Exercise 1.2 class 12 maths ncert solutions

It is given that \( f: R^{*} \rightarrow R^{*} \) defined by \( f(x)=\frac{1}{2} \)
check for one-one:
For a function to be one-one, if \( f(x)=f(y) \) then \( x=y \). \( f(x)=f(y) \) \( =\frac{1}{x}=\frac{1}{y} \)
\( \Rightarrow \) Therefore, \(f\) is one \(-\) one.
We can see that \( \mathrm{y} \in \mathrm{R} \), there exists \( x=\frac{1}{y} \in \mathrm{R} \), such that \( =f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y \)
\( \Rightarrow \mathrm{f} \) is onto.
Therefore, function f is one-one and onto.
Now, let us consider \( \mathrm{g}: \mathrm{N} \rightarrow \mathrm{R}^{*} \) defined by
\(
=\mathrm{g}(x)=\frac{1}{x}
\)
Then, we get
\(
\frac{1}{x_{1}}=\frac{1}{x_{2}}\)
\(\Rightarrow x_{1}=x_{2}
\)
\( \Rightarrow \mathrm{g} \) is one-one.
It can be observed that g is not onto as for \( 1.2 \in \mathrm{R} \) there does not exist any \(x\) in N such that
\(=\mathrm{g}(x)=\frac{1}{1.2}\)
Therefore, function g is one \(-\) one but not onto.

It is given that \( f: N \rightarrow N \) given by \( f(x)=x^{2} \)
We can see that for \( x, y \in N \),
\(
f(x)=f(y)\)
\(\Rightarrow x^{2}=y^{2}\)
\(\Rightarrow x=y
\)
\( \Rightarrow \mathrm{f} \) is injective.
Now, let \( 2 \in \mathrm{N} \). But, we can see that there does not exists any \(x\) in \(N\) such that
\(f(x)=x^{2}=2\)
\( \Rightarrow \mathrm{f} \) is not surjective.
Therefore, function f is injective but not surjective

It is given that \( f: Z \rightarrow Z \) given by \( f(x)=x^{2} \)
We can see that \( \mathrm{f}(-1)=\mathrm{f}(1)=1 \), but \( -1 \neq 1 \)
\( \Rightarrow \mathrm{f} \) is not injective.
Now, let \( -2 \in \mathrm{Z} \). But, we can see that there does not exists any \(x\) in Z such that
\( \mathrm{f}(x)=x^{2}=-2 \)
\( \Rightarrow \mathrm{f} \) is not surjective.
Therefore, function f is neither injective nor surjective.

It is given that \( f: R \rightarrow R \) given by \( f(x)=x^{2} \)
We can see that \( \mathrm{f}(-1)=\mathrm{f}(1)=1 \), but \( -1 \neq 1 \)
\( \Rightarrow \mathrm{f} \) is not injective.
Now, let \( -2 \in R \). But, we can see that there does not exists any \( x \) in \( R \) such that
\( f(x)=x^{2}=-2 \)
\( \Rightarrow \mathrm{f} \) is not surjective.
Therefore, function f is neither injective nor surjective.

It is given that \( \mathrm{f}: \mathrm{N} \rightarrow \mathrm{N} \) given by \( \mathrm{f}(x)=x^{3} \)
We can see that for \( x, \mathrm{y} \in \mathrm{N} \),
\(
f(x)=f(y)\)
\(\Rightarrow x^{3}=y^{3}\)
\(\Rightarrow x=y
\)
\( \Rightarrow \mathrm{f} \) is injective.
Now, let \( 2 \in \mathrm{N} \). But, we can see that there does not exists any \(x\) in \(N\) such that
\(
f(x)=x^{3}=2
\)
\( \Rightarrow \mathrm{f} \) is not surjective.
Therefore, function \(f\) is injective but not surjective.

It is given that \( f: Z \rightarrow Z \) given by \( f(x)=x^{3} \)
We can see that for \( x, y \in N \),
\(f(x)=f(y)\)
\(\Rightarrow x^{3}=y^{3}\)
\(\Rightarrow x=y
\)
\( \Rightarrow \mathrm{f} \) is injective.
Now, let \( 2 \in \mathrm{Z} \). But, we can see that there does not exists any \(x\) in \(Z\) such that
\(f(x)=x^{3}=2\)
\( \Rightarrow \mathrm{f} \) is not surjective.
Therefore, function f is injective but not surjective.

It is given \( f: R \rightarrow R \), given by \( f(x)=[x] \)
We can see that \( \mathrm{f}(1.2)=[1.2]=1 \)
\( f(1.9)=[1.9]=1 \)
\( \Rightarrow \mathrm{f}(1.2)=\mathrm{f}(1.9) \), but \( 1.2 \neq 1.9 \).
\( \Rightarrow \mathrm{f} \) is not one- one.
Now, let us consider \( 0.6 \in \) R.
We know that \( f(x)=[x] \) is always an integer.
\( \Rightarrow \) there does not exist any element \( x \in R \) such that \( f(x)=0.6 \)
\( \Rightarrow \mathrm{f} \) is not onto.
Therefore, the greatest integer function is neither one-one nor onto.

It is given that \( f: R \rightarrow R \), given by \( f(x)=|x| \)
We can see that \( \mathrm{f}(-1)=|-1|=1, \mathrm{f}(1)=|1|=1 \)
\( \Rightarrow \mathrm{f}(-1)=\mathrm{f}(1) \), but \( -1 \neq 1 \).
\( \Rightarrow \mathrm{f} \) is not one-one.
Now, we consider \( -1 \in \mathrm{R} \).
We know that \( f(x)=|x| \) is always positive
Therefore, there doesn't exist any element \( x \) in domain \( R \) such that \( f(x)= \) \( |x|=-1 \)
\( \Rightarrow \mathrm{f} \) is not onto.
Therefore, modulus function is neither one-one nor onto.

It is given that \( f: R \rightarrow R \), given by
\( \mathrm{f}(x)=\left\{\begin{array}{cl}1, & \text { if } x > 0 \\ 0, & \text { if } x=0 \\ -1, & \text { if } x < 0\end{array}\right. \)
We can have observed that \( \mathrm{f}(1)=\mathrm{f}(2)= \) but \( 1 \neq 2 \).
Thus, \(f\) is not one \(-\) one.
Now, as \( \mathrm{f}(x) \) takes only 3 values \( (1,0,-1) \) for the element \(-2\) in co-domain \( R \), there exists any \( x \) in domain \( R \) such that \( f(x)=-2 \)
Thus, f is not onto.
Therefore, the Signum function is neither one-one nor onto.

It is given that \( \mathrm{A}=\{1,2,3\}, \mathrm{B}=\{4,5,6,7\} \)
\( \mathrm{f}: \mathrm{A} \rightarrow \mathrm{B} \) is defined as \( \mathrm{f}=\{(1,4),(2,5),(3,6)\} \)
Therefore, \( f(1)=4, f(2)=5, f(3)=6\)
We can see that the images of distinct elements of A under f are distinct.
Therefore, function f is one- one.

It is given that \( f: R \rightarrow R \) defined by \( f(x)=3-4 x \)
Let \( x_{1}, x_{2} \in R \) such that \( \mathrm{f}\left(x_{1}\right)=\mathrm{f}\left(x_{2}\right) \)
\(
\Rightarrow 3-4 x_{1}=3-4 x_{2}\)
\(\Rightarrow-4 x_{1}=-4 x_{2}\)
\(\Rightarrow x_{1}=x_{2}\)
\(\Rightarrow \mathrm{f} \text { is one- one }
\)
For any real number \((y)\) in \(R \), there exist \( \frac{3-y}{4} \) in R such that \( \mathrm{f}\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right) \)
\( \Rightarrow \mathrm{f} \) is onto.
Therefore, function \(f\) is bijective.

It is given that \( f: R \rightarrow R \) defined by \( f(x)=1+x_{2} \)
Let \( x_{1}, x_{2} \in \mathrm{R} \) such that \( \mathrm{f}\left(x_{1}\right)=\mathrm{f}\left(x_{2}\right) \)
\( =1+x_{1}^{2}=1+x_{2}^{2} \)
\( =x_{1}^{2}=x_{1}^{2} \)
\( =x_{1}= \pm x_{2} \)
Now, \( \mathrm{f}(1)=\mathrm{f}(-1)=2 \)
\( \Rightarrow \mathrm{f}(x_{1})=\mathrm{f}(x_{2}) \) which does means that \( x_{1}=x_{2}\)
\( \Rightarrow \mathrm{f} \) is not one \(-\) one
Now consider an element \(-2\) in co- domain R.
We can see that \( \mathrm{f}(x)=1+x_{2} \) is always positive.
\( \Rightarrow \) there does not exist any \( x \) in domain \( R \) such that \( f(x)=-2 \)
\( \Rightarrow \mathrm{F} \) is not onto.
Therefore, function f is neither one-one nor onto.

It is given that \( f: A \times B \rightarrow B \times A \) is defined as \( f(a, b)=(b, a) \)
Now let us consider \( \left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right) \in A \times B \)
Such that \( f\left(a_{1}, b_{1}\right)=f\left(a_{2}, b_{2}\right) \)
\( \Rightarrow\left(\mathrm{b}_{1}, \mathrm{a}_{1}\right)=\left(\mathrm{b}_{2}, \mathrm{a}_{2}\right) \)
\( \Rightarrow \mathrm{b}_{1}=\mathrm{b}_{2} \) and \( \mathrm{a}_{1}=\mathrm{a}_{2} \)
\( \Rightarrow\left(\mathrm{a}_{1}, \mathrm{~b}_{1}\right)=\left(\mathrm{a}_{2}, \mathrm{~b}_{2}\right) \)
\( \Rightarrow \mathrm{f} \) is one-one.
Now, let \( (b, a) \in B \times A \) be any element.
Then, there exists \( (a, b) \in A \times B \) such that \( f(a, b)=(b, a) \)
\( \Rightarrow \mathrm{f} \) is onto.
Therefore, f is bijective.

It is given that
f: \( \mathrm{N} \rightarrow \mathrm{N} \) be defined by
\( \mathrm{f}(\mathrm{n})=\left\{\begin{array}{c}\frac{n+1}{2}, i f_{n} \text { is odd } \\ \quad \text {for all } n \in N \\ \frac{n}{2}, i f_{n} \text { is even }\end{array}\right. \)
We can have observed that:
\( \mathrm{f}(1) \frac{1+1}{2}=1 \) and \( \mathrm{f}(2)=\frac{2}{2}=1 \) (by using the definition of f)
Thus, \( \mathrm{f}(1)=\mathrm{f}(2) \), where \( 1 \neq 2 \).
Therefore, f is not one-one.
Now, let us consider a natural number (n) in co domain N.
Case I: When n is odd.
Then, \( \mathrm{n}=2 \mathrm{r}+1 \) for some \( \mathrm{r} \in \mathrm{N} \).
\( \Rightarrow \) there exist \( 4 \mathrm{r}+1 \in \mathrm{N} \) such that \( \mathrm{f}(4 \mathrm{r}+1)=\frac{4 r+1+1}{2}=2 \mathrm{r}+1 \)
Case II: When n is even.
Then, \( \mathrm{n}=2 \mathrm{r} \) for some \( \mathrm{r} \in \mathrm{N} \).
\( \Rightarrow \) there exist \( 4 \mathrm{r} \in \mathrm{N} \) such that \( \mathrm{f}(4 \mathrm{r})=\frac{4 r}{2}=2 \mathrm{r} \)
Therefore, f is onto.
\( \Rightarrow \) Function f is not one-one but it is onto.
Thus, Function f is not bijective function.

It is given that \( \mathrm{A}=\mathrm{R}-\{3\} \) and \( \mathrm{B}=\mathrm{R}-\{1\} \)
\(F:A \rightarrow B \) defined by
\(\mathrm{f}(x)=\left(\frac{x-2}{x-3}\right)\)
Now, let \( x, \mathrm{y} \in \mathrm{A} \) such that \( \mathrm{f}(x)=\mathrm{f}(\mathrm{y}) \)
\(\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}\)
\(\Rightarrow(x-2)(y-3)=(y-2)(x-3)\)
\(\Rightarrow x y-3 x-2 y+6=x y-3 y-2 x+6\)
\(\Rightarrow-3 x-2 y=-3 y-2 x\)
\(\Rightarrow x=y\)
\(\Rightarrow f\text { is one}- \text {one. }\)
Let \( \mathrm{y} \in \mathrm{B}=\mathrm{R}-\{1\} \)
Then, \( \mathrm{y} \neq 1 \).
The function f is onto if there exist \( x \in \mathrm{A} \) such that \( \mathrm{f}(x)=\mathrm{y} \)
Now, \( \mathrm{f}(x)=\mathrm{y} \)
\(\frac{x-2}{x-3}=\mathrm{y}\)
\(\Rightarrow x-2=\mathrm{xy}-3 \mathrm{y}\)
\(\Rightarrow x(1-\mathrm{y})=-3 \mathrm{y}+2\)
\(\Rightarrow x=\frac{2-3 y}{1-y} \in \mathrm{A}\)
\( \Rightarrow \mathrm{y} \in \mathrm{B} \), there exists \( \frac{2-3 y}{1-y} \in \mathrm{A} \) such that
\(\Rightarrow\left(\frac{2-3 y}{1-y}\right)=\frac{\left(\frac{2-3 y}{1-y}\right)}{\left(\frac{2-3 y}{1-y}\right)}=\frac{2-3 y-3+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}\)
\( \Rightarrow \mathrm{f} \) is onto.
Therefore, function f is one- one and onto.

\( f: R \rightarrow R \) be defined as \( f(x)=x^{4} \).
Let \( x, y \in R \) such that \( f(x)=f(y) \)
\(
\Rightarrow x^{4}=y^{4}\)
\(\Rightarrow x=y
\)
Therefore, \( \mathrm{f}\left(x_{1}\right)=\mathrm{f}\left(x_{2}\right) \) which does not implies \( x_{1}=x_{2} \).
For instance, \( \mathrm{f}(1)=\mathrm{f}(-1)=1 \)
Therefore, f is not one-one.
Now, an element 2 in co-domain R.
We can see that there does not exist any \(x\) in domain R such that \( \mathrm{f}(x)=2 \)
Therefore, f is not onto.
Therefore, function f is neither one-one nor onto.

It is given that \( f: R \rightarrow R \) be defined as \( f(x)=3 x \).
Let \( x, \mathrm{y} \in \mathrm{R} \) such that \( \mathrm{f}(x)=\mathrm{f}(\mathrm{y}) \).
\(
\Rightarrow 3 x=3 y\)
\(\Rightarrow x=y
\)
\( \Rightarrow \mathrm{f} \) is one-one.
Also, for any real number \( (\mathrm{y}) \) in co-domain \(R \), there exists \( \frac{y}{3} \) in R such that
\(
\mathrm{f}\left(\frac{y}{3}\right)=3\left(\frac{y}{3}\right)
\)
Therefore, f is onto.
Therefore, function f is one-one and onto.