Question 1: Find gof  \text{ and } fog \text{ when } f:R \rightarrow R \text{ and } g:R \rightarrow R are defined by:   

\text{(i) } f(x) =  2x+3     \text{ and } g(x) = x^2+5

\text{(ii) }f(x) =  2x+x^2     \text{ and } g(x) =  x^3

\text{(iii) }f(x) =  x^2+8     \text{ and } g(x) = 3x^3+1

\text{(iv) }f(x) =   x    \text{ and } g(x) = |x|

\text{(v) }f(x) =  x^2+2x-3     \text{ and } g(x) = 3x-4

\text{(vi) } f(x) =   8x^3    \text{ and } g(x) = x^{1/3}

Answer:

\text{(i) } \text{Now, } f(x) =  2x+3     \text{ and } g(x) = x^2+5

gof(x) = g(2x+3) = (2x+3)^2 + 5 = 4x^2 + 12x + 14

fog(x) = f(g(x)) = f(x^2+5) = 2(x^2+5) + 3 = 2x^2 + 13

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\text{(ii) } \text{Now, }f(x) =  2x+x^2     \text{ and } g(x) =  x^3

gof(x) = g(f(x)) = g(2x+x^2) = (2x+x^2)^3

fog(x) = f(g(x)) = f(x^3) = 2x^3 + x^6

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\text{(iii) } \text{Now, }f(x) =  x^2+8     \text{ and } g(x) = 3x^3+1

gof(x) = g(f(x)) = g(x^2 + 8) = 3(x^2+8)^3 + 1

fog(x) = f(g(x)) = f( 3x^3 + 1) = ( 3x^3+1)^2 + 8 = 9x^6 + 6x^3 + 9

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\text{(iv) } \text{Now, }f(x) =   x    \text{ and } g(x) = |x|

gof(x) = g(f(x)) = g(x) = |x|

fog(x) = = f(g(x)) = f(|x|) = |x|

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\text{(v) } \text{Now, }f(x) =  x^2+2x-3     \text{ and } g(x) = 3x-4

gof(x) = g(f(x)) = g(x^2+2x-3) = 3(x^2+2x-3) - 4 = 3x^2+ 6x-13

fog(x) = f(g(x)) = f(3x-4) = (3x-4)^2 + 2(3x-4) - 3 = 9x^2 - 18x + 5

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\text{(vi) } \text{Now, }f(x) =   8x^3    \text{ and } g(x) = x^{1/3}

\displaystyle gof(x) = g(f(x)) = g(8x^3) = ( 8x^3)^{\frac{1}{3}} = [(2x)^3]^{\frac{1}{3}}  = 2x

\displaystyle fog(x) = f(g(x)) = f\Big(x^{\frac{1}{3}}\Big)  = 8\Big(x^{\frac{1}{3}}\Big)^3 = 8x

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Question 2: Let f =\{ (3,1),(9,3),(12,4) \} \text{ and } g =\{ (1, 3),(3, 3), (4,9), (5,9) \}. Show that gof \text{ and } fog are both defined. Also, find fog \text{ and } gof.

Answer:

\text{ Given function } f = \{(3, 1), (9, 3), (12, 4)\} \text{ and } g = \{(1, 3), (3, 3),(4, 9),(5, 9)\}

f: \{3, 9, 12\} \rightarrow \{1, 3, 4\} \text{ and } g: \{1, 3, 4, 5\} \rightarrow \{3, 9 \}

\text{ Co-domain of } f \text{ is a subset of the domain of } g

\text{ Therefore, } gof \text{ exists and } gof: \{3, 9, 12\} \rightarrow \{3, 9 \}

(gof) (3) = g(f(3)) = g(1) = 3

(gof) (9) = g(f(9)) = g(3) = 3

(gof) (12) = g(f(12)) = g(4) = 9

\Rightarrow gof = \{(3, 3), (9, 3), (12, 9) \}

\text{ Co-domain of } g \text{ is a subset of the domain of } f.

\text{ Therefore, } fog \text{ exists and } fog: \{1, 3, 4, 5\} \rightarrow \{3, 9, 12\}

(fog) (1) = f(g(1)) = f(3) = 1

(fog) (3) = f(g(3)) = f(3) = 1

(fog) (4) = f(g(4)) = f(9) = 3

(fog) (5) = f(g(5)) = f(9) = 3

\Rightarrow fog = \{(1, 1), (3, 1), (4, 3), (5, 3)\}

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Question 3: Let f =\{ (1,-1), (4,-2),(9,-3),(1,6,4)\} \text{ and } g= \{ (-1, -2),(-2,-4), (-3,-6),(4,8) \}. Show that gof is defined while fog is not defined. Also, find gof.

Answer:

\text{ Given as } \\ \\ f = \{(1, -1), (4, -2), (9, -3), (16, 4)\} \text{ and } g = \{(-1, -2), (-2, -4), (-3, -6), (4, 8)\}

f: \{1, 4, 9, 16\} \rightarrow \{-1, -2, -3, 4\} \text{ and } g: \{-1, -2, -3, 4\} \rightarrow \{-2, -4, -6, 8\}

\text{ Co-domain of } f = \text{ domain of } g

\text{ Therefore, } gof \text{ exists and } gof: \{1, 4, 9, 16\} \rightarrow \{-2, -4, -6, 8\}

(gof)(1) = g(f(1)) = g(-1) = -2

(gof)(4) = g(f(4)) = g(-2) = -4

(gof)(9) = g(f(9)) = g(-3) = -6

(gof)(16) = g(f(16)) = g(4) = 8

\text{ Therefore, } gof = \{(1, -2), (4, -4), (9, -6), (16, 8)\}

\text{ Since, the co-domain of } g \text{ is not same as the domain of } f.

\text{ Therefore, } fog \text{ does not exist. }

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Question 4: Let A = \{a, b, c\}, B = \{u , v, w \} and let f and g be two functions from A to B and from B to A respectively defined as: f = \{(a, v), (b, u), (c, w)\}, g = \{(u, b), (v, a), (w, c)\}. Show that f and g both are bijections and find fog and gof.

Answer:

\text{ Given as } f = \{(a, v), (b, u), (c, w)\}, g = \{(u, b), (v, a), (w, c)\}

\text{ Also given that } A = \{a, b, c\}, B = \{uv, w\}

\text{ Let us show } f \text{ and } g \text{ both are bijective }

\text{ Consider that } f = \{(a, v), (b, u), (c, w)\} \text{ and } f: A \rightarrow B

\text{ Injectivity of } f

\text{ No two elements of } A \text{ have the same image in } B

\text{ Therefore, } f \text{ is one-one }

\text{ Surjectivity of } f:

\text{ Co-domain of } f = \{uv, w\}

\text{ Range of } f = \{uv, w\}

\text{ Both are same }

\text{ Therefore, } f \text{ is onto }

\text{ Hence, } f \text{ is a bijection. }

\text{ Now consider that } g = \{(u, b), (v, a), (w, c)\} \text{ and } g: B \rightarrow

\text{ Injectivity of } g

\text{ No two elements of } B \text{ have the same image in } A.

\text{ Therefore, } g \text{ is one-one. }

\text{ Surjectivity of } g

\text{ Co-domain of } g = \{a, b, c\}

\text{ Range of } g = \{a, b, c\}

\text{ Both are the same }

\text{ Therefore, } g \text{ is onto. }

\text{ Hence, } g \text{ is a bijection. }

\text{ Let us find } fog,

\text{ We known, Co-domain of } g \text{ is same as the domain of } f.

\text{ Therefore, } fog \text{ exists and } fog: \{uv, w\} \rightarrow \{uv, w\}

(fog)(u) = f(g(u)) = f(b) = u

(fog)(v) = f(g(v)) = f(a) = v

(fog)(w) = f(g(w)) = f(c) = w

\text{ Therefore, } fog = \{(u, u), (v, v), (w, w)\}

\text{ Let us find }gof,

\text{ Co-domain of } f \text{ is same as the domain of } g.

\text{ Therefore, } fog \text{ exists and } gof: \{a, b, c\} \rightarrow \{a, b, c\}

(gof)(a) = g(f(a)) = g(v) = a

(gof)(b) = g(f(b)) = g(u) = b

(gof)(c) = g(f(c)) = g(w) = c

\text{ Therefore, } gof = \{(a, a), (b, b), (c, c)\}

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Question 5: Find fog(2) \text{ and } gof(1) when: f:R \rightarrow  R; f (x)=x^2 +8 \text{ and } g:R \rightarrow R; g(x)=3x^3 +1

Answer:

\text{Given as } f: R \rightarrow R; f(x) = x^2 + 8 \text{ and } g: R \rightarrow R; g(x) = 3x^3 + 1.

(fog)(2) = f(g(2)) = f(3 \times 2^3 + 1) = f(3 \times 8 + 1) = f(25) = 25^2 + 8 = 633

(gof)(1) = g(f(1))  = g(1^2 + 8)  = g(9)  = 3 \times 9^3 + 1  = 2188

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Question 6: Let R^+   be the set of non-negative real numbers. If f:R^+ \rightarrow R^+  \text{ and } g: R^+ \rightarrow R^+ \text{ are defined as } f(x) = x^2  \text{ and } g(x) = + \sqrt{x} . Find fog \text{ and } gof. Are they equal functions. 

Answer:

\text{ Given that } f: R^+ \rightarrow R^+ \text{ and } g: R^+ \rightarrow R^+

\text{ Therefore, } fog: R^+ \rightarrow R^+ \text{ and } gof: R^+ \rightarrow R^+

\text{ Domains of } fog \text{ and } gof \text{ are the same. }

\text{ Let us find } fog \text{ and } gof \text{ also check whether they are equal or not, }

(fog)(x) = f(g(x)) = f(\sqrt{x}) = \sqrt{x^2} = x

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

\text{ Therefore, } (fog)(x) = (gof)(x), \forall \ x \in R^+

\text{ Hence, } fog = gof

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\displaystyle \text{Question 7: Let } f:R \rightarrow R  \text{ and } g: R \rightarrow R \text{ be defined by } f(x) = x^2  \\ \\ \text{ and } g(x)  = x+ 1 . \text{ Show the } fog \neq gof . 

Answer:

\text{ Given as } f: R \rightarrow R \text{ and } g: R \rightarrow R.

Therefore, the domains of f \text{ and } g are the same.

(fog) (x) = f(g(x))  = f(x + 1) = (x + 1)^2  = x^2 + 1 + 2x

(gof)(x) = g(f(x))  = g(x^2) = x^2 + 1

\text{ Therefore, } fog \neq gof

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\displaystyle \text{Question 8: Let } f:R \rightarrow R  \text{ and } g: R \rightarrow R \text{ be defined by } \\ \\ f(x) = x+1  \text{ and } g(x)  = x- 1 . \text{ Show the } fog = gof = I_R . 

Answer:

\text{ Given as } f: R \rightarrow R \text{ and } g: R \rightarrow R.

\Rightarrow fog : R \rightarrow \text{ and } gof : R \rightarrow R \text{ ( Also, we know that } I_R : R \rightarrow R \text{ ) }

\text{ So, the domain of all } fog , gof \text{ and } I_R \text{ are the same }

(fog)(x) = f(g(x)) = f( x-1 ) = x-1 + 1 = x = I_R(x)

(gof) (x) = g(f(x)) = g(x+1) = x+1 -1=x=I_R(x)

\therefore (fog) (x)= ( gof) (x) = I_R (x) , \forall \ x \in R

\text{ Hence, } fog = gof = I_R

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\displaystyle \text{Question 9: Verify associativity for the following three mappings: } \\ \\ f : N \rightarrow Z_0 \text{ (the set of non-zero integers), } g : Z_0 \rightarrow Q \text{ and } h : Q \rightarrow R \text{ given by } \\ \\ f(x) = 2x , g(x) = \frac{1}{x} \text{ and } h (x) = e^x.

Answer:

\text{ Given } f : N \rightarrow Z_0, \ \  g : Z_0 \rightarrow Q \text{ and } h : Q \rightarrow R

gof : N \rightarrow Q \text{ and } hog : Z_0 \rightarrow R

\Rightarrow ho(gof) : N \rightarrow R \text{ and } (hog)of : N \rightarrow R

So, both have the same domains.

\displaystyle (gof)(x) = g(f(x)) = g(2x) = \frac{1}{2x}

\displaystyle (hog)(x) = h( g(x)) = h \Big( \frac{1}{x} \Big) = e^{\frac{1}{x}}

Now

\displaystyle (ho(gof))(x) = h((gof)(x)) = h \Big( \frac{1}{2x} \Big) = e^{\frac{1}{2x}}

\displaystyle ((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = e^{\frac{1}{2x}}

\Rightarrow ho(gof) =(hog)of

Hence, the associative property has been verified.

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\displaystyle \text{Question 10: Consider } f: N \rightarrow N, g: N \rightarrow N \text{ and } h : N \rightarrow R \\ \\ \text{ defined as } f(x) = 2x , g(y) = 3y + 4 \text{ and } h(z) = \sin z \text{ for all } x, y, z \in N. \\ \\ \text{Show that } ho(gof) = ( hog)of.

Answer:

\text{ Given } f: N \rightarrow N, g: N \rightarrow N \text{ and } h : N \rightarrow R

\Rightarrow gof : N \rightarrow N \text{ and } hog : N \rightarrow R

\Rightarrow ho(gof) : N \rightarrow R \text{ and } (hog) of : N \rightarrow R

So, both have the same domains.

(gof)(x) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x + 4

(hog)(x) = h(g(x)) = h ( 3x+4) = \sin (3x+4)

Now, 

\displaystyle (ho(gof))(x) = h((gof)(x)) = h (6x+4) = \sin (6x+4)

\displaystyle ((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = \sin (6x+4)

\Rightarrow (ho(gof))(x) =((hog)of)(x) \ \forall \ x \in N

\text{ Hence,  }  ho(gof) =(hog)of

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Question 11: Give examples of two functions f: N \rightarrow N \text{ and } g : N \rightarrow N such that gof is onto but f is not onto.

Answer:

Let us consider g: N \rightarrow N given by

\displaystyle g(x) =  \Bigg\{ \begin{array}{rr}  x-1, & \text{ if } x > 1 \\  \\ 1, & \text{ if } x=1 \end{array} 

\text{ Case 1: } x > 1

(gof)(x) = g(f(x)) = g(x+1) = x+ 1 - 1 = x

\text{ Case 2: } x = 1

(gof)(x) = g(f(x)) = g(x+1) = 1

Therefore (gof)(x) = x \ \forall \ x \in N which is an identity function and hence it is onto.

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Question 12: Give examples of two functions f: N \rightarrow N \text{ and } g : Z \rightarrow Z such that gof is injective but g is not injective.

Answer:

\text{ Let } f: N \rightarrow Z \text{ as } f(x) = x \text{ and } g: Z \rightarrow  Z \text{ as } g(x) = |x|

We fist show that g is not injective.

It can be observed that

g(-1) = |-1| = 1

g(1) = |1| = 1

\therefore g(-1) = g(1), \text{ but } -1 \neq 1

Therefore g is not injective.

\text{ Now, } gof : N \rightarrow Z \text{ is defined as } gof(x) = g(f(x)) = g(x) = |x|/

\text{Let } x, y \in N \text{ such that } gof(x) = gof(y)

\Rightarrow |x| = |y|  

\text{Since } x \text{ and } y \in N, \text{ both are positive. }

\Rightarrow |x| = |y |\Rightarrow x = y

Hence, gof is injective.

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Question 13: If f : A \rightarrow B \text{ and } g:B \rightarrow C are one-one functions, show that gof is a one-one function.

Answer:

gof (x_1) = gof(x_2)

g[f(x_1)]=g[(x_2)]

f(x_1) = f(x_2)

x_1 = x_2

\text{ Since } g \text{ and } f \text{ is one-one, } gof \text{ is one-one }

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Question 14: If f : A \rightarrow B \text{ and } g:B \rightarrow C are onto functions show that gof is an onto function.

Answer:

\text{ Let there is an element } x \text{ such that } x \in C.

\text{ Since } g \text{ is onto, we can find an element } b \text{ where } b \in B \text{ such that } g ( b ) = x.

\text{ But } f \text{ is onto, so we can also find an element } a \text{ where } a \in A \text{ such that } \\ \\ f ( a ) = b.

\text{ Thus, } ( g ( f ( a ) ) = g ( b ) = x, \text{ and so } g o f  \text{ is onto. }

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