NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.3

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Chapter 1 Ex.1.3 Question 1

Let \(f:\left\{ {1,3,4} \right\} \to \left\{ {1,2,5} \right\}\) and \(g:\left\{ {1,2,5} \right\} \to \left\{ {1,3} \right\}\) be given by \(f = \left\{ {\left( {1,2} \right),\left( {3,5} \right),\left( {4,1} \right)} \right\}\) and \(g = \left\{ {\left( {1,3} \right),\left( {2,3} \right),\left( {5,1} \right)} \right\}\). Write down \(gof\).

Solution

The functions \(f:\left\{ {1,3,4} \right\} \to \left\{ {1,2,5} \right\}\) and \(g:\left\{ {1,2,5} \right\} \to \left\{ {1,3} \right\}\) are \(f = \left\{ {\left( {1,2} \right),\left( {3,5} \right),\left( {4,1} \right)} \right\}\) and \(g = \left\{ {\left( {1,3} \right),\left( {2,3} \right),\left( {5,1} \right)} \right\}\)

\(\begin{align} gof\left( 1 \right) = g\left[ {f\left( 1 \right)} \right] = g\left( 2 \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as\,f\left( 1 \right) = 2\,and\,g\left( 2 \right) = 3} \right]\\ gof\left( 3 \right) = g\left[ {f\left( 3 \right)} \right] = g\left( 5 \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as\,f\left( 3 \right) = 5\,and\,g\left( 5 \right) = 1} \right]\\ gof\left( 4 \right) = g\left[ {f\left( 4 \right)} \right] = g\left( 1 \right) = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as\,f\left( 4 \right) = 1\,and\,g\left( 1 \right) = 3} \right] \end{align}\)

\(\therefore gof = \left\{ {\left( {1,3} \right),\left( {3,1} \right),\left( {4,3} \right)} \right\}\)

Chapter 1 Ex.1.3 Question 2

Let \(f,g,h\)be functions from \(R{\text{ to }}R\).Show that

\(\begin{align} \left( {f + g} \right)oh = foh + goh\\ \left( {f.g} \right)oh = \left( {foh} \right).\left( {goh} \right) \end{align}\)

Solution

\(\left( {f + g} \right)oh = foh + goh\)

\[\begin{align} LHS &= \left[ {\left( {f + g} \right)oh} \right]\left( x \right)\\ &= \left( {f + g} \right)\left[ {h\left( x \right)} \right]= f\left[ {h\left( x \right)} \right] + g\left[ {h\left( x \right)} \right]\\ &= \left( {foh} \right)\left( x \right) + goh\left( x \right)\\ & = \left\{ {\left( {foh} \right) + \left( {goh} \right)} \right\}\left( x \right) = RHS \end{align}\]

\[\therefore \left\{ {\left( {f + g} \right)oh} \right\}\left( x \right) = \left\{ {\left( {foh} \right) + \left( {goh} \right)} \right\}\left( x \right)\,\,\,\,for\,all\,x \in R\]

Hence, \(\left( {f + g} \right)oh = foh + goh\)

\(\left( {f.g} \right)oh = \left( {foh} \right).\left( {goh} \right)\)

\[\begin{align} LHS& = \left[ {\left( {f.g} \right)oh} \right]\left( x \right)\\ &= \left( {f.g} \right)\left[ {h\left( x \right)} \right] = f\left[ {h\left( x \right)} \right].g\left[ {h\left( x \right)} \right]\\ &= \left( {foh} \right)\left( x \right).\left( {goh} \right)\left( x \right)\\ & = \left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right) = RHS \end{align}\]

\[\therefore \left[ {\left( {f.g} \right)oh} \right]\left( x \right) = \left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right)\,\,\,\,for\,all\,x \in R\]

Hence, \(\left( {f.g} \right)oh = \left( {foh} \right).\left( {goh} \right)\)

Chapter 1 Ex.1.3 Question 3

Find \(gof\) and \(fog\), if

(i) \(f\left( x \right) = \left| x \right|\)and \(g\left( x \right) = \left| {5x - 2} \right|\)

(ii) \(f\left( x \right) = 8{x^3}\)and \(g\left( x \right) = {x^{\frac{1}{3}}}\)

Solution

(i) \(f\left( x \right) = \left| x \right|\) and \(g\left( x \right) = \left| {5x - 2} \right|\)

\(\begin{align} \therefore gof\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {\left| x \right|} \right) = \left| {5\left| x \right| - 2} \right|\\ fog\left( x \right) &= f\left( {g\left( x \right)} \right) = f\left( {\left| {5x - 2} \right|} \right) = \left| {\left| {5x - 2} \right|} \right| = \left| {5x - 2} \right| \end{align}\)

(ii) \(f\left( x \right) = 8{x^3}\) and \(g\left( x \right) = {x^{\frac{1}{3}}}\)

\(\begin{align} \therefore gof\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {8{x^3}} \right) = {\left( {8{x^3}} \right)^{\frac{1}{3}}} = 2x\\ fog\left( x \right) &= f\left( {g\left( x \right)} \right) = f{\left( {{x^{\frac{1}{3}}}} \right)^3} = 8{\left( {{x^{\frac{1}{3}}}} \right)^3} = 8x \end{align}\)

Chapter 1 Ex.1.3 Question 4

If \(f\left( x \right) = \frac{{\left( {4x + 3} \right)}}{{\left( {6x - 4} \right)}},x \ne \frac{2}{3}\), show that \(fof\left( x \right) = x,\,\,\,for\,all\,x \ne \frac{2}{3}\). What is the reverse of \(f\)?

Solution

\[\begin{align} \left( {fof} \right)\left( x \right) &= f\left( {f\left( x \right)} \right) = f\left( {\frac{{4x + 3}}{{6x - 4}}} \right)\\ & = \frac{{4\left( {\frac{{4x + 3}}{{6x - 4}}} \right) + 3}}{{6\left( {\frac{{4x + 3}}{{6x - 4}}} \right) - 4}}\\& = \frac{{16x + 12 + 18x - 12}}{{24x + 18 - 24x + 16}}\\& = \frac{{34x}}{{34}}\\&= x \end{align}\]

\[\begin{align}& \therefore fof\left( x \right) = x\,\,\,\,\,for\,\,all\,\,x \ne \frac{2}{3}\\ & \Rightarrow\; fof = 1 \end{align}\]

Hence, the given function \(f\)is invertible and the inverse of \(f\)is \(f\)itself.

Chapter 1 Ex.1.3 Question 5

State with reason whether the following functions have inverse.

(i) \(f:\left\{ {1,2,3,4} \right\} \to \left\{ {10} \right\}\)with \(f = \left\{ {\left( {1,10} \right),\left( {2,10} \right),\left( {3,10} \right),\left( {4,10} \right)} \right\}\)

(ii) \(g:\left\{ {5,6,7,8} \right\} \to \left\{ {1,2,3,4} \right\}\)with \(g = \left\{ {\left( {5,4} \right),\left( {6,3} \right),\left( {7,4} \right),\left( {8,2} \right)} \right\}\)

(iii) \(h:\left\{ {2,3,4,5} \right\} \to \left\{ {7,9,11,13} \right\}\)with \(h = \left\{ {\left( {2,7} \right),\left( {3,9} \right),\left( {4,11} \right),\left( {5,13} \right)} \right\}\)

Solution

(i) \(f:\left\{ {1,2,3,4} \right\} \to \left\{ {10} \right\}\) with \(f = \left\{ {\left( {1,10} \right),\left( {2,10} \right),\left( {3,10} \right),\left( {4,10} \right)} \right\}\)

\(f\) is a many one function as \(f\left( 1 \right) = f\left( 2 \right) = f\left( 3 \right) = f\left( 4 \right) = 10\)

\(\therefore f\) is not one-one.

Function \(f\)does not have an inverse.

(ii) \(g:\left\{ {5,6,7,8} \right\} \to \left\{ {1,2,3,4} \right\}\)with \(g = \left\{ {\left( {5,4} \right),\left( {6,3} \right),\left( {7,4} \right),\left( {8,2} \right)} \right\}\)

\(g\) is a many one function as \(g\left( 5 \right) = g\left( 7 \right) = 4\)

\(\therefore g\) is not one-one.

Function \(g\)does not have an inverse.

(iii) \(h:\left\{ {2,3,4,5} \right\} \to \left\{ {7,9,11,13} \right\}\)with \(h = \left\{ {\left( {2,7} \right),\left( {3,9} \right),\left( {4,11} \right),\left( {5,13} \right)} \right\}\)

All distinct elements of the set \(\left\{ {2,3,4,5} \right\}\)have distinct images under \(h\).

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

\(h\) is onto since for every element \(y\)of the set \(\left\{ {7,9,11,13} \right\}\), there exists an element \(x\)in the set \(\left\{ {2,3,4,5} \right\}\), such that \(h\left( x \right) = y\).

\(h\)is a one-one and onto function.

Function \(h\) has an inverse.

Chapter 1 Ex.1.3 Question 6

Show that \(f:\left[ { - 1,1} \right] \to R\), given by \(f\left( x \right) = \frac{x}{{\left( {x + 2} \right)}}\) is one-one. Find the inverse of the function\(f:\left[ { - 1,1} \right] \to Range\,f\).

(Hint: For \(y \in Range\,f,y = f\left( x \right) = \frac{x}{{x + 2}}\), for some \(x\)in \(\left[ { - 1,1} \right]\), i,e.,\(x = \frac{{2y}}{{\left( {1 - y} \right)}}\)

Solution

\(f:\left[ { - 1,1} \right] \to R\), given by \(f\left( x \right) = \frac{x}{{\left( {x + 2} \right)}}\)

For one-one

\[\begin{align} &f\left( x \right) = f\left( y \right)\\ & \Rightarrow \; \frac{x}{{x + 2}} = \frac{y}{{y + 2}}\\ & \Rightarrow \; xy + 2x = xy + 2y\\ &\Rightarrow \;2x = 2y\\& \Rightarrow \; x = y \end{align}\]

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

It is clear that \(f:\left[ { - 1,1} \right] \to R\)is onto.

\(\therefore\) \(f:\left[ { - 1,1} \right] \to R\)is one-one and onto and therefore, the inverse of the function

\(f:\left[ { - 1,1} \right] \to R\)exists.

Let \(g:Range\,f \to \left[ { - 1,1} \right]\)be the inverse of \(f\).

Let \(y\) be an arbitrary element of range \(f\).

Since \(f:\left[ { - 1,1} \right] \to Range\,f\)is onto, we have:

\[\begin{align} &y = f\left( x \right)\,\,\,for\,\,same\,\,x \in \left[ { - 1,1} \right]\\& \Rightarrow\; y = \frac{x}{{x + 2}}\\ & \Rightarrow \;xy + 2y = x\\& \Rightarrow \;x\left( {1 - y} \right) = 2y\\ &\Rightarrow \;x = \frac{{2y}}{{1 - y}},y \ne 1 \end{align}\]

Now, let us define \(g:Range\,f \to \left[ { - 1,1} \right]\) as

\(g\left( y \right) = \frac{{2y}}{{1 - y}},y \ne 1\)

Now,

\[\begin{align}&\left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {\frac{x}{{x + 2}}} \right) = \frac{{2\left( {\frac{x}{{x + 2}}} \right)}}{{1 - \frac{x}{{x + 2}}}} = \frac{{2x}}{{x + 2 - x}} = \frac{{2x}}{2} = x\\&\left( {fog} \right)\left( x \right) = f\left( {g\left( y \right)} \right) = f\left( {\frac{{2y}}{{1 - y}}} \right) = \frac{{\frac{{2y}}{{1 - y}}}}{{\frac{{2y}}{{1 - y}} + 2}} = \frac{{2y}}{{2y + 2 - 2y}} = \frac{{2y}}{2} = y\\ &\therefore gof= {I_{\left[ { - 1,1} \right]}}\,\,\,\,\,\,\,and\,\,\,\,\,\,\,fog = {I_{Range\,f}}\\&\therefore {f^{ - 1}} &= g\\ &\Rightarrow\; {f^{ - 1}}\left( y \right) = \frac{{2y}}{{1 - y}},y \ne 1\end{align}\]

Chapter 1 Ex.1.3 Question 7

Consider \(f:R \to R\)given by \(f\left( x \right) = 4x + 3\). Show that \(f\) is invertible. Find the inverse of \(f\).

Solution

\(f:R \to R\)given by\(f\left( x \right) = 4x + 3\)

For one-one

\[\begin{align} &f\left( x \right) = f\left( y \right)\\& \Rightarrow \;4x + 3 = 4y + 3\\ & \Rightarrow \;4x = 4y\\ & \Rightarrow \;x = y \end{align}\]

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

For onto

\[\begin{align}& y \in R,\,\,\,let\,\,y = 4x + 3\\& \Rightarrow \;x = \frac{{y - 3}}{4} \in R \end{align}\]

Therefore, for any \(y \in R\), there exists \(x = \frac{{y - 3}}{4} \in R\) such that

\(f\left( x \right) = f\left( {\frac{{y - 3}}{4}} \right) = 4\left( {\frac{{y - 3}}{4}} \right) + 3 = y\)

\(\therefore f\)is onto.

Thus, f is one-one and onto and therefore, \({f^{ - 1}}\)exists.

Let us define \(g:R \to R\)by \(g\left( x \right) = \frac{{y - 3}}{4}\)

Now,

\[\begin{align}& \left( {gof} \right)\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {4x + 3} \right) = \frac{{\left( {4x + 3} \right) - 3}}{4} = x\\& \left( {fog} \right)\left( y \right) = f\left( {g\left( y \right)} \right) = f\left( {\frac{{y - 3}}{4}} \right) = 4\left( {\frac{{y - 3}}{4}} \right) + 3 = y - 3 + 3 = y\\ &\therefore gof = fog = {{\rm{I}}_R} \end{align}\]

Hence, \(f\)is invertible and the inverse of \(f\)is given by

\({f^{ - 1}}\left( y \right) = g\left( y \right) = \frac{{y - 3}}{4}\).

Chapter 1 Ex.1.3 Question 8

Consider \(f:{R_ + } \to [4,\infty )\)given by\(f\left( x \right) = {x^2} + 4\). Show that \(f\)is invertible with inverse \({f^{ - 1}}\)of given \(f\)by \({f^{ - 1}}\left( y \right) = \sqrt {y - 4} \), where \({R_ + }\) is the set of all non-negative real numbers.

Solution

\(f:{R_ + } \to [4,\infty )\) given by\(f\left( x \right) = {x^2} + 4\)

For one-one:

Let \(f\left( x \right) = f\left( y \right)\)

\(\begin{align}& \Rightarrow \;{x^2} + 4 = {y^2} + 4\\& \Rightarrow\; {x^2} = {y^2}\\ &\Rightarrow\; x = y \qquad \left[ {as\,\,x \in R} \right] \end{align}\)

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

For onto:

For \(y \in [4,\infty ),\,\,\,let\,\,y = {x^2} + 4\)

\[\begin{align}& \Rightarrow \; {x^2} = y - 4 \ge 0 \qquad \left[ {as\,\,y \ge 4} \right]\\ &\Rightarrow \; x = \sqrt {y - 4} \ge 0 \end{align}\]

Therefore, for any \(y \in R\), there exists \(x = \sqrt {y - 4} \in R\) such that

\(f\left( x \right) = f\left( {\sqrt {y - 4} } \right) = {\left( {\sqrt {y - 4} } \right)^2} + 4 = y - 4 + 4 = y\)

\(\therefore f\) is an onto function.

Thus, \(f\) is one-one and onto and therefore, \({f^{ - 1}}\)exists.

Let us define \(g:[4,\infty ) \to {R_ + }\)by

\(g\left( y \right) = \sqrt {y - 4} \)

Now, \(gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {{x^2} + 4} \right) = \sqrt {\left( {{x^2} + 4} \right) - 4} = \sqrt {{x^2}} = x\)

And \(fog\left( y \right) = f\left( {g\left( y \right)} \right) = f\left( {\sqrt {y - 4} } \right) = {\left( {\sqrt {y - 4} } \right)^2} + 4 = \left( {y - 4} \right) + 4 = y\)

\(\therefore gof = fog = {{\rm{I}}_R}\)

Hence, \(f\) is invertible and the inverse of \(f\) is given by

\({f^{ - 1}}\left( y \right) = g\left( y \right) = \sqrt {y - 4} \).

Chapter 1 Ex.1.3 Question 9

Consider \(f:{R_ + } \to [ - 5,\infty )\) given by\(f\left( x \right) = 9{x^2} + 6x - 5\). Show that \(f\)is invertible with \({f^{ - 1}}\left( y \right) = \left( {\frac{{\left( {\sqrt {y + 6} } \right) - 1}}{3}} \right)\).

Solution

\(f:{R_ + } \to [ - 5,\infty )\) given by \(f\left( x \right) = 9{x^2} + 6x - 5\)

Let \(y\) be an arbitrary element of \([ - 5,\infty )\).

Let \(y = 9{x^2} + 6x - 5\)

\[\begin{align} & \Rightarrow y = {\left( {3x + 1} \right)^2} - 1 - 5\\ & \Rightarrow y = {\left( {3x + 1} \right)^2} - 6\\ & \Rightarrow {\left( {3x + 1} \right)^2} = y + 6\\ & \Rightarrow 3x + 1 = \sqrt {y + 6} \qquad \left[ {as\,\,y \ge - 5 \Rightarrow y + 6 > 0} \right]\\& \Rightarrow x = \frac{{\sqrt {y + 6} - 1}}{3} \end{align}\]

\(\therefore f\) is onto, thereby range\(f = [ - 5,\infty )\).

Let us define \(g:[ - 5,\infty ) \to {R_ + }\,\,as\,\,g\left( y \right) = \frac{{\sqrt {y + 6} - 1}}{3}\)

We have,

\[\begin{align} \left( {gof} \right)\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {9{x^2} + 6x - 5} \right)\\&= g\left( {{{\left( {3x + 1} \right)}^2} - 6} \right)\\ &= \frac{{\sqrt {{{\left( {3x + 1} \right)}^2} - 6 + 6} - 1}}{3}\\&= \frac{{3x + 1 - 1}}{3} = x\\ \end{align}\]

And,

\[\begin{align} \left( {fog} \right)\left( y \right)& = f\left( {g\left( y \right)} \right) = f\left( {\frac{{\sqrt {y + 6} - 1}}{3}} \right)\\&= {\left[ {3\left( {\frac{{\sqrt {y + 6} - 1}}{3}} \right) + 1} \right]^2} - 6\\&= {\left( {\sqrt {y + 6} } \right)^2} - 6 = y + 6 - 6 = y \end{align}\]

\(\therefore gof = {{\rm{I}}_R}\,\,\,and\,\,\,fog = {{\rm{I}}_{[ - 5,\infty )}}\)

Hence, \(f\) is invertible and the inverse of \(f\) is given by

\({f^{ - 1}}\left( y \right) = g\left( y \right) = \frac{{\sqrt {y + 6} - 1}}{3}\).

Chapter 1 Ex.1.3 Question 10

Let \(f:X \to Y\) be an invertible function. Show that \(f\) has unique inverse.

(Hint: suppose \({g_1}\) and \({g_2}\) are two inverses of \(f\). Then for all \(y \in Y\), \(fo{g_1}\left( y \right) = {{\rm{I}}_Y}\left( y \right) = fo{g_2}\left( y \right)\). Use one-one ness of \(f\).

Solution

Let \(f:X \to Y\) be an invertible function.

Also suppose \(f\) has two inverses (\({g_1}\)and \({g_2}\))

Then, for all \(y \in Y\),

\[\begin{align}&fo{g_1}\left( y \right) = {{\rm{I}}_Y}\left( y \right) = fo{g_2}\left( y \right)\\& \Rightarrow f\left( {{g_1}\left( y \right)} \right) = f\left( {{g_2}\left( y \right)} \right)\\ & \Rightarrow {g_1}\left( y \right) = {g_2}\left( y \right) \qquad \left[ {f\,{\rm{is}}\,{\rm{invertible}} \Rightarrow f\,{\rm{is}}\,{\rm{one - one}}} \right]\\& \Rightarrow {g_1} = {g_2} \qquad \qquad \;\; \left[ {g\,{\rm{is}}\,{\rm{one - one}}} \right] \end{align}\]

Hence, \(f\) has unique inverse.

Chapter 1 Ex.1.3 Question 11

Consider \(f:\left\{ {1,2,3} \right\} \to \left\{ {a,b,c} \right\}\) given by \(f\left( 1 \right) = a,f\left( 2 \right) = b,f\left( 3 \right) = c\). Find \({\left( {{f^{ - 1}}} \right)^{ - 1}} = f\).

Solution

Function \(f:\left\{ {1,2,3} \right\} \to \left\{ {a,b,c} \right\}\)given by \(f\left( 1 \right) = a,f\left( 2 \right) = b,f\left( 3 \right) = c\)

If we define \(g:\left\{ {a,b,c} \right\} \to \left\{ {1,2,3} \right\}\)as\(g\left( a \right) = 1,g\left( b \right) = 2,g\left( c \right) = 3\)

\[\begin{align} \left( {fog} \right)\left( a \right) = f\left( {g\left( a \right)} \right) = f\left( 1 \right) = a\\ \left( {fog} \right)\left( b \right) = f\left( {g\left( b \right)} \right) = f\left( 2 \right) = b\\ \left( {fog} \right)\left( c \right) = f\left( {g\left( c \right)} \right) = f\left( 3 \right) = c \end{align}\]

And,

\[\begin{align} \left( {gof} \right)\left( 1 \right) = g\left( {f\left( 1 \right)} \right) = g\left( a \right) = 1\\ \left( {gof} \right)\left( 2 \right) = g\left( {f\left( 2 \right)} \right) = g\left( b \right) = 2\\ \left( {gof} \right)\left( 3 \right) = g\left( {f\left( 3 \right)} \right) = g\left( c \right) = 3 \end{align}\]

\(\therefore gof = {{\rm{I}}_X}\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,fog = {{\rm{I}}_Y}\) where \(X = \left\{ {\left( {1,2,3} \right)} \right\}\,\,\,{\rm{and}}\,\,\,Y = \left\{ {a,b,c} \right\}\)

Thus, the inverse of \(f\)exists and\({f^{ - 1}} = g\).

\(\therefore {f^{ - 1}}:\left\{ {a,b,c} \right\} \to \left\{ {1,2,3} \right\}\) is given by, \({f^{ - 1}}\left( a \right) = 1,{f^{ - 1}}\left( b \right) = 2,{f^{ - 1}}\left( c \right) = 3\)

We need to find the inverse of \({f^{ - 1}}\) i.e., inverse of \(g\).

If we define \(h:\left\{ {1,2,3} \right\} \to \left\{ {a,b,c} \right\}\)as \(h\left( 1 \right) = a,h\left( 2 \right) = b,h\left( 3 \right) = c\)

\[\begin{align} \left( {goh} \right)\left( 1 \right) = g\left( {h\left( 1 \right)} \right) = g\left( a \right) = 1\\ \left( {goh} \right)\left( 2 \right) = g\left( {h\left( 2 \right)} \right) = g\left( b \right) = 2\\ \left( {goh} \right)\left( 3 \right) = g\left( {h\left( 3 \right)} \right) = g\left( c \right) = 3 \end{align}\]

And,

\[\begin{align} \left( {hog} \right)\left( a \right) = h\left( {g\left( a \right)} \right) = h\left( 1 \right) = a\\ \left( {hog} \right)\left( b \right) = h\left( {g\left( b \right)} \right) = h\left( 2 \right) = b\\ \left( {hog} \right)\left( c \right) = h\left( {g\left( c \right)} \right) = h\left( 3 \right) = c \end{align}\]

\(\therefore goh = {{\rm{I}}_X}\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,hog = {{\rm{I}}_Y}\) where \(X = \left\{ {\left( {1,2,3} \right)} \right\}\,\,\,{\rm{and}}\,\,\,Y = \left\{ {a,b,c} \right\}\)

Thus, the inverse of \(g\)exists and\({g^{ - 1}} = h \Rightarrow {\left( {{f^{ - 1}}} \right)^{ - 1}} = h\).

It can be noted that \(h = f\).

Hence, \({\left( {{f^{ - 1}}} \right)^{ - 1}} = f\)

Chapter 1 Ex.1.3 Question 12

Let \(f:X \to Y\) be an invertible function. Show that the inverse of \({f^{ - 1}}\) is \(f\) i.e.,\({\left( {{f^{ - 1}}} \right)^{ - 1}} = f\).

Solution

Let \(f:X \to Y\) be an invertible function.

Then there exists a function \(g:Y \to X\) such that \(gof = {{\rm{I}}_X}\,\,\,{\rm{and}}\,\,\,fog = {{\rm{I}}_Y}\)

Here, \({f^{ - 1}} = g\)

Now, \(gof = {{\rm{I}}_X}\,\,\,{\rm{and}}\,\,\,fog = {{\rm{I}}_Y}\)

\( \Rightarrow {f^{ - 1}}of = {{\rm{I}}_X}\,\,\,{\rm{and}}\,\,\,fo{f^{ - 1}} = {{\rm{I}}_Y}\)

Hence, \({f^{ - 1}}:Y \to X\) is invertible and \({f^{ - 1}}\)is \(f\) i.e.,\({\left( {{f^{ - 1}}} \right)^{ - 1}} = f\).

Chapter 1 Ex.1.3 Question 13

If \(f:R \to R\) is given by \(f\left( x \right) = {\left( {3 - {x^3}} \right)^{\frac{1}{3}}}\), then \(fof\left( x \right)\)is:

(A) \(\frac{1}{{{x^3}}}\)

(B) \({x^3}\)

(C) \(x\)

(D) \(\left( {3 - {x^3}} \right)\)

Solution

\(f:R \to R\) is given by \(f\left( x \right) = {\left( {3 - {x^3}} \right)^{\frac{1}{3}}}\)

\(\begin{align} f\left( x \right)& = {\left( {3 - {x^3}} \right)^{\frac{1}{3}}}\\ \therefore fof\left( x \right) &= f\left( {f\left( x \right)} \right) = f\left( {{{\left( {3 - {x^3}} \right)}^{\frac{1}{3}}}} \right) = {\left[ {3 - {{\left( {{{\left( {3 - {x^3}} \right)}^{\frac{1}{3}}}} \right)}^3}} \right]^{\frac{1}{3}}}\\ &= {\left[ {3 - \left( {3 - {x^3}} \right)} \right]^{\frac{1}{3}}} = {\left( {{x^3}} \right)^{\frac{1}{3}}} = x\\ \therefore fof\left( x \right)& = x \end{align}\)

The correct answer is C.

Chapter 1 Ex.1.3 Question 14

If \(f:R - \left\{ { - \frac{4}{3}} \right\} \to R\) be a function defined as\(f\left( x \right) = \frac{{4x}}{{3x + 4}}\). The inverse of \(f\) is the map \(g:Range\,f \to R - \left\{ { - \frac{4}{3}} \right\}\)given by :

(A) \(g\left( y \right) = \frac{{3y}}{{3 - 4y}}\)

(B) \(g\left( y \right) = \frac{{4y}}{{4 - 3y}}\)

(C) \(g\left( y \right) = \frac{{4y}}{{3 - 4y}}\)

(D) \(g\left( y \right) = \frac{{3y}}{{4 - 3y}}\)

Solution

It is given that \(f:R - \left\{ { - \frac{4}{3}} \right\} \to R\) is defined as \(f\left( x \right) = \frac{{4x}}{{3x + 4}}\)

Let \(y\) be an arbitrary element of Range \(f\).

Then, there exists \(x \in R - \left\{ { - \frac{4}{3}} \right\}\)such that \(y = f\left( x \right)\).

\[\begin{align} & \Rightarrow \;y = \frac{{4x}}{{3x + 4}}\\& \Rightarrow\;3xy + 4y = 4x\\& \Rightarrow\; x\left( {4 - 3y} \right) = 4y\\& \Rightarrow \;x = \frac{{4y}}{{4 - 3y}} \end{align}\]

Define \(f:R - \left\{ { - \frac{4}{3}} \right\} \to R\) as \(g\left( y \right) = \frac{{4y}}{{4 - 3y}}\)

Now,

\[\begin{align}\left( {gof} \right)\left( x \right) &= g\left( {f\left( x \right)} \right) = g\left( {\frac{{4x}}{{3x + 4}}} \right)\\&= \frac{{4\left( {\frac{{4x}}{{3x + 4}}} \right)}}{{4 - 3\left( {\frac{{4x}}{{3x + 4}}} \right)}} = \frac{{16x}}{{12x + 16 - 12x}}\\ &= \frac{{16x}}{{16}} = x\end{align}\]

And

\[\begin{align} \left( {fog} \right)\left( x \right) &= \left( {g\left( x \right)} \right) = f\left( {\frac{{4y}}{{4 - 3y}}} \right)\\ &= \frac{{4\left( {\frac{{4y}}{{4 - 3y}}} \right)}}{{3\left( {\frac{{4y}}{{4 - 3y}}} \right) + 4}} = \frac{{16y}}{{12y + 16 - 12y}}\\& = \frac{{16y}}{{16}} = y \end{align}\]

\(\therefore gof = {{\rm{I}}_{R - \left\{ { - \frac{4}{3}} \right\}}}{\rm{ and }}fog = {{\rm{I}}_{Range\,f}}{\rm{ }}\)

Thus, \(g\) is the inverse of \(f\) i.e., \({f^{ - 1}} = g\)

Hence, the inverse of \(f\) is the map \(g:Range\,f \to R - \left\{ { - \frac{4}{3}} \right\}\), which is given by \(g\left( y \right) = \frac{{4y}}{{4 - 3y}}\).

The correct answer is \(B\).

  
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