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Hint: Check if the function is one-one and onto. If yes, then write \[x\] in terms of \[f\left( x \right)\], then replace \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[f\left( x \right)\] by \[x\].

(i) Here we have to find \[{{f}^{-1}}\] for \[f:A\to B\], where \[A=\left\{ 0,-1,-3,2 \right\};B=\left\{ -9,-3,0,6 \right\}\] and \[f\left( x \right)=3x\]

We know that \[f\left( x \right)\] is invertible only when \[f\left( x \right)\] is one-one and onto.

Now, we will check \[f\left( x \right)\] for one-one and onto.

Now, we will put \[\left\{ 0,-1,-3,2 \right\}\] in \[f\left( x \right)\] which is the domain \[\left( A \right)\] of the function.

Therefore, \[f\left( 0 \right)=3\left( 0 \right)=0\]

\[f\left( -1 \right)=3\left( -1 \right)=-3\]

\[f\left( -3 \right)=3\left( -3 \right)=-9\]

\[f\left( 2 \right)=3\left( 2 \right)=6\]

Therefore, \[\left\{ 0,-3,-9,6 \right\}\] is in the range of the function.

As \[A\] have different \[f\] images in \[B\], therefore the function is one – one

Also, range \[=\] co-domain, therefore function is onto.

Therefore, to find \[{{f}^{-1}}\left( x \right)\], we will write \[x\] in terms of \[f\left( x \right)\] and then replace \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[f\left( x \right)\]by \[x\].

So, \[f\left( x \right)=3x\]

\[\dfrac{f\left( x \right)}{3}=x\]

By replacing \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[f\left( x \right)\] by \[x\], we get

\[{{f}^{-1}}\left( x \right)=\dfrac{x}{3}:B\to A\] where \[B=\left\{ -9,-3,0,6 \right\}\] and \[A=\left\{ 0,-1,-3,2 \right\}\]

(ii) Here we have to find \[{{f}^{-1}}\] for \[f:A\to B\] where \[A=\left\{ 1,3,5,7,9 \right\};B=\left\{ 0,1,9,25,49,81 \right\}\] and \[f\left( x \right)={{x}^{2}}\].

First of all, we will check \[f\left( x \right)\] for one-one and onto.

Now, we will put \[\left\{ 1,3,5,7,9 \right\}\] in \[f\left( x \right)\] which is the domain of the function.

Therefore, \[f\left( 1 \right)={{\left( 1 \right)}^{2}}=1\]

\[f\left( 3 \right)={{\left( 3 \right)}^{2}}=9\]

\[f\left( 5 \right)={{\left( 5 \right)}^{2}}=25\]

\[f\left( 7 \right)={{\left( 7 \right)}^{2}}=49\]

\[f\left( 9 \right)={{\left( 9 \right)}^{2}}=81\]

Therefore, \[\left\{ 1,9,25,49,81 \right\}\] is the range of the function.

As\[A\] have different \[f\] images in \[B\], therefore the function is one – one.

Also range \[=\] co-domain. Therefore, the function is onto.

Therefore, to find \[{{f}^{-1}}\left( x \right)\], we will write \[x\] in terms of \[f\left( x \right)\] and then replace \[f\left( x \right)\] by \[x\] and \[x\] by \[{{f}^{-1}}\left( x \right)\].

So, \[f\left( x \right)={{x}^{2}}\]

\[\sqrt{f\left( x \right)}=x\]

By replacing \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[y\] by \[x\], we get

\[{{f}^{-1}}\left( x \right)=\sqrt{x}:B\to A\] where \[B:\left\{ 0,1,9,25,49,81 \right\}\] and \[A:\left\{ 0,1,3,5,7,9 \right\}\]

Note: Students must check if the function is one-one and onto before finding the inverse of the function. One – one function means different elements of the domain have different \[f\] images in the codomain. Onto function means the range of the function should be equal to its codomain.

(i) Here we have to find \[{{f}^{-1}}\] for \[f:A\to B\], where \[A=\left\{ 0,-1,-3,2 \right\};B=\left\{ -9,-3,0,6 \right\}\] and \[f\left( x \right)=3x\]

We know that \[f\left( x \right)\] is invertible only when \[f\left( x \right)\] is one-one and onto.

Now, we will check \[f\left( x \right)\] for one-one and onto.

Now, we will put \[\left\{ 0,-1,-3,2 \right\}\] in \[f\left( x \right)\] which is the domain \[\left( A \right)\] of the function.

Therefore, \[f\left( 0 \right)=3\left( 0 \right)=0\]

\[f\left( -1 \right)=3\left( -1 \right)=-3\]

\[f\left( -3 \right)=3\left( -3 \right)=-9\]

\[f\left( 2 \right)=3\left( 2 \right)=6\]

Therefore, \[\left\{ 0,-3,-9,6 \right\}\] is in the range of the function.

As \[A\] have different \[f\] images in \[B\], therefore the function is one – one

Also, range \[=\] co-domain, therefore function is onto.

Therefore, to find \[{{f}^{-1}}\left( x \right)\], we will write \[x\] in terms of \[f\left( x \right)\] and then replace \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[f\left( x \right)\]by \[x\].

So, \[f\left( x \right)=3x\]

\[\dfrac{f\left( x \right)}{3}=x\]

By replacing \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[f\left( x \right)\] by \[x\], we get

\[{{f}^{-1}}\left( x \right)=\dfrac{x}{3}:B\to A\] where \[B=\left\{ -9,-3,0,6 \right\}\] and \[A=\left\{ 0,-1,-3,2 \right\}\]

(ii) Here we have to find \[{{f}^{-1}}\] for \[f:A\to B\] where \[A=\left\{ 1,3,5,7,9 \right\};B=\left\{ 0,1,9,25,49,81 \right\}\] and \[f\left( x \right)={{x}^{2}}\].

First of all, we will check \[f\left( x \right)\] for one-one and onto.

Now, we will put \[\left\{ 1,3,5,7,9 \right\}\] in \[f\left( x \right)\] which is the domain of the function.

Therefore, \[f\left( 1 \right)={{\left( 1 \right)}^{2}}=1\]

\[f\left( 3 \right)={{\left( 3 \right)}^{2}}=9\]

\[f\left( 5 \right)={{\left( 5 \right)}^{2}}=25\]

\[f\left( 7 \right)={{\left( 7 \right)}^{2}}=49\]

\[f\left( 9 \right)={{\left( 9 \right)}^{2}}=81\]

Therefore, \[\left\{ 1,9,25,49,81 \right\}\] is the range of the function.

As\[A\] have different \[f\] images in \[B\], therefore the function is one – one.

Also range \[=\] co-domain. Therefore, the function is onto.

Therefore, to find \[{{f}^{-1}}\left( x \right)\], we will write \[x\] in terms of \[f\left( x \right)\] and then replace \[f\left( x \right)\] by \[x\] and \[x\] by \[{{f}^{-1}}\left( x \right)\].

So, \[f\left( x \right)={{x}^{2}}\]

\[\sqrt{f\left( x \right)}=x\]

By replacing \[x\] by \[{{f}^{-1}}\left( x \right)\] and \[y\] by \[x\], we get

\[{{f}^{-1}}\left( x \right)=\sqrt{x}:B\to A\] where \[B:\left\{ 0,1,9,25,49,81 \right\}\] and \[A:\left\{ 0,1,3,5,7,9 \right\}\]

Note: Students must check if the function is one-one and onto before finding the inverse of the function. One – one function means different elements of the domain have different \[f\] images in the codomain. Onto function means the range of the function should be equal to its codomain.

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