\displaystyle \textbf{Question 1: }\text{Find }g\circ f\text{ and }f\circ g\text{ when }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow\mathbb R\text{ are defined by:}
\displaystyle \text{(i) }f(x)=2x+3,\ g(x)=x^2+5
\displaystyle \text{(ii) }f(x)=2x+x^2,\ g(x)=x^3
\displaystyle \text{(iii) }f(x)=x^2+8,\ g(x)=3x^3+1
\displaystyle \text{(iv) }f(x)=x,\ g(x)=|x|
\displaystyle \text{(v) }f(x)=x^2+2x-3,\ g(x)=3x-4
\displaystyle \text{(vi) }f(x)=8x^3,\ g(x)=x^{\frac13}
\displaystyle \text{Answer:}

\displaystyle \text{(i) }g\circ f(x)=g(f(x))=g(2x+3)=(2x+3)^2+5=4x^2+12x+14.
\displaystyle f\circ g(x)=f(g(x))=f(x^2+5)=2(x^2+5)+3=2x^2+13.

\displaystyle \text{(ii) }g\circ f(x)=g(f(x))=g(2x+x^2)=(2x+x^2)^3.
\displaystyle f\circ g(x)=f(g(x))=f(x^3)=2x^3+x^6.

\displaystyle \text{(iii) }g\circ f(x)=g(f(x))=g(x^2+8)=3(x^2+8)^3+1.
\displaystyle f\circ g(x)=f(g(x))=f(3x^3+1)=(3x^3+1)^2+8=9x^6+6x^3+9.

\displaystyle \text{(iv) }g\circ f(x)=g(f(x))=g(x)=|x|.
\displaystyle f\circ g(x)=f(g(x))=f(|x|)=|x|.

\displaystyle \text{(v) }g\circ f(x)=g(f(x))=g(x^2+2x-3)=3(x^2+2x-3)-4=3x^2+6x-13.
\displaystyle f\circ g(x)=f(g(x))=f(3x-4)=(3x-4)^2+2(3x-4)-3=9x^2-18x+5.

\displaystyle \text{(vi) }g\circ f(x)=g(f(x))=g(8x^3)=(8x^3)^{\frac13}=[(2x)^3]^{\frac13}=2x.
\displaystyle f\circ g(x)=f(g(x))=f\left(x^{\frac13}\right)=8\left(x^{\frac13}\right)^3=8x.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }f=\{(3,1),(9,3),(12,4)\}\text{ and }g=\{(1,3),(3,3),(4,9),(5,9)\}.
\displaystyle \text{Show that }g\circ f\text{ and }f\circ g\text{ are both defined. Also, find }f\circ g\text{ and }g\circ f.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f=\{(3,1),(9,3),(12,4)\}\text{ and }g=\{(1,3),(3,3),(4,9),(5,9)\}.
\displaystyle \text{Range}(f)=\{1,3,4\}\subseteq\text{Domain}(g)=\{1,3,4,5\}.
\displaystyle \therefore g\circ f\text{ exists and }g\circ f:\{3,9,12\}\rightarrow\{3,9\}.
\displaystyle (g\circ f)(3)=g(f(3))=g(1)=3.
\displaystyle (g\circ f)(9)=g(f(9))=g(3)=3.
\displaystyle (g\circ f)(12)=g(f(12))=g(4)=9.
\displaystyle \therefore g\circ f=\{(3,3),(9,3),(12,9)\}.
\displaystyle \text{Range}(g)=\{3,9\}\subseteq\text{Domain}(f)=\{3,9,12\}.
\displaystyle \therefore f\circ g\text{ exists and }f\circ g:\{1,3,4,5\}\rightarrow\{1,3,4\}.
\displaystyle (f\circ g)(1)=f(g(1))=f(3)=1.
\displaystyle (f\circ g)(3)=f(g(3))=f(3)=1.
\displaystyle (f\circ g)(4)=f(g(4))=f(9)=3.
\displaystyle (f\circ g)(5)=f(g(5))=f(9)=3.
\displaystyle \therefore f\circ g=\{(1,1),(3,1),(4,3),(5,3)\}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }f=\{(1,-1),(4,-2),(9,-3),(16,4)\}\text{ and }g=\{(-1,-2),(-2,-4),(-3,-6),(4,8)\}.
\displaystyle \text{Show that }g\circ f\text{ is defined while }f\circ g\text{ is not defined. Also, find }g\circ f.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f=\{(1,-1),(4,-2),(9,-3),(16,4)\}\text{ and }g=\{(-1,-2),(-2,-4),(-3,-6),(4,8)\}.
\displaystyle \text{Range}(f)=\{-1,-2,-3,4\}\subseteq\text{Domain}(g)=\{-1,-2,-3,4\}.
\displaystyle \therefore g\circ f\text{ exists and }g\circ f:\{1,4,9,16\}\rightarrow\{-2,-4,-6,8\}.
\displaystyle (g\circ f)(1)=g(f(1))=g(-1)=-2.
\displaystyle (g\circ f)(4)=g(f(4))=g(-2)=-4.
\displaystyle (g\circ f)(9)=g(f(9))=g(-3)=-6.
\displaystyle (g\circ f)(16)=g(f(16))=g(4)=8.
\displaystyle \therefore g\circ f=\{(1,-2),(4,-4),(9,-6),(16,8)\}.
\displaystyle \text{Now, Range}(g)=\{-2,-4,-6,8\}\not\subseteq\text{Domain}(f)=\{1,4,9,16\}.
\displaystyle \therefore f\circ g\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\{a,b,c\},\ B=\{u,v,w\},\text{ and let }f:A\rightarrow B
\displaystyle \text{and }g:B\rightarrow A\text{ be defined by}
\displaystyle f=\{(a,v),(b,u),(c,w)\},\qquad g=\{(u,b),(v,a),(w,c)\}.
\displaystyle \text{Show that }f\text{ and }g\text{ are bijections and find }f\circ g\text{ and }g\circ f.
\displaystyle \text{Answer:}
\displaystyle \text{For }f:A\rightarrow B,\text{ the elements }a,b,c\text{ have distinct images }v,u,w.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Also, Range}(f)=\{u,v,w\}=B.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \text{For }g:B\rightarrow A,\text{ the elements }u,v,w\text{ have distinct images }b,a,c.
\displaystyle \therefore g\text{ is one-one.}
\displaystyle \text{Also, Range}(g)=\{a,b,c\}=A.
\displaystyle \therefore g\text{ is onto.}
\displaystyle \therefore g\text{ is a bijection.}
\displaystyle \text{Now, Range}(g)=A=\text{Domain}(f).
\displaystyle \therefore f\circ g:B\rightarrow B\text{ exists.}
\displaystyle (f\circ g)(u)=f(g(u))=f(b)=u.
\displaystyle (f\circ g)(v)=f(g(v))=f(a)=v.
\displaystyle (f\circ g)(w)=f(g(w))=f(c)=w.
\displaystyle \therefore f\circ g=\{(u,u),(v,v),(w,w)\}=I_B.
\displaystyle \text{Also, Range}(f)=B=\text{Domain}(g).
\displaystyle \therefore g\circ f:A\rightarrow A\text{ exists.}
\displaystyle (g\circ f)(a)=g(f(a))=g(v)=a.
\displaystyle (g\circ f)(b)=g(f(b))=g(u)=b.
\displaystyle (g\circ f)(c)=g(f(c))=g(w)=c.
\displaystyle \therefore g\circ f=\{(a,a),(b,b),(c,c)\}=I_A.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find }(f\circ g)(2)\text{ and }(g\circ f)(1)\text{ when }f:\mathbb R\rightarrow\mathbb R,\ f(x)=x^2+8
\displaystyle \text{and }g:\mathbb R\rightarrow\mathbb R,\ g(x)=3x^3+1.
\displaystyle \text{Answer:}
\displaystyle (f\circ g)(2)=f(g(2)).
\displaystyle =f(3\cdot2^3+1)=f(25).
\displaystyle =25^2+8=625+8=633.
\displaystyle (g\circ f)(1)=g(f(1)).
\displaystyle =g(1^2+8)=g(9).
\displaystyle =3\cdot9^3+1=3\cdot729+1=2188.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Let }\mathbb R^+\text{ be the set of non-negative real numbers. If }
\displaystyle f:\mathbb R^+\rightarrow\mathbb R^+  \ \text{and }g:\mathbb R^+\rightarrow\mathbb R^+\text{ are defined by }
\displaystyle f(x)=x^2\text{ and }g(x)=\sqrt{x}, \ \text{find }f\circ g\text{ and }g\circ f.\text{ Are they equal functions?}
\displaystyle \text{Answer:}
\displaystyle \text{Since Range}(g)\subseteq\text{Domain}(f)\text{ and Range}(f)\subseteq\text{Domain}(g),
\displaystyle \text{both }f\circ g\text{ and }g\circ f\text{ exist.}
\displaystyle (f\circ g)(x)=f(g(x))=f(\sqrt{x})=(\sqrt{x})^2=x.
\displaystyle (g\circ f)(x)=g(f(x))=g(x^2)=\sqrt{x^2}=x,\qquad x\in\mathbb R^+.
\displaystyle \therefore (f\circ g)(x)=(g\circ f)(x)=x,\ \forall x\in\mathbb R^+.
\displaystyle \therefore f\circ g=g\circ f.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow\mathbb R\text{ be defined by}
\displaystyle f(x)=x^2\text{ and }g(x)=x+1.\text{ Show that }f\circ g\ne g\circ f.
\displaystyle \text{Answer:}
\displaystyle \text{Since Range}(g)\subseteq\text{Domain}(f)\text{ and Range}(f)\subseteq\text{Domain}(g),
\displaystyle \text{both }f\circ g\text{ and }g\circ f\text{ exist.}
\displaystyle (f\circ g)(x)=f(g(x))=f(x+1)=(x+1)^2=x^2+2x+1.
\displaystyle (g\circ f)(x)=g(f(x))=g(x^2)=x^2+1.
\displaystyle \text{Since }x^2+2x+1\ne x^2+1\text{ for all }x\in\mathbb R,
\displaystyle \therefore f\circ g\ne g\circ f.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow\mathbb R\text{ be defined by}
\displaystyle f(x)=x+1\text{ and }g(x)=x-1.\text{ Show that }f\circ g=g\circ f=I_{\mathbb R}.
\displaystyle \text{Answer:}
\displaystyle \text{Since Range}(g)\subseteq\text{Domain}(f)\text{ and Range}(f)\subseteq\text{Domain}(g),
\displaystyle \text{both }f\circ g\text{ and }g\circ f\text{ exist.}
\displaystyle (f\circ g)(x)=f(g(x))=f(x-1)=x-1+1=x=I_{\mathbb R}(x).
\displaystyle (g\circ f)(x)=g(f(x))=g(x+1)=x+1-1=x=I_{\mathbb R}(x).
\displaystyle \therefore (f\circ g)(x)=(g\circ f)(x)=I_{\mathbb R}(x),\ \forall x\in\mathbb R.
\displaystyle \therefore f\circ g=g\circ f=I_{\mathbb R}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Verify associativity for the following three mappings:}
\displaystyle f:\mathbb N\rightarrow\mathbb Z_0,\ g:\mathbb Z_0\rightarrow\mathbb Q,\ \text{and }h:\mathbb Q\rightarrow\mathbb R
\displaystyle \text{defined by }f(x)=2x,\ g(x)=\frac1x,\ \text{and }h(x)=e^x.
\displaystyle \text{Answer:}
\displaystyle \text{Since Range}(f)\subseteq\text{Domain}(g)\text{ and Range}(g)\subseteq\text{Domain}(h),
\displaystyle \text{the compositions }g\circ f,\ h\circ g,\ h\circ(g\circ f)\text{ and }(h\circ g)\circ f\text{ exist.}
\displaystyle (g\circ f)(x)=g(f(x))=g(2x)=\frac1{2x}.
\displaystyle (h\circ g)(x)=h(g(x))=h\!\left(\frac1x\right)=e^{\frac1x}.
\displaystyle (h\circ(g\circ f))(x)=h\!\left((g\circ f)(x)\right)=h\!\left(\frac1{2x}\right)=e^{\frac1{2x}}.
\displaystyle ((h\circ g)\circ f)(x)=(h\circ g)(f(x))=(h\circ g)(2x)=e^{\frac1{2x}}.
\displaystyle \therefore (h\circ(g\circ f))(x)=((h\circ g)\circ f)(x),\ \forall x\in\mathbb N.
\displaystyle \therefore h\circ(g\circ f)=(h\circ g)\circ f.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Consider }f:\mathbb N\rightarrow\mathbb N,\ g:\mathbb N\rightarrow\mathbb N,\ \text{and }h:\mathbb N\rightarrow\mathbb R
\displaystyle \text{defined by }f(x)=2x,\ g(x)=3x+4,\ \text{and }h(x)=\sin x .
\displaystyle \text{Show that }h\circ(g\circ f)=(h\circ g)\circ f.
\displaystyle \text{Answer:}
\displaystyle \text{Since Range}(f)\subseteq\text{Domain}(g)\text{ and Range}(g)\subseteq\text{Domain}(h),
\displaystyle \text{the compositions }g\circ f,\ h\circ g,\ h\circ(g\circ f)\text{ and }(h\circ g)\circ f\text{ exist.}
\displaystyle (g\circ f)(x)=g(f(x))=g(2x)=6x+4.
\displaystyle (h\circ g)(x)=h(g(x))=h(3x+4)=\sin(3x+4).
\displaystyle (h\circ(g\circ f))(x)=h((g\circ f)(x))=h(6x+4)=\sin(6x+4).
\displaystyle ((h\circ g)\circ f)(x)=(h\circ g)(f(x))=(h\circ g)(2x)=\sin(6x+4).
\displaystyle \therefore (h\circ(g\circ f))(x)=((h\circ g)\circ f)(x),\ \forall x\in\mathbb N.
\displaystyle \therefore h\circ(g\circ f)=(h\circ g)\circ f.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Give examples of two functions }f:\mathbb N\rightarrow\mathbb N\text{ and }g:\mathbb N\rightarrow\mathbb N
\displaystyle \text{such that }g\circ f\text{ is onto but }f\text{ is not onto.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb N\rightarrow\mathbb N\text{ be defined by }f(x)=x+1.
\displaystyle \text{Let }g:\mathbb N\rightarrow\mathbb N\text{ be defined by}
\displaystyle g(x)=\left\{\begin{array}{ll}x-1,&\text{if }x>1\\ \\ 1,&\text{if }x=1\end{array}\right.
\displaystyle \text{Since }f(x)=x+1\ge2,\text{ we have}
\displaystyle (g\circ f)(x)=g(f(x))=g(x+1)=x+1-1=x.
\displaystyle \therefore g\circ f=I_{\mathbb N}.
\displaystyle \therefore g\circ f\text{ is onto.}
\displaystyle \text{Now, Range}(f)=\{2,3,4,\ldots\}.
\displaystyle \text{Since }1\in\mathbb N\text{ has no pre-image under }f,
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Give examples of two functions }f:\mathbb N\rightarrow\mathbb N\text{ and }g:\mathbb Z\rightarrow\mathbb Z
\displaystyle \text{such that }g\circ f\text{ is injective but }g\text{ is not injective.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb N\rightarrow\mathbb N\text{ be defined by }f(x)=x.
\displaystyle \text{Let }g:\mathbb Z\rightarrow\mathbb Z\text{ be defined by }g(x)=|x|.
\displaystyle \text{First, }g(-1)=1=g(1),\text{ although }-1\ne1.
\displaystyle \therefore g\text{ is not injective.}
\displaystyle \text{Since Range}(f)=\mathbb N\subseteq\mathbb Z=\text{Domain}(g),\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(x)=|x|=x,\qquad x\in\mathbb N.
\displaystyle \text{Let }x,y\in\mathbb N\text{ such that }(g\circ f)(x)=(g\circ f)(y).
\displaystyle \Rightarrow x=y.
\displaystyle \therefore g\circ f\text{ is injective.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }f:A\rightarrow B\text{ and }g:B\rightarrow C\text{ are one-one functions,}
\displaystyle \text{show that }g\circ f\text{ is a one-one function.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x_1,x_2\in A\text{ such that }(g\circ f)(x_1)=(g\circ f)(x_2).
\displaystyle \Rightarrow g(f(x_1))=g(f(x_2)).
\displaystyle \text{Since }g\text{ is one-one,}
\displaystyle f(x_1)=f(x_2).
\displaystyle \text{Since }f\text{ is one-one,}
\displaystyle x_1=x_2.
\displaystyle \therefore g\circ f\text{ is one-one.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }f:A\rightarrow B\text{ and }g:B\rightarrow C\text{ are onto functions,}
\displaystyle \text{show that }g\circ f\text{ is an onto function.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }c\in C\text{ be arbitrary.}
\displaystyle \text{Since }g:B\rightarrow C\text{ is onto, there exists }b\in B\text{ such that }g(b)=c.
\displaystyle \text{Since }f:A\rightarrow B\text{ is onto, there exists }a\in A\text{ such that }f(a)=b.
\displaystyle \text{Therefore,}
\displaystyle (g\circ f)(a)=g(f(a))=g(b)=c.
\displaystyle \text{Thus, every element of }C\text{ has a pre-image in }A\text{ under }g\circ f.
\displaystyle \therefore g\circ f\text{ is onto.}
\displaystyle \\

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