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

Answer:

\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.

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

Answer:

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.

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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.

Answer:

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

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\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. }  

Answer:

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) \}

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\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|.

Answer:

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|

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\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]

Answer:

\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}

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\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\}.

Answer:

\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

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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.

Answer:

\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.

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\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.

Answer:

\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. }

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\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.

Answer:

\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. }

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\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)

Answer:

\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).

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\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

Answer:

\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

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\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. }

Answer:

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/ }

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\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

Answer:

\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.

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\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.

Answer:

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. }

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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.

Answer:

\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

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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

Answer:

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. }

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\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] }  

Answer:

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.

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\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] }  

Answer:

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.

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\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). }

Answer:

\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}

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\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}.

Answer:

\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}.

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\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 } }.

Answer:

\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}} }

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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.

Answer:

\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

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\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.

Answer:

\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

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\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).

Answer:

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 \}

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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\}

Answer:

(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 \}

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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)  \} 

Answer:

(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)  \}

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\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.

Answer:

\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

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\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}

Answer:

\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}.

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

Answer:

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.

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\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} .

Answer:

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\}.

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\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]}

Answer:

\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

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\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).

Answer:

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}

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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

Answer:

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.

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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 .

Answer:

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.

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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.

Answer:

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}