Question 31: Let $A$ be a finite set. If $f : A \rightarrow A$ is a one-one function, show that $f$ is onto also.

$\text{Let } A = \{ a_1, a_2, a_3, \ldots a_n \}.$

In order to prove that f is onto function, we will have to show that every element in A (co-domain) has its pre-image in the domain $A$. In other words, range of $f = A.$

Since $f: A \rightarrow A$ is a one-one function. Therefore, $f(a_1), f(a_2), f(a_3), \ldots , f(a_n)$ are distinct elements of set $A$.

But, $A$ has only $n$ elements. Therefore, $A = \{ f(a_1), f(a_2), f(a_3), \ldots , f(a_n) \}$ i.e. Co-domain=Range.

Hence, $f : A \rightarrow A$ is onto.

$\\$

Question 32: Let $A$ be a finite set. If $f :A \rightarrow A$ is an onto function, show that $f$ is one-one also.

Let $A = \{ a_1, a_2, a_3, \ldots a_n \}.$

In order to prove that $f$ is a one-one function, we will have to show that $f(a_1), f(a_2), f(a_3), \ldots , f(a_n)$ are distinct elements of $A.$

Clearly, Range of $f = \{ f(a_1), f(a_2), f(a_3), \ldots , f(a_n) \}$

Since, $f:A \rightarrow A$ is an onto function. Therefore,

Range of $f= A \Rightarrow \{ f(a_1), f(a_2), f(a_3), \ldots , f(a_n) \} = A$

But $A$ is a a finite set consisting of $n$ elements. Therefore, $f(a_1), f(a_2), f(a_3), \ldots , f(a_n)$ are distinct elements of $A$. Hence, $f:A \rightarrow A$ is one-one.

$\\$

Question 33: Let $R$ be the set of real numbers. If $f : R \rightarrow R ; f(x) = x^2 \text{ and } g:R \rightarrow R; g(x) = 2x + 1.$ Then, find $fog \text{ and } gof.$  Also show that $fog \neq gof.$

Clearly; range of $f$ is a subset of domain of $g$ and range of $g$ is a subset of domain of $f$.

$\text{Hence, } fog \text{ and } gof \text{ both exist. }$

$\text{Now, } (gof) (x) = g(f(x)) = g(x^2) = 2 (x^2) + 1 = 2x^2 + 1$

$\text{and, } (fog)(x) = f(g(x)) = f(2x+1) = (2x+1)^2$

$\because 2x^2 + 1 \neq ( 2x+1)^2$

$\therefore gof \neq fog$

$\\$

$\text{Question 34: Let } f : \{ 2,3,4,5 \} \rightarrow \{ 3,4,5,9 \} \text{ and } g: \{ 3,4,5,9 \} \rightarrow \{ 7,11,15 \} \\ \\ \text{ be functions defined as } f (2) =3, f (3)=4, f (4) =f(5) =5 \\ \\ \text{ and, } g (3) =g (4) =7 \text{ and } g(5) =g(9) =11. \text{ Find gof. }$

We have, Range of $f = \{3,4,5 \}$

Clearly, it is a subset of domain of $g$. Hence, $gof$ exists and $gof : \{ 2, 3, 4,5 \} \rightarrow \{ 7, 11, 15 \}$ such that

$gof (2)=g(f(2)) = g(3) = 7; gof(3) = g(f(3)) = g(4) = 7$

$gof (4) = g(f(4)) = g(5) = 11 \text{ and } gof(5) = g(f(5)) = g(5) = 11$

$\text{Hence, } \\ \\ gof : \{ 2, 3,4,5 \} \rightarrow \{7,11,15 \} \text{ such that } gof = \{ (2,7),(3,7),(4,11), (5,11) \}$

$\\$

$\text{Question 35: Find } gof \text{ and fog, if } f :R \rightarrow R \text{ and } g:R \rightarrow R \text{ are given } \\ \\ f(x)= |x| \text{ and } g(x) = |5x-2|.$

Clearly,

$\displaystyle gof(x) = g(f(x)) = g(|x|) = \Big| 5|x|-2 \Big| = \Bigg\{ \begin{array}{rr} |5x-2|, & \text{ if } x \geq 0 \\ \\ |-5x-2|, & \text{ if } x < 0 \end{array}$

$\text{and } fog(x) = f(g(x)) = f(|5x-2|) = \Big| |5x-2|-2 \Big| = |5x-2|$

$\\$

$\displaystyle \text{Question 36: If the function } f:R \rightarrow R \text{ be given by } f(x)= x^2 + 2 \text{ and } \\ \\ g:R-\{1\} \rightarrow R \text{ be given by } g(x) = \frac{x}{x-1}. \text{ Find } fog \text{ and } gof.$                                                                                                                [CBSE 2014]

$\text{Clearly, range } f = \text{ domain } g \text{ and, range } g = \text{ domain } f. \\ \\ \text{ Hence, } fog \text{ and } gof \text{ both exist. }$

$\displaystyle \text{Now, } (fog)(x) = f(g(x)) = f \Big( \frac{x}{x-1} \Big) = \Big( \frac{x}{x-1} \Big)^2 + 2 = \frac{x^2}{(x-1)^2}+2$

$\displaystyle \text{and, } (gof) (x) = g(f(x)) = g( x^2 + 2) = \frac{x^2+2}{(x^2+2)- 1} = \frac{x^2+2}{x^2+1}$

$\text{Hence, } gof:R \rightarrow R \text{ and } fog:R - \{1\} \rightarrow R \text{ are given by: }$

$\displaystyle (fog)(x)= \frac{x^2}{(x-1)^2}+2 \text{ and } (gof) (x)= \frac{x^2+2}{x^2+1}$

$\\$

$\displaystyle \text{Question 37: If } f:R - \Big\{ \frac{7}{5} \Big\} \rightarrow R - \Big\{ \frac{3}{5} \Big\} \text{ be defined as } f(x) = \frac{3x+4}{5x-7} \\ \\ \text{ and } g:R - \Big\{ \frac{3}{5} \Big\} \rightarrow R - \Big\{ \frac{7}{5} \Big\} \text{ be defined as } g(x) = \frac{7x+4}{5x-3}. \text{ Show that } \\ \\ gof = I_A \text{ and } fog = I_B, \text{ where } B = R - \Big\{ \frac{3}{5} \Big\} \text{ and } A = R - \Big\{ \frac{7}{5} \Big\}.$

$\displaystyle \text{It is given that } f:A \rightarrow A \text{ and } g:B \rightarrow A. \\ \\ \text{Therefore , } gof:A \rightarrow A \text{ and } fog : B \rightarrow B.$

$\displaystyle gof(x) = g(f(x)) = g \Big( \frac{3x+4}{5x-7} \Big) = \frac{7( \frac{3x+4}{5x-7} )+4 }{5( \frac{3x+4}{5x-7} )-3 } \\ \\ \\ { \hspace{5.5cm} = \frac{21x+28+20x-28}{15x+20-15x+21} = \frac{41x}{41} = x }$

$\displaystyle \text{Therefore } gof: A \rightarrow A \text{ is such that } gof (x) = x \text{ for all } x \in A. \\ \\ \text{ Hence, } gof = I_A$

$\displaystyle fog(x) = f(g(x)) = f \Big( \frac{7x+4}{5x-3} \Big) = \frac{3( \frac{7x+4}{5x-3} )+4 }{5( \frac{7x+4}{5x-3} )-7 } \\ \\ \\ { \hspace{5.5cm} = \frac{21x+12+20x-12}{35x+20-35x+21} = \frac{41x}{41} = x }$

$\displaystyle \text{Therefore } fog: B \rightarrow B \text{ is such that }fog (x) = x \text{ for all } x \in B. \\ \\ \text{ Hence, } fog = I_B$

$\\$

Question 38: If $f ,g : R \rightarrow R$ are defined respectively by $f(x)=x^2 + 3x+1, g(x) = 2x-3$, find $\text{(i)} \ fog \ \text{(ii)} \ gof \ \text{(iii)} \ fof \ \text{(iv)} \ gog.$

$\text{Clearly, Range } f=\text{ Domain } g \text{ and, } \text{ Range } g =\text{ Domain } f. \\ \text{Therefore, } fog, gof, fof \text{ and } gog \text{ all exist. }$

$\text{(i) For any } x \in R, \text{ we have }$

$(fog) (x) = f (g(x)) = f (2x - 3) = (2x - 3)^2 + 3(2x- 3) +1 = 4x^2 -6x +1$

$\text{Therefore, } fog:R \rightarrow R \text{ is defined by } (fog)(x) = 4x^2 -6x+1 \text{ for all } x \in R.$

$\text{(ii) For any } x \in R, \text{ we have }$

$(gof)(x) = g(f(x)) = g( x^2 +3x+1) = 2(x^2 + 3x+1) -3 = 2x^2 +6x-1$

$\text{Therefore, } gof : R \rightarrow R \text{ is defined by } (gof)(x)= 2x^2 + 6x -1 \text{ for all } x \in R$

$\text{(iii) For any } x \in R, \text{ we have }$

$(fog)(x) = f (f (x)) = f (x^2+3x+1) = (x^2+3x+1)^2+ 3(x^2+ 3x+1)+1$

$\Rightarrow (fof)(x) = x^4 +6x^3 +14x^2 +15x+5$

$\text{Therefore, } fof : R \rightarrow R \text{ is defined by } (fof)(x)=x^4+6x^3+14x^2+15x+5 \\ \text{ for all } x \in R$

$\text{(iv) For any } x \in R, \text{ we have }$

$(gog)(x) = g(g(x)) = g(2x-3) = 2(2x-3)-3 = 4x-9$

$\text{Therefore, } gog: R \rightarrow R \text{ is defined by } (gog) (x) =4x -9.$

$\\$

$\text{Question 39: Let } f :Z \rightarrow Z \text{ be defined by } f (x) = x + 2. \\ \\ \text{ Find } g :Z \rightarrow Z \text{ such that } gof = I_Z.$

$\text{We have, } gof = I_Z$

$\Rightarrow gof(x) =I_Z(x) \text{ for all } x \in Z$

$\Rightarrow g(f(x)) = x \text{ for all } x \in Z$

$\Rightarrow g(x+2)=x \text{ for all } x \in Z$

$\Rightarrow g(y) =y-2 \text{ for all } y \in Z, \text{ where } x+2=y$

$\Rightarrow g(x) = x -2 \text{ for all } x \in Z$

$\text{Hence, } g: Z \rightarrow Z \text{ defined by } g (x) = x - 2 \text{ for all } x \in Z, \\ \text{is the required function. }$

$\\$

$\text{Question 40: Let } f :R \rightarrow R \text{ be defined by } f (x) = 2x. \\ \\ \text{Find } g :R \rightarrow R \text{ such that } gof = I_R.$

$\text{We have, } gof = I_R$

$\Rightarrow gof(x) =I_R(x) \text{ for all } x \in R$

$\Rightarrow g(f(x)) = x \text{ for all } x \in R$

$\Rightarrow g(2x)=x \text{ for all } x \in R$

$\displaystyle \Rightarrow g(y) =\frac{y}{2} \text{ for all } y \in R, \text{ where } 2x=y$

$\displaystyle \Rightarrow g(x) = \frac{x}{2} \text{ for all } x \in R$

$\displaystyle \text{Hence, } g: R \rightarrow R \text{ defined by } g (x) = \frac{x}{2} \text{ for all } x \in R, \\ \text{is the required function. }$

$\\$

$\text{Question 41: Let } f , g \text{ and } h \text{ be functions from } R \text{ to } R. \\ \\ \text{ Show that: } \text{ (i) } (f+ g)oh=foh+ goh \text{ (ii) } (fg) oh = (foh) (goh)$

$\text{(i) Since } f, g \text{ and } h \text{ are functions from } R \text{ to } R. \text{ Therefore, }$

$(f+g) oh:R \rightarrow R \text{ and } foh + goh: R \rightarrow R$

$\text{For any } x \in R$

$((f + g) oh) (x) = (f + g) (h(x)) = f (h (x)) + g (h(x)) = foh (x) + goh(x)$

$\text{Therefore, } (f + g) oh=foh+ goh$

$\text{ (ii) Clearly, } (fg) oh: R \rightarrow R \text{ and } (foh) (goh) : R \rightarrow R \text{ such that }$

$\{(fg) oh \} (x) =(fg) (h (x)) = f (h (x)) g (h (x)) = (foh) (x) (goh) (x)$

$\{ (fg) oh \} (x) = \{ (foh) .(goh) \} (x) \text{ for all } x \in R$

$\text{Therefore } (fg) oh=(foh).(goh).$

$\\$

$\text{Question 42: Let } A =\{ x \in R : 0 \leq x \leq1 \}. \text{ If } f : A \rightarrow A \text{ is defined by }$

$\displaystyle f(x) = \Bigg\{ \begin{array}{rr} x, & \text{ if } x \in Q \\ \\ 1-x, & \text{ if } x \notin Q \end{array}$

$\text{then prove that } fof(x) = x \text{ for all } x \in A$

$\text{Let } x \in A. \text{ Then, either } x \text{ is rational or } x \text{ is irrational. Hence two cases arise. }$

$\text{Case 1: When } x \in Q$

$\text{In this case, we have } f(x) = x$

$\text{Therefore } fof(x) = f(f(x)) = f(x) = x$

$\text{Case 2: When } x \notin Q$

$\text{In this case, we have } f(x) = 1- x$

$\text{Therefore } fof(x) = f(f(x))$

$\Rightarrow fof(x) = f( 1-x)$

$\Rightarrow fof(x) = 1 - ( 1-x) = x$

$\text{Therefore,} fof(x) = x \text{ whether } x \in Q \text{ or } x \notin Q$

$\text{Hence, } fof(x) = x \text{ for all } x \in A$

$\\$

$\text{Question 43: lf } f:R \rightarrow R \text{ given by }$

$\displaystyle f(x) = \sin^2 x + \sin^2 \Big(x + \frac{\pi}{3} \Big) + \cos x \cos \Big(x + \frac{\pi}{3} \Big) \text{ for all } x \in R, \text{ and } g : R \rightarrow R$

$\displaystyle \text{be such that } g \Big(\frac{5}{4} \Big) = 1, \text{ then prove that } gof : R \rightarrow R \text{ is a constant function. }$

We have,

$\displaystyle f(x) = \sin^2 x + \sin^2 \Big(x + \frac{\pi}{3} \Big) + \cos x \cos \Big(x + \frac{\pi}{3} \Big)$

$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ 2 \sin^2 x + 2 \sin^2 \Big(x + \frac{\pi}{3} \Big) + 2 \cos x \cos \Big(x + \frac{\pi}{3} \Big) \Bigg\}$

$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ 1 - \cos 2x + 1 - \cos \Big(2x + \frac{2\pi}{3} \Big) + \cos \Big(2x + \frac{\pi}{3} \Big) + \cos \frac{\pi}{3} \Bigg\}$

$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - \cos 2x - \cos \Big(2x + \frac{2\pi}{3} \Big) + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$

$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - \Big[ \cos 2x + \cos \Big(2x + \frac{2\pi}{3} \Big) \Big] + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$

$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - 2 \cos \Big(2x + \frac{\pi}{3} \Big) \cos \frac{\pi}{3} + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$

$\displaystyle \Rightarrow f(x) = \frac{1}{2} \Bigg\{ \frac{5}{2} - \cos \Big(2x + \frac{\pi}{3} \Big) + \cos \Big(2x + \frac{\pi}{3} \Big) \Bigg\}$

$\displaystyle \Rightarrow f(x) = \frac{5}{4} \text{ for all } x \in R$

Therefore, for any $x \in R,$ we have

$\displaystyle gof(x) = g(f(x)) = g \Big( \frac{5}{4} \Big) = 1$

$\text{Thus, } gof (x) =1 \text{ for all } x \in R. \text{ Hence, } gof :R \rightarrow R \text{ is a constant function/ }$

$\\$

$\text{Question 44: Let } f : Z \rightarrow Z \text{ be defined by } f (n) = 3n \text{ for all } n \in Z \\ \\ \text{ and } g: Z \rightarrow Z \text{ be defined by }$

$\displaystyle g(n) = \Bigg\{ \begin{array}{rr} \frac{n}{3}, & \text{ if } n \text{ is a multiple of } 3 \\ \\ 0, & \text{ if } n \text{ is not a multiple of } 3 \end{array} \text{ for all } n \in Z.$

$\text{Show that } gof = I_Z \text{ and } fog \neq I_Z$

$\text{Since } f:Z \rightarrow Z \text{ and } g: Z \rightarrow Z. \text{ Therefore, } gof :Z \rightarrow Z \text{ and } fog:Z \rightarrow Z.$

$\text{For any } n \in Z, \text{ we have }$

$gof (n) = g(f (n))$

$\Rightarrow gof (n) = g(3n)$

$\displaystyle \Rightarrow gof (n) = \frac{3n}{3} = n$

$\Rightarrow gof (n) = n \text{ for all } n \in Z$

$\Rightarrow gof = I_Z$

$\text{For any } n \in Z, \text{ we have }$

$fog(n) = f(g(n))$

$\displaystyle \Rightarrow fog(n) = \Bigg\{ \begin{array}{rr} f(\frac{n}{3}), & \text{ if } n \text{ is a multiple of } 3 \\ \\ f(0), & \text{ if } n \text{ is not a multiple of } 3 \end{array}$

$\displaystyle \Rightarrow fog(n) = \Bigg\{ \begin{array}{rr} 3(\frac{n}{3}), & \text{ if } n \text{ is a multiple of } 3 \\ \\ 3 \times 0, & \text{ if } n \text{ is not a multiple of } 3 \end{array}$

$\displaystyle \Rightarrow fog(n) = \Bigg\{ \begin{array}{rr} n, & \text{ if } n \text{ is a multiple of } 3 \\ \\ 0, & \text{ if } n \text{ is not a multiple of } 3 \end{array}$

$\text{ Clearly, } fog(n) \neq n \text{ for all } n \in Z. \text{ In fact, } fog(n) = n \text{ only for multiple of } 3. \\ \\ \text{ Hence, } fog \neq I_Z.$

$\\$

$\text{Question 45: Let } f :R \rightarrow R \text{ be a function given by } f (x) = ax+b \\ \\ \text{ for all } x \in R. \text{ Find the constants } a \text{ and } b \text{ such that } fof = l_R.$

We have, $fof = I_R$

$\Rightarrow fof (x) = I_R (x) \text{ for all } x \in R$

$\Rightarrow f(f(x)) = x \text{ for all } x \in R$

$\Rightarrow f(ax+b) = x \text{ for all } x \in R$

$\Rightarrow a(ax+b)+b=x \text{ for all } x \in R$

$\Rightarrow (a^2 -1) x + ab + b = 0 \text{ for all } x \in R$

$\Rightarrow a^2-1=0 \text{ and } ab +b =0$

$\Rightarrow a \pm 1 \text{ and } b(a+1) = 0$

$\text{When } a = 1$

$b(a+1)=0 \Rightarrow 2b=0 \Rightarrow b=0$

$a=-1 \text{ and } b=0.$

$\text{When } a = -1$

$b(a+1) = 0 \text{ for all } b \in R$

$\text{Therefore, } a = -1 \text{ and } b \text{ can take any real value. }$

$\text{Hence, either } a =1 \text{ and } b = 0, \text{ or } a =-1 \text{ and } b \text{ can take any real value. }$

$\\$

Question 46: Let $f : A \rightarrow A$ be a function such that $fof = f$ .Show that $f$ is onto if and only if $f$ is one-one. Describe $f$ in this case.

$\text{We have, } fof =f$

Let $f: A \rightarrow A$ be onto. Then, we have to prove that $f$ is one-one.

$\text{Let } x, y \in A. \text{ Then, as } f : A \rightarrow A \text{ is onto there exist } \alpha, \beta \in A \text{ such that. }$

$f(\alpha) = x \text{ and } f(\beta) = y$

$\text{Now, } f(x) = f(y)$

$\Rightarrow f(f(\alpha)) = f(f(\beta))$

$\Rightarrow fof(\alpha) = fof(\beta)$

$\Rightarrow f(\alpha) = f(\beta)$

$\Rightarrow x = y$

$\text{Hence, } f \text{ is one-one. }$

$\text{Therefore, } f :A \rightarrow A \text{ is onto } \Rightarrow f :A \rightarrow A \text{ is one-one. }$

$\text{Conversely, let } f : A \rightarrow A \text{ be one-one. Then, we have to prove that } f \text{ is onto. }$

$\text{Let } y \text{ be an arbitrary element in } A. \text{ Then, }$

$fof = f$

$\Rightarrow fof(y) = f(y)$

$\Rightarrow f(f(y)) = f(y)$

$\Rightarrow f(y) = y$

$\text{Therefore, for all } y \in A, \text{ there exists } y \in A \text{ such that } f (y) = y. \\ \text{Hence, } f \text{ is onto. }$

$\text{Now, } fof = f$

$\Rightarrow fof(x) = f(x) \text{ for all } x \in A$

$\Rightarrow f(f(x)) = f(x) \text{ for all } x \in A$

$\Rightarrow f(\alpha) = \alpha \text{ for all } \alpha = f(x) \in A$

$\text{Therefore, } f(x) = x \text{ for all } x \in A$

$\\$

Question 47:  lf $f :R\rightarrow R \text{ and } g: R\rightarrow R$ be functions defined by $f (x)=x^2+1 \text{ and } g(x)= \sin x,$ then find $fog \text{ and } gof$

We have,

$f (x) = x^2+1 \text{ and } g(x) = \sin x$

$\text{Now , } x^2 \geq 1 \text{ for all } x \in R$

$\Rightarrow x^2+1 \geq 1 \text{ for all } x \in R$

$\Rightarrow f (x)\geq1 \text{ for all } x \in R$

$\Rightarrow \text{ Range }(f) = [1, \infty)$

$\text{Also, } -1\leq \sin x \leq 1 \text{ for all } x \in R$

$\Rightarrow \text{ Range }(g) = [-1,1]$

$\text{Clearly, } \\ \text{Range} (f ) =[1, \infty) \subseteq \text{ Domain } (g) \text{ and, Range} (g) =[-1. 1] \subseteq \text{ Domain } (f )$

$\text{Hence, } gof : R \rightarrow R \text{ and } fog: R \rightarrow R \text{ are given by }$

$gof (x) = g(f(x)) = g(x^2 +1) = \sin (x^2+1)$

$\text{and, } fog(x) = f (g(x)) =f ( \sin x) = \sin^2 x+1 \text{ respectively. }$

$\\$

$\text{Question 48: If } f (x) = e^x \text{ and } g (x) = \log_e x (x > 0), \text{ find } fog \text{ and } gof. \\ \\ \text{ Is } fog = gof? \hspace{5.0cm} \text{[CBSE 2002] }$

We observe that

$\text{Domain } (f) =R, \text{ Range }(f) =(0, \infty), \text{ Domain } (g) =(0,\infty) \\ \text{and, Range }(g) =R.$

$\text{Computation of } fog: \text{ We observe that }$

$\text{Range }( g) =\text{ Domain } (f)$

$\therefore fog \text{ exists and } fog: \text{ Domain } (g) \rightarrow R \text{ i.e. } fog: (0, \infty) \rightarrow R \text{ such that }$

$\displaystyle fog(x) = f (g(x)) \rightarrow f (\log_e x) = e^{\log_e x} = x$

$\text{Thus, } fog : (0, \infty) \rightarrow R \text{ is defined as } fog (x) = x.$

$\text{Computation of } gof : \text{ We have, }$

$\text{Range }(f) = (0, \infty) = \text{ Domain } (g)$

$\therefore gof \text{ exists and } gof: \text{ Domain } (f ) \rightarrow R \text{ i.e. } gof : R \rightarrow R \text{ such that }$

$\displaystyle gof (x) = g(f(x)) = g(e^x) = \log_e e^x = x \log_e e = x$

$\text{Thus, } gof : R \rightarrow R \text{ is defined as } gof (x) = x$

$\text{We observe that Domain } (gof) \neq \text{ Domain } (fog).$

$\therefore gof \neq fog.$

$\\$

$\text{Question 49: If } f (x)=\sqrt{x} (x > 0) \text{ and } g(x)=x^2 -1 \\ \\ \text{ are two real functions, find } fog \text{ and } gof. \text{ Is } fog = gof? \hspace{1.0cm} \text{[CBSE 2002] }$

We observe that

$\text{Domain } (f ) =[0, \infty), \text{ Range } (f ) : [0, \infty), \text{ Domain } (g) = R$

$\text{and, Range } (g) = [-1, \infty)$

$\text{Computation of } gof: \text{ We observe that : Range }(f) = [0, \infty) \subseteq \text{ Domain } (g).$

$\text{Therefore, } gof \text{ exists and } gof: [0, \infty) \rightarrow R \text{ such that }$

$gof (x) = g(f(x)) =g(\sqrt{x})=( \sqrt{x})^2 - 1 = x - 1$

$\text{Thus, } gof (x) = [0, \infty) \rightarrow R \text{ is defined as } gof(x) = x - 1$

$\text{Computation of } fog: \text{ We observe that }$

$\text{Range } (g) = [-1, \infty) \nsubseteq \text{ Domain } (f)$

$\therefore \text{Domain } (fog) = \{ x: x \in \text{ Domain } (g) \text{ and } g(x) \in \text{ Domain } (f ) \}$

$\Rightarrow \text{Domain } (fog) = \{ x: x \in R \text{ and } g (x) \in [0, \infty) \}$

$\Rightarrow \text{Domain } (fog) =\{ x: x \in R \text{ and } x^2 -1 \in [0. \infty) \}$

$\Rightarrow \text{Domain } (fog) = \{ x: x \in R \text{ and } x^2 -1 \geq 0 \}$

$\Rightarrow \text{Domain } (fog) = \{ x: x \in R \text{ and } x \leq -1 \text{ or, } x \geq 1 \}$

$\Rightarrow \text{Domain } (fog) =\{ x : x \leq-1 \text{ or } x \geq 1 \}$

$\Rightarrow \text{Domain } (fog) = (-\infty, -1] \cup [1, \infty )$

$\text{Also, } fog(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1}$

$\text{Thus, } fog: ( -\infty, -1) \cup [1, \infty) \rightarrow R \text{ is defined as } fog( x) = \sqrt{x^2 -1 }$

We find that $fog$ and $gof$ have distinct domains. Also, their formulas are not same.

$\text{Hence } fog \neq gof.$

$\\$

$\displaystyle \text{Question 50: Let} f \text{ and } g \text{ be real functions defined by } f(x) = (x){x+1} \\ \\ \text{ and } g(x) = \frac {1}{x+3}. \text{ Describe the functions } gof \text{ and } fog \text{ (if they exist). }$

$\displaystyle \text{We have } f(x) = \frac{x}{x+1} \text{ and } g(x) = \frac {1}{x+3}$

$\text{Clearly, Domain } (f) = R - \{ -1 \} \text{ and, Range } (f ) = R - \{ 1 \}$

$\text{Also, Domain } (g) = R - \{ - 3 \text{ and, Range } (g) = R - \{ 0 \}.$

$\text{Computation of } gof : \text{ We observe that }$

$\text{Range } (f) \not\subset \text{ Domain } (g)$

$\text{Therefore, Domain } (gof ) = \{ x : x \in \text{ Domain } (f) \text{ and } f (x) \in \text{ Domain } (g) \}$

$\displaystyle \Rightarrow \text{Domain } (gof ) = \Big\{ x: x \in R - \{ -1 \} \text{ and } \frac{x}{x+1} \in R - \{ -3 \} \Big\}$

$\displaystyle \Rightarrow \text{Domain } (gof ) = \Big\{ x \in R : x \neq -1 \text{ and } \frac{x}{x+1} \neq -3 \Big\}$

$\displaystyle \Rightarrow \text{Domain } (gof ) = \Big\{ x \in R : x \neq -1 \text{ and } x \neq - \frac{3}{4} \Big\}$

$\displaystyle \Rightarrow \text{Domain } (gof ) = R - \Big\{ -\frac{3}{4} , -1 \Big\}$

$\displaystyle \text{Hence, } gof : R - \Big\{ -\frac{3}{4} , -1 \Big\} \rightarrow R \text{ is defined as } gof(x) = \frac{x+1}{4x+3}$

Computation of $fog:$ We observe that: Range $(g) \not\subset \text{ Domain } (f )$

$\text{Therefore, Domain } (fog) = \{ x: x \in \text{ Domain } (g) \text{ and } g (x) \in \text{ Domain } (f ) \}$

$\displaystyle \Rightarrow \text{Domain } (fog ) = \Big\{ x: x \in R - \{ -3 \} \text{ and } \frac{1}{x+3} \in R - \{ -1 \} \Big\}$

$\displaystyle \Rightarrow \text{Domain } (fog ) = \Big\{ x : x \neq -3 \text{ and } \frac{1}{x+3} \neq -1 \Big\}$

$\displaystyle \Rightarrow \text{Domain } (fog ) = \Big\{ x : x \neq -3 \text{ and } x \neq - 4 \Big\}$

$\displaystyle \Rightarrow \text{Domain } (fog ) = \{ x \in R : x \neq -3, -4 \}$

$\displaystyle \Rightarrow \text{Domain } (fog ) = R - \{ -3 , -4 \}$

$\displaystyle \text{Also, } fog(x) = f(g(x)) = f \Big( \frac{1}{x+3} \Big) = \frac{\frac{1}{x+3}}{\frac{1}{x+3} + 1} = \frac{1}{x+4}$

$\displaystyle \text{Hence, } fog : R - \{ -3, -4 \} \rightarrow R \text{ is defined as } fog(x) = \frac{1}{x+4}$

$\\$

$\displaystyle \text{Question 51: If } f(x) = \frac{1}{2x+1}, x \neq -\frac{1}{2}, \text{ then show that } f(f(x)) = \frac{2x+1}{2x+3}, \\ \\ \text{ provided that } x \neq -\frac{1}{2}, -\frac{3}{2}.$

$\displaystyle \text{We have, } f(x) = \frac{1}{2x+1}$

$\displaystyle \text{Clearly, Domain } (f) = R - \Big\{ -\frac{1}{2} \Big\}$

$\displaystyle \text{Let } y = \frac{1}{2x+1}. \text{ Then, }$

$\displaystyle y = \frac{1}{2x+1} \Rightarrow 2x + 1 = \frac{1}{y} \Rightarrow x = \frac{1-y}{2y}$

$\displaystyle \text{Since, } x \text{ is a real number distinct from } -\frac{1}{2}. \\ \\ \text{Therefore, } y \text{ can take any non-zero real value. }$

$\displaystyle \text{Hence, Range } (f) = R - \{ 0 \}$

$\displaystyle \text{We observe that Range } (f) = R - \{ 0 \} \nsubseteq \text{ Domain } (f) = R - \Big\{ -\frac{1}{2} \Big\}$

$\displaystyle \text{Therefore, Domain } (fof) = \{ x: x \in \text{ Domain } (f) \text{ and } f (x) \in \text{ Domain } (f) \}$

$\displaystyle \Rightarrow \text{Domain } (fof) = \Bigg\{ x:x \in R - \Big\{ -\frac{1}{2} \Big\} \text{ and } f(x) \in R - \Big\{ -\frac{1}{2} \Big\} \Bigg\}$

$\displaystyle \Rightarrow \text{Domain } (fof) = \Big\{ x:x \neq -\frac{1}{2} \text{ and } f(x) \neq -\frac{1}{2} \Big\}$

$\displaystyle \Rightarrow \text{Domain } (fof) = \Big\{ x:x \neq -\frac{1}{2} \text{ and } \frac{1}{2x+1} \neq -\frac{1}{2} \Big\}$

$\displaystyle \Rightarrow \text{Domain } (fof) = \Big\{ x:x \neq -\frac{1}{2} \text{ and } x \neq -\frac{3}{2} \Big\} = R - \Big\{ -\frac{1}{2}, -\frac{3}{2} \Big\}$

$\displaystyle \text{Thus, } fof : R - \Big\{ -\frac{1}{2}, -\frac{3}{2} \Big\} \rightarrow \text{ is defined by } fof(x) = \frac{2x+1}{2x+3}.$

$\displaystyle \text{Hence, } f(f(x)) = \frac{2x+1}{2x+3} \text{ for all } x \in R , x \neq -\frac{1}{2}, -\frac{3}{2}.$

$\\$

$\displaystyle \text{Question 52: Let } f(x) = \frac{x}{\sqrt{1+x^2 } } . \text{ Then, show that } (fofof)(x) = \frac{x}{\sqrt{1+3x^2 } }.$

$\displaystyle \text{We have, } f(x) = \frac{x}{\sqrt{1+x^2 } }.$

Clearly, domain $(f ) = R.$

In order to find the range of $f$ , we proceed as follows:

Let $f(x)=y$. Then,

$\displaystyle y = f(x) \Rightarrow \frac{x}{\sqrt{1+x^2 } } = y \Rightarrow \frac{x^2}{1+x^2} = y^2 \Rightarrow x = \pm \frac{y}{\sqrt{1-y^2}}$

Since $x$ takes real values. Therefore

$1- y^2 > 0 \Rightarrow y^2 - 1 < 0 \Rightarrow y \in ( -1 , 1).$

$\text{Hence, Range } (f) = ( - 1, 1)$

$\text{Clearly, Range } (f) \subset \text{ Domain } f. \\ \\ \text{Therefore, } fof :R \rightarrow R \text{ and } fofof :R \rightarrow R.$

$\text{Now, } (fofof)(x) = ((fof)of) (x) = ( fof) (f(x))$

$\displaystyle \Rightarrow (fofof)(x) = (fof) \Bigg( \frac{x}{\sqrt{1+x^2 } } \Bigg) = f \Bigg( f \Big( \frac{x}{\sqrt{1+x^2 } } \Big) \Bigg)$

$\displaystyle \Rightarrow (fofof)(x) = f \Bigg( \frac{\frac{x}{\sqrt{1+x^2 } }}{\sqrt{1+\frac{x^2}{1+x^2}}} \Bigg) = f \Big( \frac{x}{\sqrt{1+2x^2}} \Big) \\ \\ \\ { \hspace{3.5cm} = \frac{\frac{x}{\sqrt{1+2x^2 } }}{\sqrt{1 + \frac{x^2}{1+2x^2}}} = \frac{x}{\sqrt{1+3x^2}} }$

$\\$

Question 53: Let $f$ be a real function defined by $f (x)=\sqrt{x-1}.$ Find $(fofof)(x).$ Also, show that $fof \neq f^2.$

$\text{We have, } f(x) = \sqrt{x-1}$

$\text{Clearly, Domain } (f) = [1, \infty) \text{ and Range } (f) = [0, \infty)$

$\text{We observe that range } (f ) \text{ is not a subset of domain of } f$

$\text{Domain } (fof) = \{ x : x \in \text{ Domain } (f) \text{ and } f(x) \in \text{ Domain } (f) \}$

$= \{ x : x \in [1, \infty) \text{ and } \sqrt{x-1} \in [1, \infty) \}$

$= \{ x : x \in [1, \infty) \text{ and } \sqrt{x-1} \geq 1 \}$

$= \{ x : x \in [1, \infty) \text{ and } x \geq 2 \}= [2, \infty)$

$\text{Clearly, Range } (f) = [ 0, \infty) \not\subset \text{ Domain } (fof)$

$\text{Therefore, Domain } ( ( fof)of) = \{ x: x \in \text{ Domain } (f) \text{ and } f(x) \in \text{ Domain } (fof) \}$

$= \{ x: x \in [1, \infty) \text{ and } f(x) \in [2, \infty) \}$

$= \{ x: x \in [1, \infty) \text{ and } \sqrt{x-1} \in [2, \infty) \}$

$= \{ x: x \geq 1 \text{ and } \sqrt{x-1} \geq 2 \}$

$= \{ x :x \geq 1 \text{ and } x-1 \geq 4 \}$

$\Rightarrow \text{Domain } ((fof)of) - \{ x \geq 1 \text{ and } x \geq 5 \} = [ 5, \infty )$

$\text{Now, } (fof)(x) = f(f(x)) = f(\sqrt{x-1} ) = \sqrt{\sqrt{x-1}-1}$

$\text{and } (fofof)(x) = ((fof)of)(x)$

$= (fof)(f(x))$

$= (fof)(\sqrt{x-1})$

$= f(f(\sqrt{x-1}) )= f(\sqrt{\sqrt{x-1}-1}) = \sqrt{\sqrt{\sqrt{x-1}-1}-1}$

$\text{Thus, } fof: [2, \infty) \rightarrow R \text{ and } fofof : [5, \infty) \text{ are defined as }$

$fof(x) = \sqrt{\sqrt{x-1}-1} \text{ and } (fofof)(x) = \sqrt{\sqrt{\sqrt{x-1}-1}-1}$

$\text{Now, } f^2(x) = [f(x)]^2 = ( \sqrt{x-1})^2 = x-1$

$\text{Therefore, } f^2:[1, \infty) \rightarrow \text{ is given by } f^2(x) = x-1$

$\text{Clearly, } fof \neq f^2$

$\\$

$\displaystyle \text{Question 54: If } f(x) = \frac{x-1}{x+1}, x \neq -1, \text{ then show that } f(f(x)) = -\frac{1}{x} \\ \\ \text{ provided that } x \neq 0, -1.$

$\displaystyle \text{We have, } f(x) = \frac{x-1}{x+1}$

$\text{Clearly, } f(x) \text{ is defined for all } x \in R \text{ except } x+1 = 0 \text{ i.e. } x = - 1$

$\text{Therefore, Domain } (f) = R - \{ -1 \}$

$\text{Let us now find the range of } f.$

$\text{Let } y = f(x). \text{ Then, }$

$\displaystyle y = \frac{x-1}{x+1} \Rightarrow x = \frac{y+1}{1-y}$

$\text{As } x \text{ takes all real values other than } - 1. \\ \\ \text{Therefore, } y \text{ also takes all real values other than } 1.$

$\text{Therefore, Range } (f) = R - \{ 1 \}$

$\text{We observe that Range } (f) \not\subset \text{ Domain } (f)$

$\text{Therefore, Domain } (fof) = \{ x : x \in \text{ Domain } (f) \text{ and } f(x) \in \text{ Domain } (f) \}$

$\displaystyle = \Big\{ x : x \in R - \{-1 \} \text{ and } \frac{x-1}{x+1} \in R - \{ - 1 \} \Big\}$

$\displaystyle = \Big\{ x : x\neq -1 \text{ and } \frac{x-1}{x+1}\neq -1 \Big\}$

$= \{ x : x \neq -1 \text{ and } x \neq 0 \}$

$= R - \{ -1, 0 \}$

$\displaystyle \text{Now, } fof(x) = f(f(x)) = f \Big( \frac{x-1}{x+1} \Big) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} = \frac{-2}{2x}= -\frac{1}{x}$

$\text{Thus, } fof : R - \{ -1, 0 \} \rightarrow \text{ is defined as }$

$\displaystyle fof(x) = -\frac{1}{x} \text{ or , } f(f(x)) = -\frac{1}{x}$

$\displaystyle \text{Hence, } f(f(x)) = - \frac{1}{x} \text{ for all } x \neq 0, -1$

$\\$

$\text{Question 55: If the function } f: R \rightarrow \text{ be defined by } f(x) = x^2 + 5x + 9 , \\ \\ \text{ find } f^{-1} ( 8) \text{ and } f^{-1}(9).$

Let $f^{-1} (8) = x.$ Then,

$\displaystyle f(x) = 8 \Rightarrow x^2 + 5x + 9 = 8 \Rightarrow x = \frac{-5\pm\sqrt{21}}{2} \text{ which are in } R.$

$\displaystyle \therefore f^{-1}(8) = \Bigg\{ \frac{-5 +\sqrt{21}}{2}, \frac{-5- \sqrt{21}}{2} \Bigg\}$

$\text{Now, Let } f^{-1} (9) = x$

$\Rightarrow f(x) = 9$

$\Rightarrow x^2 + 5x + 9 = 9 \Rightarrow x^2 + 5x = 0 \Rightarrow x(x+5) = 0 \Rightarrow x = 0, -5, \text{ which are in } R$

$\text{Therefore, } f^{-1} (9) = \{ 0, -5 \}$

$\\$

Question 56:  Let $f:R \rightarrow R$ be defined as $f(x) = x^2 + 1.$ Find:

$(i) f^{-1}(5) \hspace{1.0cm} (i) f^{-1}(26) \hspace{1.0cm} (i) f^{-1}\{10, 37\}$

(i) Let $f^{-1} (-5) = x.$ Then,

$f(x) = - 5 \Rightarrow x^2 + 1 = - 5 \Rightarrow x^2 = - 6 \Rightarrow = \pm \sqrt{-6}, \text{ which is not in R. }$

(ii) Let $f^{-1} (26) = x.$ Then,

$f(x) = 26 \Rightarrow x^2 + 1 = 26 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5 \text{ which are real numbers. }$

(iii) $f^{-1}\{10, 37\} = \{ x\in R : f(x) = 10 \text{ or } f(x) = 37 \}$

$= \{ x \in R : x^2 + 1 = 10 \text{ or } x^2 + 1 = 37 \}$

$= \{ x \in R : x^2 = 9 \text{ or } x^2 = 36 \} = \{ 3, -3, 6, -6 \}$

$\\$

Question 57:  Let $S =\{ 1,2, 3\}.$ Determine whether the function $f : S \rightarrow S$ defined as below have inverse. Find $f^{-1},$ if it exists

$\text{(i) } f = \{ (1,1),(2,2), (3,3) \} \hspace{1.0cm} \text{(ii) } f = \{ (1,2), (2,1), (3, 1) \} \hspace{1.0cm} \\ \\ \text{(iii) } f = \{ (1, 3), (3, 2), (2, 1) \}$

(i) Clearly, $f : S \rightarrow S$ is a bijection. Hence, $f$ is invertible and its inverse is given by $f^{-1} =\{ (1,1),(2,2), (3,3) \}.$

(ii) Clearly, $f (2) = f ( 3) = 1 .$ Therefore, $f$ is many-one and hence it is not invertible.

(iii) Clearly, $f : S \rightarrow S$ is a bijection and hence invertible. The inverse of $f$ is given by $f^{-1} = \{ (3,1), (2, 3), (1, 2) \}$

$\\$

$\text{Question 58: Consider } f :\{1,2,3\} \rightarrow \{ a,b, c \} \text{ given by } \\ \\ f (1) =a, f (2)=b \text{ and } f (3) =c. \text{ Find the inverse } (f^{-1})^{-1} \text{ of } f^{-1}. \\ \\ \text{Show that } (f^{-1})^{-1}=f.$

$\text{We have } f= \{ (1, a), (2, b), (3, c) \}$

Clearly, $f$ is a bijection and hence invertible. The inverse of $f$ is given by

$f^{-1} = \{ (a, 1), (b, 2), (c, 3) \}$

$\Rightarrow (f^{-1})^{-1} = \{ (1, a), (2, a), (3, c) \}$

$\text{Therefore } (f^{-1})^{-1}=f$

$\\$

$\text{Question 59: Let } f : N \cup \{ 0 \} \rightarrow N \cup \{0\} \text{ be defined by}$

$\displaystyle f(n) = \Bigg\{ \begin{array}{rr} n+1, & \text{ if } n \text{ is even } \\ \\ n-1, & \text{ if } n \text{ is odd } \end{array}$

Show that $f$ is invertible and $f = f^{-1}$

$\text{In order to find } f^{-1} , \text{ let } n, m \in N \cup \{0\} \text{ such that}$

$f(n) = m$

$\Rightarrow n+1 = m, \text{ if } n \text{ is even}$

$n-1 = m, \text{ if } n \text{ is odd}$

$\displaystyle \Rightarrow n = \Bigg\{ \begin{array}{rr} m-1, & \text{ if } m \text{ is odd } \\ \\ m+1, & \text{ if } m \text{ is odd } \end{array}$

$\displaystyle \Rightarrow f^{-1}(m) = \Bigg\{ \begin{array}{rr} m-1, & \text{ if } m \text{ is odd } \\ \\ m+1, & \text{ if } m \text{ is odd } \end{array}$

$\displaystyle \text{Hence, } f^{-1}(n) = \Bigg\{ \begin{array}{rr} n+1, & \text{ if } n \text{ is even } \\ \\ n-1, & \text{ if } n \text{ is odd } \end{array}$

Therefore, $f = f^{-1}.$

$\\$

Question 60:  Prove that the function $f : R \rightarrow R$ defined as $f (x) =2x - 3$ is invertible. Also, find $f^{-1}.$

In order to prove that $f$  is invertible, it is sufficient to show that $f$ is a bijection. $f \text{ is one-one: Let } x, y \in R. \text{ Then, }$

$f(x) = f(y) \Rightarrow 2x-3 = 2y - 3 \Rightarrow x = y$

$\text{Thus, } f(x) = f(y) \Rightarrow x=y \text{ for all } x, y \in R$

Hence, $f \text{ is one-one }$

$f \text{ is onto: Let } y \text{ be an arbitrary element in } R ( \text{ co-domain of } f). \text{ Then, }$

$\displaystyle f(x) = y \Rightarrow 2x-3 = y \Rightarrow x = \frac{y+3}{2}$

$\displaystyle \text{Clearly, } x = \frac{y+3}{2} \in R \text{ (domain) for all } y \in R \text{ (co-domain). } \\ \\ \text{Thus, for each } y \in R \text{ there exists } x \in R \text{ such that } f(x) = y. \\ \\ \text{ Hence, } f \text{ is onto. }$

$\text{Since } f:R \rightarrow R \text{ is one-one and onto both. } \\ \\ \text{Hence, it is a bijection and hence invertible. }$

$fof^{-1} (x) = x$

$\Rightarrow f(f^{-1}(x)) = x$

$\Rightarrow 2f^{-1}(x) - 3 = x$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{x+3}{2}$

$\displaystyle \text{Thus } f^{-1}: R \rightarrow R \text{ is given by } f(x) =\frac{x+3}{2} \text{ for all } x \in R.$

$\\$

$\displaystyle \text{Question 61: Show that } f:R - \{ -1\} \rightarrow R - \{ 1 \} \text{ given by } f(x) = \frac{x}{x+1} \\ \\ \text{ is invertible. Also, find } f^{-1} .$

In order to prove the invertibility of $f(x)$, it is sufficient to show that it is a bijection. $f$ is one-one: For any $x, y \in R - \{ -1 \}.$

$\displaystyle f(x) = f(y) \Rightarrow \frac{x}{x+1} = \frac{y}{y+1} \Rightarrow xy + x = xy + y \Rightarrow x = y$

$\text{Hence, } f \text{ is one-one. }$

$f \text{ is onto : Let } y \in R - \{ 1 \}. \text{ Then }$

$\displaystyle f(x) = y \Rightarrow \frac{x}{x+1} = y \Rightarrow x = \frac{y}{1-y}$

$\text{Clearly, } x \in R \text{ for all } y \in R - \{ 1 \}. \text{ Also, } x \neq -1. \text{ Because, }$

$\displaystyle x = - 1 \Rightarrow \frac{y}{1-y} = - 1 \Rightarrow y = -1 + y, \text{ which is not possible. }$

$\displaystyle \text{Thus, for each } y \in R - \{ 1 \} \text{ there exists } x = \frac{y}{1-y} \in R - \{ - 1 \} \text{ such that }$

$\displaystyle f(x) = \frac{x}{x+1} = \frac{ \frac{y}{1-y}}{ \frac{y}{1-y}+1} = y$

$\text{So, } f \text{ is onto. }$

$\text{Thus, } f \text{ is both one-one and onto. Consequently it is invertible. }$

Now,

$fof^{-1}(x) = x \text{ for all } x \in R - \{ 1 \}$

$\displaystyle \Rightarrow f(f^{-1}(x)) = x \Rightarrow \frac{f^{-1}(x)}{f^{-1}(x) + 1} = x \Rightarrow f^{-1}(x) = \frac{x}{1-x} \text{ for all } x \in R - \{1\}.$

$\\$

$\text{Question 62: Let } f: R \rightarrow R \text{ be defined as } f(x) = 10x + 7. \\ \\ \text{ Find the function } g: R \rightarrow R \text{ such that } gof = fog = I_R \hspace{1.0cm} \text{[CBSE 2011]}$

$\text{We have } fog = I_R$

$fog(x) = I_R(x) \text{ for all } x \in R$

$\Rightarrow f(g(x)) = x \text{ for all } x \in R$

$\Rightarrow 10g(x) + 7 = x \text{ for all } x \in R$

$\displaystyle \Rightarrow g(x) = \frac{x-7}{10} \text{ for all } x \in R$

$\\$

$\text{Question 63: If the function } f:[1, \infty ) \rightarrow [1, \infty ) \text{ defined by } f(x) = 2^{x(x-1)} \\ \\ \text{ is invertible, find } f^{-1}(x).$

It is given that f is invertible with $f^{-1}$ as its inverse.

$\displaystyle \text{Therefore } (fof^{-1}) (x) = x \text{ for all } x \in [1, \infty)$

$\Rightarrow f(f^{-1}(x)) = x$

$\displaystyle \Rightarrow 2^{f^{-1}(x)\{ f^{-1}(x)-1\}} = x$

$\displaystyle \Rightarrow f^{-1}(x) \{ f^{-1} (x)- 1 \} = \log_2 x$

$\displaystyle \Rightarrow \{ f^{-1}(x) \}^2 - f^{-1}(x) - \log_2 x = 0$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{1 \pm \sqrt{1 + 4 \log_2 x}}{2}$

$\displaystyle \because f^{-1} (x) \in [1, \infty ) \ \ \therefore f^{-1} (x) \geq 1$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{1 + \sqrt{1 + 4 \log_2 x}}{2}$

$\\$

Question 64:  Let $f :N \rightarrow Y$ be a function defined as $f (x)=4x+ 3$,where $Y =\{ y \in N : y = 4x + 3 \text{ for some } x \in N).$ Show that f is invertible. Find its inverse

In order to prove that $f$ is invertible, it is sufficient to show that it is a bijection. $f$ is one-one: For any $x, y \in N$, we find that

$f(x) = f(y) \Rightarrow 4x+3 = 4y+ 3 \Rightarrow x = y$

$\text{Hence, } f:N \rightarrow Y \text{ is one-one. }$

$f$ is onto : Let $y$ be an arbitrary element of $Y$. Then there exists $x \in N$ such that $y=4x+3$

$\Rightarrow y = f(x)$

Thus, for each $y \in N$ there exists $x \in N$ such that $f (x) = y$. So, $f : N \rightarrow Y$ is onto.

Thus, $f : N \rightarrow Y$ is both one and onto. Consequently, it is invertible. Let $f^{-1}$ be the inverse of $f$.

Then, $\displaystyle fof^{-1}(x) = x \text{ for all } x \in Y$

$\displaystyle \Rightarrow f(f^{-1}(x)) = x \text{ for all } x \in Y$

$\displaystyle \Rightarrow 4 f^{-1}(x) + 3 = x \text{ for all } x \in Y$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{x-3}{4} \text{ for all } x \in Y$

$\displaystyle \text{Hence, } f^{-1} : Y \rightarrow N \text{ is given by } f^{-1} (x) = \frac{x-3}{4} \text{ for all } x \in Y.$

$\\$

Question 65:  Let $Y = \{ n^2 : n \in N \} \subset N.$ Consider $f : N \rightarrow Y$ given by $f(n) = n^2$. Show that $f$ is invertible. Find the inverse of $ff$.

In order to prove that $f$ is invertible, it is sufficient to show that it is a bijection.

$f \text{ is one-one: For any } n,m \in N, \text{ we find that }$

$f (n)=f (m)$

$\Rightarrow n^2 = m^2$

$\because n, m \in N$

$\Rightarrow n = m$

$\text{Hence, } f :N \rightarrow Y \text{ is one-one. }$

$f \text{ is onto: Let } y \text{ be an arbitrary element of } Y. \text{ Then there exists } \\ \\ n \in N \text{ such that } y = n^2$

$\Rightarrow y = f(n)$

$\text{Thus, for each } y \in Y \text{there exists } n \in N \text{such that } y = f (n). \\ \\ \text{Hence, } f : N \rightarrow Y \text{is onto. }$

$\text{Hence, } f: N \rightarrow Y \text{is a bijection. Consequently, it is invertible. }$

$\text{Let } f^{-1} \text{denote the inverse of } f. \text{Then, }$

$fof^{-1}(x)=x \text{for all } x \in Y$

$\Rightarrow f(f^{-1}(x)) = x \text{ for all } x \in Y$

$\Rightarrow \{ f^{-1}(x) \}^2 = x \text{ for all } x \in Y$

$\Rightarrow f^{-1} (x) = \sqrt{x} \text{ for all } x \in Y$

$\text{Hence, } f^{-1} : Y \rightarrow N \text{ is given by } f^{-1}(x) = \sqrt{x} \text{ for all } x \in Y.$

$\\$

Question 66: Let $f:N \rightarrow R$ be a function defined as $f(x) = 4x^2 + 12 x + 15.$ Show that $f:N \rightarrow \text{ Range } (f)$ is invertible. Find the inverse of $f.$

In order to prove that $f$ is invertible, it is sufficient to show that $f : N \rightarrow \text{ Range } (f )$ is a bijection.

$f \text{ is one-one: } \text{For any } x, y \in N, \text{ we find that } f(x)=f(y)$

$4x^2 +12x+15=4y^2 +12y +15$

$4(x^2 -y^2) +12(x -y) =0$

$(x-y)(4x+4y + 3)=0$

$x-y=0$

$x=y$

$\text{Hence, } f:N \rightarrow \text{ Range } (f) \text{ is one-one. }$

$\text{Clearly, } f:N \rightarrow \text{ Range } (f) \text{ is onto. Hence, } f:N \rightarrow \text{ Range } (f) \text{ is invertible. }$

$\text{Let } f^{-1} \text{ denote the inverse of } f. \text{ Then, }$

$fof^{-1} (x) = x \text{ for all } x \in \text{ Range } (f)$

$\Rightarrow f(f^{-1}(x)) = x \text{ for all } x \in \text{ Range } (f)$

$\Rightarrow 4 \{ f^{-1} (x) \}^2 + 12 f^{-1} (x) + 15 = x \text{ for all } x \in \text{ Range } (f)$

$\Rightarrow 4 \{ f^{-1} (x) \}^2 + 12 f^{-1} (x) + 15 - x = 0$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{-12 \pm \sqrt{144-16(15-x)}}{8}$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{-12 \pm \sqrt{16x-96}}{8}$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{-3 \pm \sqrt{x-6}}{2}$

$\because f^{-1}(x) \in N, \therefore f^{-1}(x) > 0$

$\displaystyle \Rightarrow f^{-1}(x) = \frac{-3 + \sqrt{x-6}}{2}$