\displaystyle \textbf{Question 1: }\text{Give an example of a function}
\displaystyle \text{(i) which is one-one but not onto,}
\displaystyle \text{(ii) which is not one-one but onto,}
\displaystyle \text{(iii) which is neither one-one nor onto.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }f:\mathbb N\rightarrow\mathbb N\text{ defined by }f(x)=x^2.
\displaystyle \text{To check injectivity, let }x,y\in\mathbb N\text{ and suppose }f(x)=f(y).
\displaystyle x^2=y^2
\displaystyle x^2-y^2=0
\displaystyle (x-y)(x+y)=0
\displaystyle \text{Since }x,y\in\mathbb N,\ x+y>0.
\displaystyle \therefore x-y=0\Rightarrow x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{However, }2\in\mathbb N\text{ has no pre-image in }\mathbb N,
\displaystyle \text{because }x^2=2\Rightarrow x=\sqrt2\notin\mathbb N.
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \text{Hence, }f:\mathbb N\rightarrow\mathbb N,\ f(x)=x^2,\text{ is one-one but not onto.}
\displaystyle \\

\displaystyle \text{(ii) Consider }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x^3-x.
\displaystyle f(-1)=(-1)^3-(-1)=0
\displaystyle f(0)=0^3-0=0
\displaystyle \text{Thus, }-1\ne0\text{ but }f(-1)=f(0).
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \text{To check surjectivity, let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y\Rightarrow x^3-x-y=0.
\displaystyle \text{The equation }x^3-x-y=0\text{ is an odd-degree polynomial equation}
\displaystyle \text{with real coefficients and hence has at least one real root, say }\alpha.
\displaystyle \alpha^3-\alpha-y=0
\displaystyle \alpha^3-\alpha=y
\displaystyle f(\alpha)=y.
\displaystyle \text{Thus, for every }y\in\mathbb R,\text{ there exists }\alpha\in\mathbb R\text{ such that }f(\alpha)=y.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Hence, }f:\mathbb R\rightarrow\mathbb R,\ f(x)=x^3-x,\text{ is not one-one but onto.}
\displaystyle \\

\displaystyle \text{(iii) Consider }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x^2.
\displaystyle f(-1)=(-1)^2=1
\displaystyle f(1)=1^2=1
\displaystyle \text{Thus, }-1\ne1\text{ but }f(-1)=f(1).
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \text{Also, }f(x)=x^2\geq0\text{ for every }x\in\mathbb R.
\displaystyle \text{Therefore, negative real numbers have no pre-images in }\mathbb R.
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \text{Hence, }f:\mathbb R\rightarrow\mathbb R,\ f(x)=x^2,\text{ is neither one-one nor onto.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which of the following functions from }A\text{ to }B\text{ are one-one and onto?}
\displaystyle \textbf{(i) }\;f_1=\{(1,3),(2,5),(3,7)\};\ A=\{1,2,3\},\ B=\{3,5,7\}
\displaystyle \textbf{(ii) }\;f_2=\{(2,a),(3,b),(4,c)\};\ A=\{2,3,4\},\ B=\{a,b,c\}
\displaystyle \textbf{(iii) }\;f_3=\{(a,x),(b,x),(c,z),(d,z)\};\ A=\{a,b,c,d\},\ B=\{x,y,z\}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }f_1=\{(1,3),(2,5),(3,7)\},\ A=\{1,2,3\},\ B=\{3,5,7\}.
\displaystyle \text{Check for Injectivity:}
\displaystyle \text{Distinct elements of }A\text{ have distinct images in }B.
\displaystyle \therefore f_1\text{ is one-one.}
\displaystyle \text{Check for Surjectivity:}
\displaystyle \text{Every element of }B\text{ has a pre-image in }A.
\displaystyle \therefore f_1\text{ is onto.}
\displaystyle \\

\displaystyle \text{(ii) Given, }f_2=\{(2,a),(3,b),(4,c)\},\ A=\{2,3,4\},\ B=\{a,b,c\}.
\displaystyle \text{Check for Injectivity:}
\displaystyle \text{Distinct elements of }A\text{ have distinct images in }B.
\displaystyle \therefore f_2\text{ is one-one.}
\displaystyle \text{Check for Surjectivity:}
\displaystyle \text{Every element of }B\text{ has a pre-image in }A.
\displaystyle \therefore f_2\text{ is onto.}
\displaystyle \\

\displaystyle \text{(iii) Given, }f_3=\{(a,x),(b,x),(c,z),(d,z)\},\ A=\{a,b,c,d\},\ B=\{x,y,z\}.
\displaystyle \text{Check for Injectivity:}
\displaystyle f_3(a)=x=f_3(b)\text{ and }f_3(c)=z=f_3(d).
\displaystyle \therefore \text{Distinct elements of }A\text{ do not have distinct images.}
\displaystyle \therefore f_3\text{ is not one-one.}
\displaystyle \text{Check for Surjectivity:}
\displaystyle \text{The element }y\in B\text{ has no pre-image in }A.
\displaystyle \therefore f_3\text{ is not onto.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that the function }f:\mathbb N\rightarrow\mathbb N\text{ defined by}
\displaystyle f(x)=x^2+x+1\text{ is one-one but not onto.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb N\rightarrow\mathbb N\text{ defined by }f(x)=x^2+x+1.
\displaystyle \text{Check for Injectivity:}
\displaystyle \text{Let }x,y\in\mathbb N\text{ such that }f(x)=f(y).
\displaystyle x^2+x+1=y^2+y+1
\displaystyle x^2-y^2+x-y=0
\displaystyle (x-y)(x+y+1)=0
\displaystyle \text{Since }x,y\in\mathbb N,\ x+y+1>0.
\displaystyle \therefore x-y=0
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Surjectivity:}
\displaystyle \text{Consider }2\in\mathbb N.
\displaystyle f(x)=2
\displaystyle \Rightarrow x^2+x+1=2
\displaystyle \Rightarrow x^2+x-1=0.
\displaystyle \text{This equation has no solution in }\mathbb N.
\displaystyle \text{Hence, }2\text{ has no pre-image in the domain.}
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \therefore f:\mathbb N\rightarrow\mathbb N,\ f(x)=x^2+x+1\text{ is one-one but not onto.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\{-1,0,1\}\text{ and }f=\{(x,x^2):x\in A\}.
\displaystyle \text{Show that }f:A\rightarrow A\text{ is neither one-one nor onto.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{-1,0,1\}\text{ and }f=\{(x,x^2):x\in A\}.
\displaystyle \text{Thus, }f(-1)=1,\quad f(0)=0,\quad f(1)=1.
\displaystyle \text{Check for Injectivity:}
\displaystyle f(-1)=1=f(1).
\displaystyle \text{Since }-1\ne1\text{ but }f(-1)=f(1),\ f\text{ is not one-one.}
\displaystyle \text{Check for Surjectivity:}
\displaystyle \text{The range of }f\text{ is }\{0,1\}.
\displaystyle \text{Therefore, }-1\in A\text{ has no pre-image in the domain }A.
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \therefore f:A\rightarrow A\text{ is neither one-one nor onto.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \textbf{(i) }\;f:\mathbb N\rightarrow\mathbb N\text{ defined by }f(x)=x^2.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb N\rightarrow\mathbb N,\ f(x)=x^2.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb N\text{ such that }f(x)=f(y).
\displaystyle x^2=y^2
\displaystyle \therefore x=y,\text{ since }x,y\in\mathbb N.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb N\text{ be arbitrary.}
\displaystyle f(x)=y\Rightarrow x^2=y
\displaystyle \Rightarrow x=\sqrt{y}.
\displaystyle \text{For }y=3,\ x=\sqrt3\notin\mathbb N.
\displaystyle \text{Hence, }3\text{ has no pre-image in }\mathbb N.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is an injection but not a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \textbf{(ii) }\;f:\mathbb Z\rightarrow\mathbb Z\text{ defined by }f(x)=x^2.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Z\rightarrow\mathbb Z,\ f(x)=x^2.
\displaystyle \text{Check for Injection:}
\displaystyle f(1)=1=f(-1).
\displaystyle \text{Since }1\ne-1\text{ but }f(1)=f(-1),
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Since }x^2\ge0\text{ for every }x\in\mathbb Z,
\displaystyle \text{the integer }-1\in\mathbb Z\text{ has no pre-image.}
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \textbf{(iii) }\;f:\mathbb N\rightarrow\mathbb N\text{ defined by }f(x)=x^3.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb N\rightarrow\mathbb N,\ f(x)=x^3.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb N\text{ such that }f(x)=f(y).
\displaystyle x^3=y^3
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb N\text{ be arbitrary.}
\displaystyle f(x)=y\Rightarrow x^3=y
\displaystyle \Rightarrow x=\sqrt[3]{y}.
\displaystyle \text{For }y=3,\ x=\sqrt[3]{3}\notin\mathbb N.
\displaystyle \text{Hence, }3\text{ has no pre-image in }\mathbb N.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is an injection but not a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \textbf{(iv) }\;f:\mathbb Z\rightarrow\mathbb Z\text{ defined by }f(x)=x^3.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Z\rightarrow\mathbb Z,\ f(x)=x^3.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb Z\text{ such that }f(x)=f(y).
\displaystyle x^3=y^3
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb Z\text{ be arbitrary.}
\displaystyle f(x)=y\Rightarrow x^3=y
\displaystyle \Rightarrow x=\sqrt[3]{y}.
\displaystyle \text{For }y=3,\ x=\sqrt[3]{3}\notin\mathbb Z.
\displaystyle \text{Hence, }3\text{ has no pre-image in }\mathbb Z.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is an injection but not a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following functions as injection, surjection or bijection:}
\displaystyle \text{(v) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=|x|.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=|x|.
\displaystyle \text{Check for Injection:}
\displaystyle f(1)=|1|=1
\displaystyle f(-1)=|-1|=1
\displaystyle \text{Thus, }1\ne-1\text{ but }f(1)=f(-1).
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle |x|\geq0\text{ for every }x\in\mathbb R.
\displaystyle \text{Therefore, }-1\in\mathbb R\text{ has no pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(vi) }f:\mathbb Z\rightarrow\mathbb Z\text{ defined by }f(x)=x^2+x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Z\rightarrow\mathbb Z\text{ defined by }f(x)=x^2+x.
\displaystyle \text{Check for Injection:}
\displaystyle f(2)=2^2+2=6
\displaystyle f(-3)=(-3)^2+(-3)=6
\displaystyle \text{Thus, }2\ne-3\text{ but }f(2)=f(-3).
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle f(x)=x^2+x=x(x+1).
\displaystyle \text{For every }x\in\mathbb Z,\ x(x+1)\geq0.
\displaystyle \text{Therefore, }-1\in\mathbb Z\text{ has no pre-image in }\mathbb Z.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following functions as injection, surjection or bijection:}
\displaystyle \text{(vii) }f:\mathbb Z\rightarrow\mathbb Z\text{ defined by }f(x)=x-5.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Z\rightarrow\mathbb Z\text{ defined by }f(x)=x-5.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb Z\text{ such that }f(x)=f(y).
\displaystyle x-5=y-5
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb Z\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow x-5=y
\displaystyle \Rightarrow x=y+5.
\displaystyle \text{Since }y+5\in\mathbb Z,\ x\in\mathbb Z.
\displaystyle \therefore \text{Every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is a surjection.}
\displaystyle \therefore f\text{ is both an injection and a surjection.}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(viii) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=\sin x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=\sin x.
\displaystyle \text{Check for Injection:}
\displaystyle f(0)=\sin0=0
\displaystyle f(\pi)=\sin\pi=0
\displaystyle \text{Thus, }0\ne\pi\text{ but }f(0)=f(\pi).
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{The range of }f\text{ is }[-1,1].
\displaystyle \text{The co-domain is }\mathbb R.
\displaystyle \text{Since }[-1,1]\ne\mathbb R,\text{ not every real number has a pre-image.}
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following functions as injection, surjection or bijection:}
\displaystyle \text{(ix) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x^3+1.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x^3+1.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle x^3+1=y^3+1
\displaystyle \Rightarrow x^3=y^3
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow x^3+1=y
\displaystyle \Rightarrow x=\sqrt[3]{\,y-1\,}\in\mathbb R.
\displaystyle \therefore \text{Every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is a surjection.}
\displaystyle \therefore f\text{ is both an injection and a surjection.}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(x) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x^3-x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x^3-x.
\displaystyle \text{Check for Injection:}
\displaystyle f(1)=1^3-1=0
\displaystyle f(-1)=(-1)^3-(-1)=0
\displaystyle \text{Thus, }1\ne-1\text{ but }f(1)=f(-1).
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow x^3-x-y=0.
\displaystyle \text{The equation }x^3-x-y=0\text{ is an odd-degree polynomial equation}
\displaystyle \text{with real coefficients and hence has at least one real root, say }\alpha.
\displaystyle \therefore f(\alpha)=y.
\displaystyle \therefore f\text{ is a surjection.}
\displaystyle \therefore f\text{ is not an injection but is a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following functions as injection, surjection or bijection:}
\displaystyle \text{(xi) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=\sin^2x+\cos^2x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=\sin^2x+\cos^2x.
\displaystyle \text{Using }\sin^2x+\cos^2x=1,
\displaystyle f(x)=1\text{ for every }x\in\mathbb R.
\displaystyle \text{Check for Injection:}
\displaystyle f(0)=1=f(\pi).
\displaystyle \text{Since }0\ne\pi\text{ but }f(0)=f(\pi),
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{The range of }f\text{ is }\{1\},\text{ whereas its co-domain is }\mathbb R.
\displaystyle \text{Therefore, every real number other than }1\text{ has no pre-image.}
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(xii) }f:\mathbb Q-\{3\}\rightarrow\mathbb Q\text{ defined by }f(x)=\frac{2x+3}{x-3}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Q-\{3\}\rightarrow\mathbb Q\text{ defined by }f(x)=\frac{2x+3}{x-3}.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb Q-\{3\}\text{ such that }f(x)=f(y).
\displaystyle \frac{2x+3}{x-3}=\frac{2y+3}{y-3}
\displaystyle \Rightarrow (2x+3)(y-3)=(2y+3)(x-3)
\displaystyle \Rightarrow 2xy-6x+3y-9=2xy-6y+3x-9
\displaystyle \Rightarrow 9y=9x
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb Q\text{ and suppose }f(x)=y.
\displaystyle \frac{2x+3}{x-3}=y
\displaystyle \Rightarrow 2x+3=xy-3y
\displaystyle \Rightarrow x(2-y)=-3(y+1)
\displaystyle \Rightarrow x=\frac{3(y+1)}{y-2},\quad y\ne2.
\displaystyle \text{For }y=2,\text{ the equation becomes}
\displaystyle \frac{2x+3}{x-3}=2
\displaystyle \Rightarrow 2x+3=2x-6,
\displaystyle \text{which is impossible.}
\displaystyle \text{Hence, }2\in\mathbb Q\text{ has no pre-image in }\mathbb Q-\{3\}.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is an injection but not a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following functions as injection, surjection or bijection:}
\displaystyle \text{(xiii) }f:\mathbb Q\rightarrow\mathbb Q\text{ defined by }f(x)=x^3+1.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb Q\rightarrow\mathbb Q\text{ defined by }f(x)=x^3+1.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb Q\text{ such that }f(x)=f(y).
\displaystyle x^3+1=y^3+1
\displaystyle \Rightarrow x^3=y^3
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb Q\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow x^3+1=y
\displaystyle \Rightarrow x=\sqrt[3]{\,y-1\,}.
\displaystyle \text{For }y=8,\ x=\sqrt[3]{7}\notin\mathbb Q.
\displaystyle \text{Hence, }8\text{ has no pre-image in }\mathbb Q.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is an injection but not a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(xiv) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=5x^3+4.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=5x^3+4.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle 5x^3+4=5y^3+4
\displaystyle \Rightarrow x^3=y^3
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow 5x^3+4=y
\displaystyle \Rightarrow x^3=\frac{y-4}{5}
\displaystyle \Rightarrow x=\sqrt[3]{\frac{y-4}{5}}\in\mathbb R.
\displaystyle \therefore \text{Every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is a surjection.}
\displaystyle \therefore f\text{ is both an injection and a surjection.}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following functions as injection, surjection or bijection:}
\displaystyle \text{(xv) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=3-4x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=3-4x.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle 3-4x=3-4y
\displaystyle \Rightarrow -4x=-4y
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow 3-4x=y
\displaystyle \Rightarrow x=\frac{3-y}{4}\in\mathbb R.
\displaystyle \therefore \text{Every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is a surjection.}
\displaystyle \therefore f\text{ is both an injection and a surjection.}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(xvi) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=1+x^2.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=1+x^2.
\displaystyle \text{Check for Injection:}
\displaystyle f(1)=1+1^2=2
\displaystyle f(-1)=1+(-1)^2=2
\displaystyle \text{Thus, }1\ne-1\text{ but }f(1)=f(-1).
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Consider }y=0\in\mathbb R.
\displaystyle f(x)=0
\displaystyle \Rightarrow 1+x^2=0
\displaystyle \Rightarrow x^2=-1,
\displaystyle \text{which has no solution in }\mathbb R.
\displaystyle \text{Hence, }0\text{ has no pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Classify the following function as injection, surjection or bijection:}
\displaystyle \text{(xvii) }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=\frac{x}{x^2+1}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=\frac{x}{x^2+1}.
\displaystyle \text{Check for Injection:}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle \frac{x}{x^2+1}=\frac{y}{y^2+1}
\displaystyle \Rightarrow x(y^2+1)=y(x^2+1)
\displaystyle \Rightarrow xy^2+x=x^2y+y
\displaystyle \Rightarrow xy^2-x^2y+x-y=0
\displaystyle \Rightarrow (x-y)(1-xy)=0.
\displaystyle \text{Thus, }x=y\text{ or }xy=1.
\displaystyle \text{For example,}
\displaystyle f(2)=\frac{2}{2^2+1}=\frac25
\displaystyle f\left(\frac12\right)=\frac{\frac12}{\frac14+1}=\frac25.
\displaystyle \text{Since }2\ne\frac12\text{ but }f(2)=f\left(\frac12\right),
\displaystyle \therefore f\text{ is not an injection.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Consider }y=1\in\mathbb R.
\displaystyle f(x)=1
\displaystyle \Rightarrow \frac{x}{x^2+1}=1
\displaystyle \Rightarrow x=x^2+1
\displaystyle \Rightarrow x^2-x+1=0.
\displaystyle \text{Here, the discriminant is}
\displaystyle D=(-1)^2-4(1)(1)=-3<0.
\displaystyle \text{Therefore, the equation has no real solution.}
\displaystyle \text{Hence, }1\text{ has no pre-image in }\mathbb R.
\displaystyle \therefore f\text{ is not a surjection.}
\displaystyle \therefore f\text{ is neither an injection nor a surjection.}
\displaystyle \therefore f\text{ is not a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }f:A\rightarrow B\text{ is an injection such that the range of }f=\{a\},\text{ determine the number of elements in }A.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:A\rightarrow B\text{ is an injection.}
\displaystyle \text{Also, the range of }f=\{a\}.
\displaystyle \text{Hence, every element of }A\text{ has the same image }a.
\displaystyle \text{Since }f\text{ is an injection, distinct elements of }A\text{ cannot have the same image.}
\displaystyle \therefore A\text{ can contain only one element.}
\displaystyle \therefore \text{The number of elements in }A=1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Show that the function }f:\mathbb R-\{3\}\rightarrow\mathbb R-\{1\}
\displaystyle \text{given by }f(x)=\frac{x-2}{x-3}\text{ is a bijection.}\quad\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R-\{3\}\rightarrow\mathbb R-\{1\}\text{ defined by}
\displaystyle f(x)=\frac{x-2}{x-3}.
\displaystyle \text{We shall show that }f\text{ is one-one and onto.}
\displaystyle \text{Check for One-One:}
\displaystyle \text{Let }x,y\in\mathbb R-\{3\}\text{ such that }f(x)=f(y).
\displaystyle \frac{x-2}{x-3}=\frac{y-2}{y-3}
\displaystyle \Rightarrow (x-2)(y-3)=(y-2)(x-3)
\displaystyle \Rightarrow xy-3x-2y+6=xy-3y-2x+6
\displaystyle \Rightarrow -x+y=0
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Let }y\in\mathbb R-\{1\}\text{ be arbitrary.}
\displaystyle \text{We have to find }x\in\mathbb R-\{3\}\text{ such that }f(x)=y.
\displaystyle \frac{x-2}{x-3}=y
\displaystyle \Rightarrow x-2=xy-3y
\displaystyle \Rightarrow xy-x=3y-2
\displaystyle \Rightarrow x(y-1)=3y-2
\displaystyle \Rightarrow x=\frac{3y-2}{y-1}.
\displaystyle \text{Since }y\ne1,\ x\text{ is a real number.}
\displaystyle \text{Also,}
\displaystyle x-3=\frac{3y-2}{y-1}-3
\displaystyle =\frac{3y-2-3y+3}{y-1}
\displaystyle =\frac{1}{y-1}\ne0.
\displaystyle \therefore x\ne3,\text{ and hence }x\in\mathbb R-\{3\}.
\displaystyle \text{Thus, every element of the co-domain has a pre-image in the domain.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Since }f\text{ is both one-one and onto,}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }A=[-1,1].\text{ Discuss whether the following functions from }
\displaystyle A\text{ to itself are one-one, onto or bijective:}
\displaystyle \text{(i) }f(x)=\frac{x}{2}\qquad \text{(ii) }g(x)=|x|\qquad \text{(iii) }h(x)=x^2
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }f:A\rightarrow A\text{ defined by }f(x)=\frac{x}{2}.
\displaystyle \text{Check for One-One:}
\displaystyle \text{Let }x,y\in A\text{ such that }f(x)=f(y).
\displaystyle \frac{x}{2}=\frac{y}{2}
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{The range of }f\text{ is }\left[-\frac12,\frac12\right].
\displaystyle \text{Since }\left[-\frac12,\frac12\right]\ne[-1,1],
\displaystyle f\text{ is not onto.}
\displaystyle \text{For example, }1\in A\text{ has no pre-image in }A,
\displaystyle \text{because }\frac{x}{2}=1\Rightarrow x=2\notin A.
\displaystyle \therefore f\text{ is one-one but not onto.}
\displaystyle \therefore f\text{ is not bijective.}
\displaystyle \\

\displaystyle \text{(ii) Given }g:A\rightarrow A\text{ defined by }g(x)=|x|.
\displaystyle \text{Check for One-One:}
\displaystyle g(1)=|1|=1
\displaystyle g(-1)=|-1|=1.
\displaystyle \text{Since }1\ne-1\text{ but }g(1)=g(-1),
\displaystyle \therefore g\text{ is not one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Since }|x|\ge0\text{ for every }x\in A,
\displaystyle \text{the range of }g\text{ is }[0,1].
\displaystyle \text{Thus, }-1\in A\text{ has no pre-image in }A.
\displaystyle \therefore g\text{ is not onto.}
\displaystyle \therefore g\text{ is neither one-one nor onto.}
\displaystyle \therefore g\text{ is not bijective.}
\displaystyle \\

\displaystyle \text{(iii) Given }h:A\rightarrow A\text{ defined by }h(x)=x^2.
\displaystyle \text{Check for One-One:}
\displaystyle h(1)=1^2=1
\displaystyle h(-1)=(-1)^2=1.
\displaystyle \text{Since }1\ne-1\text{ but }h(1)=h(-1),
\displaystyle \therefore h\text{ is not one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Since }x^2\ge0\text{ for every }x\in A,
\displaystyle \text{the range of }h\text{ is }[0,1].
\displaystyle \text{Thus, }-1\in A\text{ has no pre-image in }A.
\displaystyle \therefore h\text{ is not onto.}
\displaystyle \therefore h\text{ is neither one-one nor onto.}
\displaystyle \therefore h\text{ is not bijective.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Are the following sets of ordered pairs functions? If so, examine whether the mapping is injective or surjective.}
\displaystyle \text{(i) }\{(x,y):x\text{ is a person and }y\text{ is the mother of }x\}
\displaystyle \text{(ii) }\{(a,b):a\text{ is a person and }b\text{ is an ancestor of }a\}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }f=\{(x,y):x\text{ is a person and }y\text{ is the mother of }x\}.
\displaystyle \text{Each person has exactly one mother.}
\displaystyle \therefore f\text{ is a function.}
\displaystyle \text{Check for Injection:}
\displaystyle \text{A mother may have more than one child.}
\displaystyle \therefore \text{Distinct persons may have the same image.}
\displaystyle \therefore f\text{ is not injective.}
\displaystyle \text{Check for Surjection:}
\displaystyle \text{Every mother in the co-domain is the mother of at least one person.}
\displaystyle \therefore \text{Every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is surjective.}
\displaystyle \\
\displaystyle \text{(ii) Let }g=\{(a,b):a\text{ is a person and }b\text{ is an ancestor of }a\}.
\displaystyle \text{A person generally has more than one ancestor.}
\displaystyle \text{Hence, a single element of the domain is associated with more than one element of the co-domain.}
\displaystyle \therefore g\text{ is not a function.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Let }A=\{1,2,3\}.\text{ Write all one-one functions from }A\text{ to itself.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{1,2,3\}.
\displaystyle \text{Since }|A|=3,\text{ the number of one-one functions from }A\text{ to itself is }3!=6.
\displaystyle \text{The six one-one functions are:}
\displaystyle \text{(i) }\{(1,1),(2,2),(3,3)\}
\displaystyle \text{(ii) }\{(1,1),(2,3),(3,2)\}
\displaystyle \text{(iii) }\{(1,2),(2,1),(3,3)\}
\displaystyle \text{(iv) }\{(1,2),(2,3),(3,1)\}
\displaystyle \text{(v) }\{(1,3),(2,1),(3,2)\}
\displaystyle \text{(vi) }\{(1,3),(2,2),(3,1)\}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }f:\mathbb R\rightarrow\mathbb R\text{ is the function defined by}
\displaystyle f(x)=4x^3+7,\text{ show that }f\text{ is a bijection.}\qquad\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=4x^3+7.
\displaystyle \text{Check for One-One:}
\displaystyle \text{Let }x,y\in\mathbb R\text{ such that }f(x)=f(y).
\displaystyle 4x^3+7=4y^3+7
\displaystyle \Rightarrow 4x^3=4y^3
\displaystyle \Rightarrow x^3=y^3
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow 4x^3+7=y
\displaystyle \Rightarrow x^3=\frac{y-7}{4}
\displaystyle \Rightarrow x=\sqrt[3]{\frac{y-7}{4}}\in\mathbb R.
\displaystyle \therefore \text{Every element of the co-domain has a pre-image.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Since }f\text{ is both one-one and onto,}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Show that the exponential function }f:\mathbb R\rightarrow\mathbb R
\displaystyle \text{defined by }f(x)=e^x\text{ is one-one but not onto. What happens if the co-domain}
\displaystyle \text{is replaced by }\mathbb R_0^+\text{, the set of all positive real numbers?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=e^x.
\displaystyle \text{Check for One-One:}
\displaystyle \text{Let }x_1,x_2\in\mathbb R\text{ such that }f(x_1)=f(x_2).
\displaystyle e^{x_1}=e^{x_2}
\displaystyle \therefore x_1=x_2.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Since }e^x>0\text{ for every }x\in\mathbb R,
\displaystyle \text{the range of }f\text{ is }(0,\infty)=\mathbb R_0^+.
\displaystyle \text{But the co-domain is }\mathbb R.
\displaystyle \text{Thus, no non-positive real number has a pre-image.}
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \therefore f:\mathbb R\rightarrow\mathbb R\text{ is one-one but not onto.}
\displaystyle \text{Now, let the co-domain be replaced by }\mathbb R_0^+.
\displaystyle \text{Let }y\in\mathbb R_0^+\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow e^x=y
\displaystyle \Rightarrow x=\log y\in\mathbb R.
\displaystyle \therefore \text{Every element of }\mathbb R_0^+\text{ has a pre-image in }\mathbb R.
\displaystyle \therefore f:\mathbb R\rightarrow\mathbb R_0^+\text{ is onto.}
\displaystyle \text{Since }f\text{ is both one-one and onto,}
\displaystyle \therefore f:\mathbb R\rightarrow\mathbb R_0^+\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Show that the logarithmic function }f:\mathbb R_0^+\rightarrow\mathbb R
\displaystyle \text{defined by }f(x)=\log_a x,\ a>0,\ a\ne1,\text{ is a bijection.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R_0^+\rightarrow\mathbb R\text{ defined by }f(x)=\log_a x,\ a>0,\ a\ne1.
\displaystyle \text{Check for One-One:}
\displaystyle \text{Let }x_1,x_2\in\mathbb R_0^+\text{ such that }f(x_1)=f(x_2).
\displaystyle \log_a x_1=\log_a x_2
\displaystyle \therefore x_1=x_2.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Let }y\in\mathbb R\text{ be arbitrary.}
\displaystyle f(x)=y
\displaystyle \Rightarrow \log_a x=y
\displaystyle \Rightarrow x=a^y.
\displaystyle \text{Since }a^y>0,\ x\in\mathbb R_0^+.
\displaystyle \therefore \text{Every element of the co-domain has a pre-image in the domain.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Since }f\text{ is both one-one and onto,}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }A=\{1,2,3\},\text{ show that a one-one function }f:A\rightarrow A\text{ must be onto.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{1,2,3\}\text{ and }f:A\rightarrow A\text{ is one-one.}
\displaystyle \text{Since }f\text{ is one-one, distinct elements of }A\text{ have distinct images.}
\displaystyle \text{Thus, the three elements of }A\text{ have three distinct images.}
\displaystyle \text{But the co-domain }A\text{ also contains exactly three elements.}
\displaystyle \text{Hence, the range of }f\text{ contains all the elements of the co-domain.}
\displaystyle \therefore \text{Range}(f)=A.
\displaystyle \therefore f\text{ is onto.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }A=\{1,2,3\},\text{ show that an onto function }f:A\rightarrow A\text{ must be one-one.}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose }f\text{ is not one-one.}
\displaystyle \text{Then, two distinct elements of }A,\text{ say }1\text{ and }2,\text{ have the same image.}
\displaystyle \text{The third element }3\text{ can have only one image.}
\displaystyle \text{Hence, the range of }f\text{ can contain at most two distinct elements.}
\displaystyle \text{But the co-domain }A\text{ has three elements.}
\displaystyle \text{Therefore, }f\text{ cannot be onto, which is a contradiction.}
\displaystyle \therefore f\text{ must be one-one.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the number of all onto functions from the set}
\displaystyle A=\{1,2,3,\ldots,n\}\text{ to itself.}
\displaystyle \text{Answer:}
\displaystyle \text{The set }A\text{ contains }n\text{ elements.}
\displaystyle \text{Let }f:A\rightarrow A\text{ be an onto function.}
\displaystyle \text{Since the domain and co-domain have the same finite number of elements,}
\displaystyle \text{every onto function }f:A\rightarrow A\text{ is also one-one.}
\displaystyle \text{Thus, every onto function from }A\text{ to itself is a permutation of }A.
\displaystyle \text{The number of permutations of }n\text{ elements is}
\displaystyle n(n-1)(n-2)\cdots3\cdot2\cdot1=n!.
\displaystyle \therefore \text{The number of onto functions from }A\text{ to itself is }n!.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Give examples of two one-one functions }f_1\text{ and }f_2\text{ from }\mathbb R\text{ to }\mathbb R
\displaystyle \text{such that }f_1+f_2:\mathbb R\rightarrow\mathbb R,\text{ defined by}
\displaystyle (f_1+f_2)(x)=f_1(x)+f_2(x),\text{ is not one-one.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f_1:\mathbb R\rightarrow\mathbb R\text{ and }f_2:\mathbb R\rightarrow\mathbb R\text{ be defined by}
\displaystyle f_1(x)=x\qquad\text{and}\qquad f_2(x)=-x.
\displaystyle \text{Both }f_1\text{ and }f_2\text{ are one-one functions.}
\displaystyle \text{Now,}
\displaystyle (f_1+f_2)(x)=f_1(x)+f_2(x)
\displaystyle =x+(-x)=0.
\displaystyle \therefore f_1+f_2\text{ is a constant function.}
\displaystyle \text{For example,}
\displaystyle (f_1+f_2)(0)=0=(f_1+f_2)(1),
\displaystyle \text{although }0\ne1.
\displaystyle \therefore f_1+f_2\text{ is not one-one.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Give examples of two surjective functions }f_1\text{ and }f_2\text{ from }\mathbb Z\text{ to }\mathbb Z
\displaystyle \text{such that }f_1+f_2\text{ is not surjective.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f_1:\mathbb Z\rightarrow\mathbb Z\text{ be defined by }f_1(x)=x.
\displaystyle \text{Let }f_2:\mathbb Z\rightarrow\mathbb Z\text{ be defined by }f_2(x)=-x.
\displaystyle \text{Both }f_1\text{ and }f_2\text{ are surjective functions.}
\displaystyle \text{Now,}
\displaystyle (f_1+f_2)(x)=f_1(x)+f_2(x)=x+(-x)=0.
\displaystyle \text{Thus, }f_1+f_2\text{ is the constant function }0.
\displaystyle \text{Since, for example, }1\in\mathbb Z\text{ has no pre-image,}
\displaystyle \therefore f_1+f_2\text{ is not surjective.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Show that if }f_1\text{ and }f_2\text{ are one-one maps from }\mathbb R\text{ to }\mathbb R,
\displaystyle \text{then the product }f_1\times f_2:\mathbb R\rightarrow\mathbb R\text{ defined by}
\displaystyle (f_1\times f_2)(x)=f_1(x)f_2(x)\text{ need not be one-one.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f_1:\mathbb R\rightarrow\mathbb R\text{ be defined by }f_1(x)=x.
\displaystyle \text{Let }f_2:\mathbb R\rightarrow\mathbb R\text{ be defined by }f_2(x)=x.
\displaystyle \text{Clearly, both }f_1\text{ and }f_2\text{ are one-one functions.}
\displaystyle \text{Now,}
\displaystyle (f_1\times f_2)(x)=f_1(x)f_2(x)=x\cdot x=x^2.
\displaystyle (f_1\times f_2)(-1)=1=(f_1\times f_2)(1).
\displaystyle \text{Since }-1\ne1\text{ but they have the same image,}
\displaystyle \therefore f_1\times f_2\text{ is not one-one.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Suppose }f_1\text{ and }f_2\text{ are non-zero one-one functions from }\mathbb R\text{ to }\mathbb R.
\displaystyle \text{Is }\frac{f_1}{f_2}\text{ necessarily one-one? Justify your answer, where}
\displaystyle \left(\frac{f_1}{f_2}\right)(x)=\frac{f_1(x)}{f_2(x)}\text{ for all }x\in\mathbb R.
\displaystyle \text{Answer:}
\displaystyle \text{No, }\frac{f_1}{f_2}\text{ need not be one-one.}
\displaystyle \text{Consider }f_1:\mathbb R\rightarrow\mathbb R\text{ defined by }f_1(x)=x,
\displaystyle \text{and }f_2:\mathbb R\rightarrow\mathbb R\text{ defined by }f_2(x)=e^x.
\displaystyle \text{Both }f_1\text{ and }f_2\text{ are one-one functions, and }e^x\neq0\text{ for all }x\in\mathbb R.
\displaystyle \text{Now,}
\displaystyle \left(\frac{f_1}{f_2}\right)(x)=\frac{x}{e^x}=xe^{-x}.
\displaystyle \text{The function }xe^{-x}\text{ is not one-one on }\mathbb R.
\displaystyle \therefore \frac{f_1}{f_2}\text{ need not be one-one.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Given }A=\{2,3,4\},\ B=\{2,5,6,7\}.\text{ Construct an example of each of the following:}
\displaystyle \text{(i) an injective map from }A\text{ to }B
\displaystyle \text{(ii) a mapping from }A\text{ to }B\text{ which is not injective}
\displaystyle \text{(iii) a mapping from }A\text{ to }B.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{2,3,4\},\ B=\{2,5,6,7\}.
\displaystyle \text{(i) Let }f:A\rightarrow B\text{ be defined by}
\displaystyle f=\{(2,5),(3,6),(4,7)\}.
\displaystyle \text{Since distinct elements of }A\text{ have distinct images, }f\text{ is injective.}
\displaystyle \text{(ii) Let }g:A\rightarrow B\text{ be defined by}
\displaystyle g=\{(2,2),(3,2),(4,5)\}.
\displaystyle \text{Since }g(2)=g(3)=2,\ g\text{ is not injective.}
\displaystyle \text{(iii) One mapping from }A\text{ to }B\text{ is}
\displaystyle h=\{(2,2),(3,5),(4,5)\}.
\displaystyle \text{Thus, }h\text{ is a mapping from }A\text{ to }B.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Show that }f:\mathbb R\rightarrow\mathbb R,\text{ given by }f(x)=x-[x],\text{ is neither one-one nor onto.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R\text{ defined by }f(x)=x-[x].
\displaystyle \text{Check for One-One:}
\displaystyle f(1)=1-[1]=1-1=0.
\displaystyle f(2)=2-[2]=2-2=0.
\displaystyle \text{Since }1\ne2\text{ but }f(1)=f(2),
\displaystyle \therefore f\text{ is not one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{For every }x\in\mathbb R,\ 0\le x-[x]<1.
\displaystyle \therefore \text{Range}(f)=[0,1).
\displaystyle \text{Since the co-domain is }\mathbb R,\text{ elements such as }-1\text{ and }2\text{ have no pre-images.}
\displaystyle \therefore f\text{ is not onto.}
\displaystyle \therefore f\text{ is neither one-one nor onto.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Let }f:\mathbb N\rightarrow\mathbb N\text{ be defined by}
\displaystyle f(n)=\left\{\begin{array}{ll}n+1,&\text{if }n\text{ is odd}\\ \\ n-1,&\text{if }n\text{ is even}\end{array}\right.
\displaystyle \text{Show that }f\text{ is a bijection.}
\displaystyle \text{Answer:}
\displaystyle \text{Check for One-One:}
\displaystyle \text{Let }f(x)=f(y).
\displaystyle \text{If }f(x)=m,\text{ then }m\text{ has a unique pre-image.}
\displaystyle \text{If }m\text{ is even, then }m=f(m-1),\text{ where }m-1\text{ is odd.}
\displaystyle \text{If }m\text{ is odd, then }m=f(m+1),\text{ where }m+1\text{ is even.}
\displaystyle \text{Hence every element of the range has exactly one pre-image.}
\displaystyle \therefore x=y.
\displaystyle \therefore f\text{ is one-one.}
\displaystyle \text{Check for Onto:}
\displaystyle \text{Let }n\in\mathbb N\text{ be arbitrary.}
\displaystyle \text{Case 1: If }n\text{ is even, then }n=f(n-1),\text{ since }n-1\text{ is odd.}
\displaystyle \text{Case 2: If }n\text{ is odd, then }n=f(n+1),\text{ since }n+1\text{ is even.}
\displaystyle \text{Thus every element of }\mathbb N\text{ has a pre-image.}
\displaystyle \therefore f\text{ is onto.}
\displaystyle \text{Since }f\text{ is both one-one and onto,}
\displaystyle \therefore f\text{ is a bijection.}
\displaystyle \\

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