can also be ignored under a similar argument.
Therefore, both n and m must be either odd or even.
Now, if both n and m are odd, then we get:
\(
\mathrm{f}(\mathrm{n})=\mathrm{f}(\mathrm{m})\)
\(\Rightarrow \mathrm{n}-1=\mathrm{m}-1\)
\(\Rightarrow \mathrm{n}=\mathrm{m}
\)
Again, if both n and m are even, the we get:
\(
\mathrm{f}(\mathrm{n})=\mathrm{f}(\mathrm{m})
\)
\(
\Rightarrow \mathrm{n}+1=\mathrm{m}+1\)
\(\Rightarrow \mathrm{n}=\mathrm{m}\)
\(\Rightarrow \mathrm{f} \text { is one }- \text { one. }
\)
Now, it is clear that any odd number \( 2 \mathrm{r}+1 \) 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 \( 2 r+1 \) in domain \( N \).
\( \Rightarrow \mathrm{f} \) is onto.
\( \Rightarrow \mathrm{f} \) is an invertible function.
Now, let us define \( \mathrm{g}: \mathrm{W} \rightarrow \mathrm{W} \) be defined as
\(
g(m)=\left\{\begin{array}{l}
m+1, \text { if m is even } \\
m-1, \text { if m is odd }
\end{array}\right.
\)
Now, when n is odd:
\( \operatorname{g\circ f}(\mathrm{n})=\mathrm{g}(\mathrm{f}(\mathrm{n}))=\mathrm{g}(\mathrm{n}-1)=\mathrm{n}-1+1=\mathrm{n}( \)when n is odd, then \( \mathrm{n}-1 \) is even\( ) \)
And when n is even:
\( \operatorname{g\circ f}(\mathrm{n})=\mathrm{g}(\mathrm{f}(\mathrm{n}))=\mathrm{g}(\mathrm{n}+1)=\mathrm{n}+1-1=\mathrm{n} \) (when n is even, then \( \mathrm{n}+1 \) is odd)
Similarly, when \( m \) is odd:
\(
f \circ g(m)=f(g(m))=f(m-1)=m-1+1=m
\)
And when n is even:
\(
f \circ g(m)=f(g(m))=f(m+1)=m+1-1=m
\)
Therefore, \(g\circ f =\mathrm{IW} \) and \( \mathrm{f\circ g}=\mathrm{IW} \)
Therefore, \( f \) is invertible and the inverse of \( f \) is given by \( f^{-1}=g \), which is the same as f .
Thus, the inverse of f is f itself.

It is given that \( f: R \rightarrow R \) is defined by \( f(x)=x^{2}-3 x+2 \).
\(
f(f(x))=f\left(x^{2}-3 x+2\right)\)
\(=\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2\)
\(=x^{4}+9 x^{2}+4-6 x^{3}-12 x+4 x^{2}-3 x^{2}+9 x-6+2\)
\(=x^{4}-6 x^{3}+10 x^{2}-3 x
\)

It is given that \( \mathrm{f}: \mathrm{R} \rightarrow\{x \in \mathrm{R}:-1 < x < 1\} \) defined by \( f(x)=\frac{x}{1+|X|}, x \in \mathrm{R} \)
Now, suppose that \( f(x)=f(y) \), where \( x, y \in R \).
\(
=\frac{X}{1+|X|}=\frac{Y}{1+|y|}
\)
We can see that if x is positive and y is negative, then we get:
\(
\frac{x}{1+x}=\frac{y}{1+y}=2 x y=x-y
\)
Since, \( x \) is positive, and \( y \) is negative.
Then, \( 2 x y \neq x-y \).
Thus, the case of \( x \) being positive and \( y \) being negative can be ruled out.
Similarly, \(x\) being negative and \(y\) being positive can also be ruled out.
Therefore, \(x\) and \(y\) have to be either positive or negative.
When \(x\) and \(y\) are both positive, we get:
\(
\mathrm{f}(x)=\mathrm{f}(\mathrm{y})\)
\(=\frac{x}{1+x}=\frac{y}{1+y}=x+x\mathrm{y}=\mathrm{y}+x\mathrm{y}=x=\mathrm{y}
\)
And when \(x\) and \( y\) are both negative, we get:
\( \mathrm{f}(x)=\mathrm{f}(\mathrm{y}) \)
\( =\frac{x}{1-x}=\frac{y}{1-y}=x-x y=y-x y=x=y \)
\( \Rightarrow \mathrm{f} \) is one- one.
Now, let \( y \in R \) such that \( -1 < \mathrm{y} < 1 \).
It y is negative, then there exists \( x=\frac{y}{1+y} \in \mathrm{R} \), such that
\(
\mathrm{f}(x)=\mathrm{f}\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left(\frac{y}{1-y}\right)}=\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}=\frac{y}{1-y+y}
\)
\( \Rightarrow \mathrm{f} \) is onto.
Therefore, f is one - one and onto.

Let \( f: R \rightarrow R \) given by \( f(x)=x^{3} \).
Suppose \( f(x)=f(y) \), where \( x, y \in R \).
\(
\Rightarrow x^{3}=y^{3} \quad \ldots\ldots\text{(1)}\)
Now, we need to show that \( x=y \).
Suppose \( x \neq y \), their cubes will also not be equal.
\(
\Rightarrow x^{3} \neq y^{3}
\)
However, this will be contraction to (1).
Thus, \( x=y \)
Therefore, f is injective.

Define \( \mathrm{f}: \mathrm{N} \rightarrow \mathrm{Z} \) as \( \mathrm{f}(x)=x \) and \( \mathrm{g}: \mathrm{Z} \rightarrow \mathrm{Z} \) as \( \mathrm{g}(x)=|x| \)
Now, we can see that
\(
\mathrm{G}(-1)=|-1|=1\)
\(\mathrm{g}(1)=|1|=1\)
\(\Rightarrow \mathrm{g}(-1)=\mathrm{g}(1), \text { but }-1 \neq 1
\)
\( \Rightarrow \mathrm{g} \) is not injective.
Now, \(g\circ f: N \rightarrow Z \) is defined as \( g\circ f(x)=\mathrm{g}(\mathrm{f}(x))=\mathrm{g}(x)=|x| \)
Let \( x, y \in N \) such that \( \operatorname{g\circ f}(x)=\operatorname{g\circ f}(\mathrm{y}) \).
\(\Rightarrow|x|=|y|\)
\(\Rightarrow x=y
\)
Therefore, \(g\circ f\) is injective.

It is given that \( f: N \rightarrow N \) by, \( f(x)=x+1 \)
And, \( \mathrm{g}: \mathrm{N} \rightarrow \mathrm{N} \) by,
\( \mathrm{g}(x)=\left\{\begin{array}{c}x-l\text { if } x > 1 \\ l\text { if } x=1\end{array}\right. \)
Now, consider element 1 in co-domain N. So, it is clear that this element is not an image of any of the elements in domain N.
\( \Rightarrow \mathrm{f} \) is onto.
Now, \( g\circ f:\mathrm{N} \rightarrow \mathrm{N} \) is defined as:
\(
\operatorname{g\circ f}(x)=g(f(x))=g(x+1)=(x+1)-1[x \in N= > (x+1) > 1]
\)
Then we can see that for \( y \in N \), there exists \( x=y \in N \) such that \( \operatorname{g\circ f}(x)= y\).
Therefore, \(g\circ f\) is onto.

We know that every set is a subset of itself, ARA for all \( A \in P(X) \).
\( \Rightarrow \mathrm{R} \) is reflexive.
This cannot be implied to \( \mathrm{B} \subset \mathrm{A} \).
So, if \( \mathrm{A}=\{1,2\} \) and \( \mathrm{B}=\{1,2,3\} \), then it cannot be implied that B is related to A.
\( \Rightarrow \mathrm{R} \) is not symmetric.
So, if ARB and BRC, then \( \mathrm{A} \subset \mathrm{B} \) and \( \mathrm{B} \subset \mathrm{C} \).
\( \Rightarrow \mathrm{A} \subset \mathrm{C} \)
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, R is not an equivalence relation since it is not symmetric.

It is given that \( *: \mathrm{P}(X) \times \mathrm{P}(X) \rightarrow \mathrm{P}(X) \) given by
\( \mathrm{A} * \mathrm{B}=\mathrm{A} \cap \mathrm{B} \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(X) \).
As we know that,
\(
\Rightarrow \mathrm{A} * X=\mathrm{A}=X * \mathrm{A} \forall \mathrm{A} \in \mathrm{P}(X)
\)
Thus, \(X\) is the identity element for the given binary operation \(*\).
Now, an element \( A \in P(X) \) is invertible if there exists \( B \in P(X) \) such that
\( \mathrm{A} * \mathrm{B}=X=\mathrm{B} * \mathrm{A} \) (As \(X\) is the identity element)
\( \mathrm{A} \cap \mathrm{B}=X=\mathrm{B} \cap \mathrm{A} \)
This can be possible only when \( \mathrm{A}=X=\mathrm{B} \).
Therefore, \(X\) is the only invertible element in \( \mathrm{P}(X) \) w.r.t. given operation \(*\).
Hence Proved.

Onto function from the set \( \{1,2,3, \ldots, \mathrm{n}\} \) to itself is simply a permutation on n symbols \( 1,2,3, \ldots, \mathrm{n} \).
Therefore, the total number of onto maps from \( \{1,2,3, \ldots, \mathrm{n}\} \) to itself is the same as the total number of permutations on \( n \) symbols \( 1,2,3, \ldots, n \), which is \(n !\)

It is given that \( \mathrm{S}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \) and \( \mathrm{T}=\{1,2,3\} \)
\( \mathrm{F}: \mathrm{S} \rightarrow \mathrm{T} \) is defined as:
\(
\mathrm{F}=\{(\mathrm{a}, 3),(\mathrm{b}, 2),(\mathrm{c}, 1)\}\)
\(= > \mathrm{F}(\mathrm{a})=3, \mathrm{F}(\mathrm{b})=2, \mathrm{F}(\mathrm{c})=1
\)
Therefore, \( \mathrm{F}^{-1}: \mathrm{T} \rightarrow \mathrm{S} \) is given by:
\(
\mathrm{F}^{-1}=\{(3, \mathrm{a}),(2, \mathrm{b}),(1, \mathrm{c})\}
\)

It is given that \( \mathrm{S}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \) and \( \mathrm{T}=\{1,2,3\} \)
\( \mathrm{F}: \mathrm{S} \rightarrow \mathrm{T} \) is defined as:
\( \mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\} \)
\( = > \mathrm{F}(\mathrm{b})=1, \mathrm{~F}(\mathrm{c})=1, \mathrm{~F} \) is not one-one.
Therefore, \( F \) is not invertible
\( \Rightarrow \mathrm{F}^{-1} \) does not exist.

It is given that \( *: \mathrm{R} \times \mathrm{R} \rightarrow \mathrm{R} \) and \(\circ : \mathrm{R} \times \mathrm{R} \rightarrow \mathrm{R} \) defined as \( \mathrm{a} * \mathrm{b}=|\mathrm{a}-\mathrm{b}| \) and \(\mathrm{a}\circ \mathrm{b}=\mathrm{a}, \forall \mathrm{a}, \mathrm{b} \in R \).
For \( a, b \in R \), we get:
\(
a* b=|a-b|\)
\(b*a=|b-a|=|-(a-b)|=|a-b|
\)
Therefore, \( a * b=b * a \)
\( \Rightarrow \) the operation \(*\) is commutative.
We can see that
\(
(1 * 2) * 3=(|1-2|) * 3=|1-3|=2\)
\(1 *(2 * 3)=1 *(|2-3|)=1 * 1=1
\)
Therefore, the operation \(*\) is not associative.
Now, consider the operation \(\circ\):
We can have observed that \( 1\circ 2=1 \) and \( 2 \circ 1=2 \).
\( \Rightarrow 1 \circ 2 \neq 2\circ 1 \) (where \( 1,2 \in \mathrm{R} \))
\( \Rightarrow \) the operation \(\circ\) is not commutative.
Let \( a, b, c \in R \). Then, we get:
\((a \circ b) \circ\mathrm{c}=a\circ c=\mathrm{a} \)
\( a \circ(\mathrm{b} \circ c)=\mathrm{a} \circ \mathrm{b}=\mathrm{a} \)
\( \Rightarrow(\mathrm{a} \circ \mathrm{b}) \circ\mathrm{c}=\mathrm{a} \circ(b\circ c) \)
\( \Rightarrow \) the operation \(\circ\) is associative.
Now, let \( a, b, c \in R \), then we have:
\( \mathrm{a} *(\mathrm{b} \circ c)=\mathrm{a} * \mathrm{b}=|\mathrm{a}-\mathrm{b}| \)
\( (\mathrm{a} * \mathrm{b}) \circ(\mathrm{a} * \mathrm{c})=(|\mathrm{a}-\mathrm{b}|) \circ(|\mathrm{a}-\mathrm{c}|=|\mathrm{a}-\mathrm{b}|) \)
Thus, \( a *(b \circ c )=(\mathrm{a} * \mathrm{b}) \circ(\mathrm{a} * \mathrm{c}) \)
Now,
\( 1 \circ (2 * 3)=1 \circ (|2-3|)=1 \circ 1=1 \)
\( (1 \circ 2 ) *(1 \circ 3)=1 * 1=|1-1|=0 \)
Therefore, \( 1 \circ(2 * 3) \neq(1 \circ 2) *(1 \circ 3) \) (where \(1, 2,3 \in \mathrm{R} \))
Therefore, the operation \(\circ\) does not distribute over \(*\).

It is given that \( *: \mathrm{P}(X) \times \mathrm{P}(X) \rightarrow \mathrm{P}(X) \) be defined as
\(
A * B=(A-B) \cup(B-A), A, B \in P(X)
\)
Now, let \( \mathrm{A} \in \mathrm{P}(X) \). Then, we get,
\(
\mathrm{A} * \phi =(\mathrm{A}-\phi ) \cup(\phi -\mathrm{A})=\mathrm{A} \cup \phi =\mathrm{A}\)
\(\phi * \mathrm{A}=(\phi -\mathrm{A}) \cup(\mathrm{A}-\phi )=\phi \cup \mathrm{A}=\mathrm{A}\)
\(\Rightarrow \mathrm{A} * \phi =\mathrm{A}=\phi * \mathrm{A}, \mathrm{A} \in \mathrm{P}(X)
\)
Therefore, \( \phi \) is the identity element for the given operation \(*\).
Now, an element \( A \in P(X) \) will be invertible if there exists \( B \in P(X) \) such that
\( \mathrm{A} * \mathrm{B}=\phi=\mathrm{B} * \mathrm{A} \). (as \( \phi \) is an identity element.)
Now, we can see that \( \mathrm{A} * \mathrm{A}=(\mathrm{A}-\mathrm{A}) \cup(\mathrm{A}-\mathrm{A})=\phi \cup \phi=\phi \mathrm{A} \in \mathrm{P}(X) \).
Therefore, all the element \( A \) of \( P(X) \) are invertible with \( A^{-1}=A \).

Let \( X=\{0,1,2,3,4,5\} \)
The operation \(*\) on \(X\) is defined as
\( \mathrm{a} * \mathrm{b}=\left\{\begin{array}{c}a+b,\text{ if }a+b<6 \\ a+b-6,\text{ if }a+b \geq 6\end{array}\right. \)
An element \( e \in x \) is the identity element for the operation \(*\),
If \( a * e=a=e * a \forall a \in X \)
For \( a \in X \), we can see that:
Therefore, \( o \) is the identity element for the given operation \(*\).
An element \( a \in X \) is invertible if there exists \( b \in X \) such that
\(a * b = 0 = b * a\).
\( =\left\{\begin{array} {lll} a+b =0=b+a, \text { if }a+b < 6 \\ a+b-6 =0=b+a-6, \text { if } a \geq b \geq 6\end{array}\right. \)
\( a=-b \) or \( b=6-a \)
But, \( X=\{0,1,2,3,4,5\} \) and \( \mathrm{a}, \mathrm{b} \in X \). Then, \( \mathrm{a} \neq-\mathrm{b} \).
Therefore, \( b=6-a \) is the inverse of \( a \in X \).
Thus, the inverse of an element \( a \in X, a \neq 0 \) is \( 6-a, a-1=6-a \).

It is given that \( A=\{-1,0,1,2\}, B=\{-4,-2,0,2\} \)
And also, it is given that \( f, g: A \rightarrow B \) be functions defined by \( f(x)=x^{2}- \) \( x, x \in \mathrm{A} \) and \( \mathrm{g}(x)=2\left|x-\frac{1}{2}\right|-1, x \in \mathrm{A} \)
We can see that
\(
f(-1)=(-1)^{2}-(-1)=1+1=2\)
\(g(-1)=2\left|(-1)-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2\)
\(\Rightarrow f(-1)=\mathrm{g}(-1)\)
\(\mathrm{F}(0)=(0) 2-0=0+0=0\)
\(\mathrm{g}(-1)=2\left|0-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0\)
\(\Rightarrow \mathrm{f}(0)=\mathrm{g}(0)\)
\(\mathrm{f}(1)=(1)^{2}-1=1-1=0\)
\(\mathrm{~g}(1)=2\left|1-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0\)
\(\Rightarrow \mathrm{f}(1)=\mathrm{g}(1)\)
\(\mathrm{F}(2)=(2)^{2}-2=4-2=2\)
\(\mathrm{g}(2)=2\left|2-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2\)
\(\mathrm{f}(2)=\mathrm{g}(2)
\)
Thus, \( \mathrm{f}(\mathrm{a})=\mathrm{g}(\mathrm{a}) \forall . \mathrm{a} \in \mathrm{A} \)
Therefore, the functions \( f \) and \( g \) are equal.

This is because relation \( R \) is reflexive as \( (1,1),(2,2),(3,3) \in R \).
Relation \(R\) is symmetric as \( (1,2),(2,1) \in \mathrm{R} \) and \( (1,3),(3,1) \in \mathrm{R} \).
But relation \( R \) is not transitive as \( (3,1),(1,2) \in R \) but \( (3,2) \in R \).
Now, if we add any one of the two pairs \( (3,2) \) and \( (2,3) \) (or both) to relation \( R \),
Then, relation R will become transitive.
Therefore, the total number of desired relations is one.

It is given that \( \mathrm{A}=\{1,2,3\} \).
An equivalence relation is reflexive, symmetric and transitive.
The smallest equivalence relations containing \( (1,2) \) is equal to
\(
\mathrm{R}_{1}=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}
\)
Now, only four pairs are left \( (2,3),(3,2),(1,3) \) and \( (3,1) \).
So, if we add one pair to R, then for symmetry we must add \( (3,2) \).
Also, for transitivity we required to add \( (1,3) \) and \( (3,1) \).
Thus, the only equivalence relation is the universal relation.
Therefore, the total number of equivalence relations containing \( (1,2) \) is \(2 \).

It is given that
\( f: R \rightarrow R \) be the Signum Function defined as
\( \mathrm{f}(x)=\left\{\begin{array}{c}1, x > 0 \\ 0, x=0 \\ -1, x < 0\end{array}\right. \)
Also, \( \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} \) is defined as \( \mathrm{g}(x)=[x] \), where \( [x] \) is the greatest integer less than or equal to \( x \).
Now, let \( x \in(0,1] \)
Then, we get,
\( [x]=1 \) if \( x=1 \) and \( [x]=0 \) if \( 0 < x < 1 \)
Therefore, \( \operatorname{f\circ g}(x)=f(g(x))=f([x]) \)
\( \begin{array}{l}=\left\{\begin{array}{c}f(1),\text{ if }x=1 \\ f(0),\text{ if }x \in(0,1)\end{array}=\left\{\begin{array}{c}1, \text { if } x=1 \\ 0,\text{ if }x \in(0,1)\end{array}\right.\right. \\ \operatorname{g\circ f}(x)=\mathrm{g}(\mathrm{f}(x)) \\ =\mathrm{g}(1)[x>0] \\ =[1]=1\end{array} \)
Then, when \( x \in(0,1) \), we get \( f\circ g(x)=0 \) and \( g\circ f(x)=1 \)
But \(f\circ g (1) \neq g\circ f(1)\)
Therefore, \(f\circ g\) and \(g\circ f\) do not coincide in \( (0,1] \).

A binary option \(*\) on \( \{a, b\} \) is a function from
\(
\{\mathrm{a}, \mathrm{b}\} \times\{\mathrm{a}, \mathrm{b}\} \rightarrow\{\mathrm{a}, \mathrm{b}\}
\)
i.e. y
Therefore, the total number of binary operations on the set \( \{a, b\} \) is 24 that is 16.