Question 1: Give an example  of a function

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto

(i)     $\text{ Now, } Let, f:N \rightarrow N \text{ given by } f(x) = x^2$

Check for Injectivity:

$\text{ Let } x,y \text{ be elements belongs to } N \text{ i.e } x, y \in N \text{ such that }$

$\text{ So, from definition }$

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

$\Rightarrow x^2 = y^2$

$\Rightarrow x^2 - y^2 = 0$

$\Rightarrow (x - y)(x + y) = 0$

$\text{ As } x, y \in N \text{ therefore } x + y>0$

$\Rightarrow x - y = 0$

$\Rightarrow x = y$

$\text{ Hence } f \text{ is One - One function }$

Check for Surjectivity:

$\text{ Let } y \text{ be element belongs to } N \text{ i.e } y \in N \text{ be arbitrary, then }$

$\Rightarrow f(x) = y$

$\Rightarrow x^2 = y$

$\Rightarrow x = \sqrt{y}$

$\Rightarrow \sqrt{y } \text{ not belongs to } N \text{ for non - perfect square value of } y.$

$\text{ Therefore no non-perfect square value of } y \text{ has a pre image in domain } N.$

$\text{ Hence, } f:N \rightarrow N \text{ given by } f(x) = x^2 \text{ is One - One but not onto. }$

$\\$

$\text{ (ii) Now, Let, } f: R \rightarrow R \text{ given by } f(x) = x^3 - x$

Check for Injectivity:

$\text{ Let } x,y \text{ be elements belongs to } R \text{ i.e } x, y \in R \text{ such that }$

So, from definition

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

$\Rightarrow x^3 - x = y^3 - y$

$\Rightarrow x^3 - y^3 - (x - y) = 0$

$\Rightarrow (x - y)(x^2 + xy + y^2 - 1) = 0$

$As x^2 + xy + y^2 \geq 0$

$\Rightarrow x^2 + xy + y^2 - 1\geq - 1$

$\Rightarrow x - y\neq 0$

$\Rightarrow x \neq y \text{ for some } x, y \in R$

$\text{ Hence } f \text{ is not One - One function }$

Check for Surjectivity:

$\text{ Let } y \text{ be element belongs to } R \text{ i.e } y \in R \text{ be arbitrary, then }$

$\Rightarrow f(x) = y$

$\Rightarrow x^3 - x = y$

$\Rightarrow x^3 - x - y = 0$

$\text{ Now, we know that for } 3 \text{ degree equation has a real root }$

$\text{ So, let } x = \alpha \text{ be that root }$

$\Rightarrow\alpha^3 - \alpha = y$

$f(\alpha) = y$

$\text{ Thus for clearly } y \in R, \text{ there exist } \alpha \in R \text{ such that } f(x) = y$

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

$\text{ Hence, } f: R \rightarrow R \text{ given by } f(x) = x^3 - x \text{ is not One - One but onto }$

$\\$

(iii) which is neither one-one nor onto

$\\$

Question 2:  Which of the following functions from A to B are one-one and onto?

$\textbf{(i) } f_1 = \{(1, 3), (2, 5), (3, 7 )\} ; A = \{1, 2, 3\}, B = \{3, 5, 7\}$

$\textbf{(ii) } f_2 = \{(2, a), (3,b), (4, c) \} ; A = \{2, 3, 4\}, B =\{a,b, c\}$

$\textbf{(iii) } f_3 = \{(a, x), (b, x), (c, z), (d, z)\} ; A = \{a, b, c, d\}, B = \{x, y, z\}$

(i) Now, As given,

$f_1 = \{(1, 3), (2, 5), (3, 7)\}$

$A = \{1, 2, 3\}, B = \{3, 5, 7\}$

Thus we can see that,

Check for Injectivity:

Every element of $A$ has a different image from $B$

Hence $f$ is a One – One function

Check for Surjectivity:

Also, each element of $B$ is an image of some element of $A$

Hence $f$ is Onto.

$\\$

(ii) Now, As given,

$f_2 = \{(2, a), (3, b), (4, c)\}$

$A = \{2, 3, 4\}, B = \{a, b, c\}$

Thus we can see that

Check for Injectivity:

Every element of $A$ has a different image from $B$

Hence $f$ is a One – One function

Check for Surjectivity:

Also, each element of $B$ is an image of some element of $A$

Hence $f$ is Onto.

$\\$

(iii) Now, As given,

$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}$

$A = \{a, b, c, d\}, B = \{x, y, z\}$

Thus we can clearly see that

Check for Injectivity:

Every element of $A$ does not have different image from $B$

$\text{Since, } f_3(a) = x = f_3(b) \text{ and } f_3(c) = z = f_3(d)$

Therefore $f$ is not One – One function

Check for Surjectivity:

Also each element of $B$ is not image of any element of $A$

Hence $f$ is not Onto.

$\\$

$\textbf{Question 3: Prove that the function } f:N \rightarrow N , \textbf{ defined by } \\ \\ f(x) =x^2 + x+ 1 \textbf{ is one-one but not onto. }$

$\text{ Now, } f: N \rightarrow N \text{ given by } f(x) = x^2 + x + 1$

Check for Injectivity:

$\text{ Let } x,y \text{ be elements belongs to } N \text{ i.e } x, y \in N \text{ such that }$

So, from definition

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

$\Rightarrow x^2 + x + 1 = y^2 + y + 1$

$\Rightarrow x^2 - y^2 + x - y = 0$

$\Rightarrow ( x - y )( x + y + 1) = 0$

$\text{ As } x, y \in N \text{ therefore } x + y + 1>0$

$\Rightarrow x - y = 0$

$\Rightarrow x = y$

$\text{ Hence } f \text{ is One - One function }$

Check for Surjectivity:

$y \text{ be element belongs to } N \text{ i.e } y \in N \text{ be arbitrary }$

$\text{ Since for } y > 1, \text{ we do not have any pre image in domain } N.$

$\text{ Hence, } f \text{ is not Onto function. }$

$\\$

$\textbf{ Question 4: Let } A = \{-1,0,1\} \textbf{ and } f =\{ (x, x^2) : x \in A \}. \textbf{ Show that } \\ \\ f :A \rightarrow A \textbf{ is neither one-one nor onto. }$

$\text{ Now, We have, } A = \{-1, 0, 1\} \text{ and } f = \{(x, x2) : x \in A\}.$

$\text{ To Prove: } f : A \rightarrow A \text{ is neither One - One nor onto function }$

Check for Injectivity:

$\text{ We can clearly see that } f(1) = 1 \text{ and } f( - 1) = 1$

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

$\text{ Therefore every element of } A \text{ does not have different image from } A$

$\text{ Hence } f \text{ is not One - One function }$

Check for Surjectivity:

$\text{ Since , } y = - 1 \text{ be element belongs to } A$

$\text{ i.e. } -1 \in A \text{ in co-domain does not have any pre image in domain } A.$

$\text{ Hence, } f \text{ is not Onto function. }$

$\\$

Question 5: Classify the following functions as injection, surjection or bijection:

$\textbf{(i) } f: N \rightarrow N \textbf{ given by } f(x) = x^2$

$\textbf{(ii) } f: Z \rightarrow Z \textbf{ given by } f(x) = x^2$

$\textbf{(iii) } f: N \rightarrow N \textbf{ given by } f(x) = x^3$

$\textbf{(iv) } f: Z \rightarrow Z \textbf{ given by } f(x) = x^3$

$\textbf{(v) } f: R \rightarrow R \textbf{ defined by } f(x) = |x|$

$\textbf{(vi) } f: Z \rightarrow Z \textbf{ defined by } f(x) = x^2+x$

$\textbf{(vii) }f: Z \rightarrow Z \textbf{ defined by } f(x) = x-5$

$\textbf{(viii) } f: R \rightarrow R \textbf{ defined by } f(x) = \sin x$

$\textbf{(ix) }f: R \rightarrow R \textbf{ defined by } f(x) = x^3 + 1$

$\textbf{(x) }f: R \rightarrow R \textbf{ defined by } f(x) = x^3 - x$

$\textbf{(xi) } f: R \rightarrow R \textbf{ defined by } f(x) = \sin^2 x+ \cos^2 x$

$\displaystyle \textbf{(xii) }f: Q - \{3 \} \rightarrow Q \textbf{ defined by } f(x) = \frac{2x+3}{x-3}$

$\textbf{(xiii) } f: Q \rightarrow Q \textbf{ defined by } f(x) = x^3+1$

$\textbf{(xiv) } f: R \rightarrow R \textbf{ defined by } f(x) = 5x^3+4$

$\textbf{(xv) } f: R \rightarrow R \textbf{ defined by } f(x) = 3-4x$

$\textbf{(xvi) } f: R \rightarrow R \textbf{ defined by } f(x) = 1+x^2$

$\displaystyle \textbf{(xvii) } f: R \rightarrow R \textbf{ defined by } f(x) = \frac{x}{x^2+1}$

$\text{ (i) Given as } f: N \rightarrow N, \text{ given by } f(x) = x^2$

Injection condition:

$\text{ Let } x \text{ and } y \text{ be any two elements in domain } (N), \text{ such that } f(x) = f(y).$

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

$\Rightarrow x^2 = y^2$

$\Rightarrow x = y \text{ (We do not get } \pm \text{ because } x \text{ and } y \text{ are in } N \text{ that is natural numbers) }$

$\text{ So, } f \text{ is an injection. }$

Surjection condition:

$\text{ Let }y \text{ be any element in the co-domain } (N), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } N \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^2= y$

$\Rightarrow x = \sqrt{ y }, \text{ which may not be in } N.$

$\text{ For example, if }y = 3, x = \sqrt{ 3 } \text{ is not in }N.$

$\text{ So, }f \text{ is not a surjection. }$

$\text{ Also } f \text{ is not a bijection. }$

$\\$

$\text{ (ii) Given } f: Z \rightarrow Z, \text{ given by } f(x) = x^2$

Injection condition:

$\text{ Let } x \text{ and } y \text{ be any two elements in domain } (Z), \text{ such that } f(x) = f(y).$

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

$\Rightarrow x^2 = y^2$

$\Rightarrow x = \pm y$

$\text{ Therefore }, f \text{ is not an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (Z), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } Z \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^2 = y$

$\Rightarrow x = \pm \sqrt{ y } \text{ which may not be in }Z.$

$\text{ For example, if }y = 3, x = \pm \sqrt{ 3 } \text{ is not in }Z.$

$\text{ So, } f \text{ is not a surjection. }$

$\text{ Also } f \text{ is not bijection. }$

$\\$

$\text{ (iii) Given } f: N \rightarrow N \text{ given by } f(x) = x^3$

Injection condition:

$\text{ Let } x \text{ and } y \text{ be any two elements in domain } (N), \text{ such that } f(x) = f(y).$

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

$\Rightarrow x^3 = y^3$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection }$

Surjection condition:

$\text{ Let } y \text{ be any element in the co-domain } (N), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } N \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^3= y$

$\Rightarrow x = \sqrt[3]{ y } \text{ which may not be in } N.$

$\text{ For example, if } y = 3, X = \sqrt[3]{ 3 } \text{ is not in } N.$

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (iv) Given } f: Z \rightarrow Z \text{ given by } f(x) = x^3$

Injection condition:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (Z), \text{ such that } f(x) = f(y)$

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

$\Rightarrow x^3 = y^3$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection. }$

Surjection condition:

$\text{ Let } y \text{ be any element in the co-domain } (Z), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } Z \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^3 = y$

$\Rightarrow x = \sqrt[3]{ y } \text{ which may not be in } Z$

$\text{ For example, if } y = 3, x = \sqrt[3]{ 3 } \text{ is not in } Z.$

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (v) Given } f: R \rightarrow R, \text{ defined by } f(x) = |x|$

Injection condition:

$\text{ Let } x \text{ and } y \text{ be any two elements in domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow |x| = |y|$

$\Rightarrow x = \pm y$

$\text{ Therefore, } f \text{ is not an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in co-domain } (R), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow |x| = y$

$\Rightarrow x = \pm y \in Z$

$\text{ Therefore, } f \text{ is a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (vi) Given as } f: Z \rightarrow Z, \text{ defined by } f(x) = x^2 + x$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in domain } (Z), \text{ such that } f(x) = f(y).$

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

$\Rightarrow x^2+ x = y^2 + y$

$\text{ Here, we cannot say that } x = y.$

$\text{ For example, } x = 2 \text{ and } y = - 3$

$\text{ Then, } x^2 + x = 2^2 + 2 = 6$

$y^2 + y = (-3)^2 - 3 = 6$

$\text{ Therefore, we have two numbers } 2 \text{ and } -3 \text{ in the domain } Z \\ \\ \text{ whose image is same as } 6.$

$\text{ So, } f \text{ is not an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (Z), \text{ such that } f(x) = y \text{ for some element } \\ \\ x \text{ in } Z \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^2 + x = y$

$\text{ Here, we cannot say } x \in Z.$

$\text{ For example, } y = - 4$

$\Rightarrow x^2 + x = - 4$

$\Rightarrow x^2 + x + 4 = 0$

$\displaystyle \Rightarrow x = \frac{-1 \pm \sqrt{-5} }{2} = \frac{-1 \pm i\sqrt{5}}{2} \text{ which is not in } Z.$

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{(vii) Given } f: Z \rightarrow Z, \text{ defined by } f(x) = x - 5$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (Z), \text{ such that } f(x) = f(y)$

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

$\Rightarrow x - 5 = y - 5$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (Z), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } Z \text{ (domain) }$

$\Rightarrow f(x) = y$

$\Rightarrow x - 5 = y$

$\Rightarrow x = y + 5, \text{ which is in } Z.$

$\text{ So, } f\text{ is a surjection and } f \text{ is a bijection }$

$\\$

$\text{ (viii) Given } f: R \rightarrow R, \text{ defined by } f(x) = \sin x$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow \sin x = \sin y$

$\text{ Here, } x \text{ may not be equal to } y \text{ because } \sin 0 = \sin \pi .$

$\text{ So, } 0 \text{ and } \pi \text{ have the same image 0. }$

$\text{ So, } f \text{ is not an injection. }$

Surjection test:

$\text{ Range of } f = [-1, 1]$

$\text{ Co-domain of } f = R$

Both are not same.

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (ix) Given } f: R \rightarrow R, \text{ defined by } f(x) = x^3 + 1$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow x^3+1 = y^3+ 1$

$\Rightarrow x^3 = y^3$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (R), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^3 + 1 = y$

$\Rightarrow x = \sqrt[3]{y - 1} \in R$

$\text{ So, } f \text{ is a surjection. Thus, } f \text{ is a bijection. }$

$\\$

$\text{ (x) Given } f: R \rightarrow R, \text{ defined by } f(x) = x^3 - x$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow x^3 - x = y^3 - y$

$\text{ Here, we cannot say } x = y.$

$\text{ For example, } x = 1 \text{ and } y = -1$

$\Rightarrow x^3 - x = 1 - 1 = 0$

$\Rightarrow y^3 - y = (-1)^3 - (-1) - 1 + 1 = 0$

$\text{ So, } 1 \text{ and } -1 \text{ have the same image 0. }$

$\text{ Thus, } f \text{ is not an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in co-domain } (R), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^3 - x = y$

$\text{ By observation we can say that there exist some } x \text{ in } R, \text{ such that } \\ \\ x^3 - x = y.$

$\text{ Therefore, } f \text{ is a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (xi) Given as } f: R \rightarrow R, \text{ defined by } f(x) = \sin^2 x + \cos^2 x$

Injection condition:

$f(x) = \sin^2 x + \cos^2 x$

$\sin^2 x + \cos^2 x = 1 \text{ (by formula) }$

$\text{ So, } f(x) = 1 \text{ for every } x \text{ in } R.$

Therefore, for all elements in the domain, the image is 1.

$\text{ So, } f \text{ is not an injection. }$

Surjection condition:

$\text{ Range of } f = \{ 1 \}$

$\text{ Co-domain of } f = R$

Both are not same.

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\displaystyle \text{ (xii) Given } f: Q - \{3 \} \rightarrow Q, \text{ defined by } f (x) = \frac{2x + 3}{x - 3}$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (Q - \{ 3 \}), \\ \\ \text{ such that } f(x) = f(y)$

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

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

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

$\Rightarrow 2xy - 6x + 3y - 9 = 2xy- 6y + 3x - 9$

$\Rightarrow 9x = 9y$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (Q - \{ 3 \} ), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } Q \text{ (domain). }$

$\Rightarrow f(x) = y$

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

$\Rightarrow 2x + 3 = xy - 3y$

$\Rightarrow 2x - xy = -3y - 3$

$\Rightarrow x(2 - y) = -3(y + 1)$

$\displaystyle x = \frac{3(y + 1)}{y - 1} \text{ which is not defined at } y = 2.$

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (xiii) Given } f: Q \rightarrow Q, \text{ defined by } f(x) = x^3 + 1$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (Q), \text{ such that } f(x) = f(y)$

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

$\Rightarrow x^3 + 1 = y^3 + 1$

$\Rightarrow x^3 = y^3$

$\Rightarrow x = y$

$\text{ Therefore, } f \text{ is an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in co-domain } (Q), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } Q \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow x^3 + 1 = y$

$\Rightarrow x = \sqrt[3]{y-1}, \text{ which may not be in } Q.$

$\text{ For example, if } y= 8,$

$\Rightarrow x^3 + 1 = 8$

$\Rightarrow x^3 = 7$

$\Rightarrow x = \sqrt[3]{7}, \text{ which is not in } Q.$

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\text{ (xiv) Given } f: R \rightarrow R, \text{ defined by } f(x) = 5x^3 + 4$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow 5x^3 + 4 = 5y^3 + 4$

$\Rightarrow 5x^3 = 5y^3$

$\Rightarrow x^3 = y^3$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (R), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow 5x^3 + 4 = y$

$\displaystyle \Rightarrow x^3 = \frac{y-4}{5} \in R$

$So, f \text{ is a surjection and } f \text{ is a bijection. }$

$\\$

$\text{(xv) } f: R \rightarrow R \text{ defined by } f(x) = 3-4x$

Injection test:

$\text{ Let } x \text{ and } y \text{ be any two elements in the domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow 3-4x = 3-4y$

$\Rightarrow -4x=-4y$

$\Rightarrow x = y$

$\text{ So, } f \text{ is an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (R), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow 3-4x = y$

$\displaystyle \Rightarrow 4x = 3-y \Rightarrow x = \frac{3-y}{4} \in R$

$So, f \text{ is a surjection and } f \text{ is a bijection. }$

$\\$

$\text{ (xvi) Given as } f: R \rightarrow R, \text{ defined by } f(x) = 1 + x^2$

Injection condition:

$\text{ Let } x \text{ and } y \text{ be any two elements in domain } (R), \text{ such that } f(x) = f(y)$

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

$\Rightarrow 1 + x^2 = 1 + y^2$

$\Rightarrow x^2 = y^2$

$\Rightarrow x = \pm y$

$\text{ Therefore, } f \text{ is not an injection. }$

Surjection test:

$\text{ Let } y \text{ be any element in the co-domain } (R), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R \text{ (domain). }$

$\Rightarrow f(x) = y$

$\Rightarrow 1 + x^2 = y$

$\Rightarrow x^2 = y - 1$

$\Rightarrow x = \pm \sqrt{-1} = \pm i \text{ is not in } R.$

$\text{ So, } f \text{ is not a surjection and } f \text{ is not a bijection. }$

$\\$

$\displaystyle \text{ (xvii) Given } f: R \rightarrow R, \text{ defined by } f(x) = \frac{x}{x^2+1}$

Injection condition:

$\displaystyle \text{ Let } x \text{ and } y \text{ be any two elements in domain } (R), \text{ such that } f(x) = f(y)$

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

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

$\displaystyle \Rightarrow xy^2+ x = x^2y + y$

$\displaystyle \Rightarrow xy^2 - x^2y + x - y = 0$

$\displaystyle \Rightarrow -xy(-y + x) + 1(x - y) = 0$

$\displaystyle \Rightarrow (x - y) (1 - xy) = 0$

$\displaystyle \Rightarrow x = y \text{ or } x = \frac{1}{y}$

$\displaystyle \text{ So, } f \text{ is not an injection. }$

Surjection test:

$\displaystyle \text{ Let } y \text{ be any element in co-domain } (R), \text{ such that } f(x) = y \text{ for some element } x \text{ in } R \text{ (domain). }$

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

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

$\displaystyle \Rightarrow yx^2 - x + y = 0$

$\displaystyle \Rightarrow x = \frac{1 \pm \sqrt{1-4y^2}}{2y} \text{, which may not be in R}$

$\displaystyle \text{For example, if } y = 1, \text{ then } \frac{1 \pm \sqrt{1-4y^2}}{2y} = \frac{1 \pm i\sqrt{3}}{2} , \text{ which is not in } R$

$\text{Therefore, } f \text{ is not surjection and } f \text{ is not bijection. }$

Question 6: If $f: A \rightarrow B$ is an injection such that range of $f =\{ a \}.$ Determine the number of elements in $A$.

Given that $f: A \rightarrow B$ is an injection

And also given the range of $f = \{ a \}$

Therefore, the number of images of  $f = 1$

Since, $f$  is an injection, there will be exactly one image for each element of $f .$

Therefore, number of elements in $A = 1$

$\\$

$\displaystyle \textbf{ Question 7: Show that the function } f: R - \{ 3 \} \rightarrow R - \{ 1 \} \textbf{ given by } \\ \\ f(x) = \frac{x-2}{x-3} \textbf{ is a bijection. [CBSE 2012]}$

$\displaystyle \text{ Given } f: R - \{3\} \rightarrow R - \{2\} \text{ given by } f (x) = \frac{x-2}{x-3}$

Let us show that, given function is one-one and on-to

Injectivity:

$\displaystyle \text{ Let } x \text{ and } y \text{ be any two elements in domain } (R - \{3\}), \text{ such that } f(x) = f(y)$

$\displaystyle f(x) = f(y)$

$\displaystyle \Rightarrow \frac{x-2}{x-3} = \frac{y-2}{y-3}$

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

$\displaystyle \Rightarrow xy - 3x -2y + 6 = xy - 3y -2x + 6$

$\displaystyle \Rightarrow x = y$

$\displaystyle \text{ Therefore, } f \text{ is one-one. }$

Surjectivity:

$\displaystyle \text{ Let } y \text{ be any element in co-domain } (R - \{ 2 \}), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } R -\{ 3 \} (domain).$

$\displaystyle f(x) = y$

$\displaystyle \Rightarrow (x - 2)/(x - 3) = y$

$\displaystyle \Rightarrow x - 2 = xy - 3y$

$\displaystyle \Rightarrow xy - x = 3y - 2$

$\displaystyle \Rightarrow x(y - 1) = 3y - 2$

$\displaystyle \Rightarrow x = \frac{3y-2}{y-1}, \text{ which is in } R - \{3\}$

Therefore, for every element in the co-domain, there exists some pre-image in the domain.

$\displaystyle \Rightarrow f \text{ is onto. }$

$\displaystyle \text{ Since, } f \text{ is both one-one and onto, it is a bijection. }$

$\\$

$\displaystyle \textbf{ Question 8: Let } A = [-1, 1] . \textbf{ Then, discuss whether the following functions form } \\ \\ A \textbf{ to itself are one-one, onto or bijective: }$

$\displaystyle \text{(i) } f(x) = \frac{x}{2} \hspace{1.0cm} \text{(ii) } g(x) = |x| \hspace{1.0cm} \text{(iii) } h(x) = x^2$

$\displaystyle \text{ (i) Given as } f: A \rightarrow A, \text{ given by } f(x) = \frac{x}{2}$

Let us show that the given function is one-one and on-to

Injection test:

$\displaystyle \text{ Let } x \text{ and } y \text{ be any two elements in domain } (A), \text{ such that } f(x) = f(y)$

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

$\displaystyle \Rightarrow \frac{x}{2} = \frac{y}{2}$

$\displaystyle \Rightarrow x = y$

$\displaystyle \text{ Therefore, } f \text{ is one-one. }$

Surjection test:

$\displaystyle \text{ Let } y \text{ be any element in the co-domain } (A), \text{ such that } f(x) = y \\ \\ \text{ for some element } x \text{ in } A (domain)$

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

$\displaystyle \Rightarrow \frac{x}{2} = y$

$\displaystyle \Rightarrow x = 2y, \text{ which may not be in } A.$

$\displaystyle \text{ For example, if } y = 1, \text{ then } x = 2, \text{ which is not in } A.$

$\displaystyle \text{ So,} f \text{ is not onto.}$

$\displaystyle \text{ Therefore,} f \text{ is not bijective.}$

$\\$

$\displaystyle \text{ (ii) Given } f: A \rightarrow A, \text{ given by } g (x) = |x|$

We have to show that the given function is one-one and on-to

Injection test:

$\displaystyle \text{ Let } x \text{ and } y \text{ be any two elements in the domain } (A), \text{ such that } f(x) = f(y)$

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

$\displaystyle \Rightarrow |x| = |y|$

$\displaystyle \Rightarrow x = \pm y$

$\displaystyle \text{ Therefore, } f \text{ is not one-one. }$

Surjection test:

$\displaystyle \text{ For } y = -1, \text{ there is no value of } x \text{ in } A. \text{ So, } f \text{ is not onto. So, } f \text{ is not bijective. }$

$\\$

$\displaystyle \text{ (iii) Given } f: A \rightarrow A, \text{ given by } h (x) = x^2$

Let us show that the given function is one-one and on-to

Injection test:

$\displaystyle \text{ Let } x \text{ and } y \text{ be any two elements in the domain } (A), \text{ such that } f(x) = f(y)$

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

$\displaystyle \Rightarrow x^2 = y^2$

$\displaystyle \Rightarrow x = \pm y$

$\displaystyle \text{ Therefore, } f \text{ is not one-one. }$

Surjection test:

$\displaystyle \text{ For } y = -1, \text{ there is no value of } x \text{ in }A.$

$\displaystyle \text{ So, } f \text{ is not onto. Therefore, } f \text{ is not bijective. }$

$\\$

Question 9: Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:

$\textbf{(i) } \{ (x, y) : x \textbf{ is a person } y \textbf{ is the mother of } x \}$

$\textbf{(ii) } \{ (a, b) : a \textbf{ is a person }, b \textbf{ is an ancestor of } a \}$

(i) Let $f = \{ (x, y): x \text{ is a person, } y \text{ is the mother of } x \}$

Since, for each element $x$ in domain set, there is a unique related element y in co-domain set.

Therefore, $f$ is the function.

Injection test:

As, $y$ can be mother of two or more persons

Therefore, $f$ is not injective.

Surjection test:

For every mother $y$ defined by $(x, y) ,$ there exists a person $x$ for whom $y$ is mother. So, $f$ is surjective.

Therefore, $f$ is surjective function.

(ii) Let $g = \{ (a, b): a \text{ is a person, } b \text{ is an ancestor of } a \}$

As, the ordered map $(a, b)$ does not map $'a' - a$ person to a living person.

Therefore, $g$ is not a function.

$\\$

Question 10: Let $A = \{1, 2, 3 \}.$ Write all one-one from $\bf A$ to itself.

Given as $A = \{ 1, 2, 3 \}$

The number of elements in  $A = 3$

The number of one-one functions = number of ways of arranging 3 elements $= 3! = 6$

$\text{ (i) } \{(1, 1), (2, 2), (3, 3) \}$

$\text{ (ii) } \{(1, 1), (2, 3), (3, 2) \}$

$\text{ (iii) } \{(1, 2 ), (2, 2), (3, 3 ) \}$

$\text{ (iv) } \{(1, 2), (2, 1), (3, 3) \}$

$\text{ (v) } \{(1, 3), (2, 2), (3, 1) \}$

$\text{ (vi) } \{(1, 3), (2, 1), (3,2 ) \}$

$\\$

$\textbf{Question 11: If } f : R \rightarrow R \textbf{ be the function defined by } \\ \\ f (x) =4x^3 +7 , \textbf{ show that } f \textbf{ is a bijection. [CBSE 2011] } .$

$\displaystyle \text{ Given as } f: R \rightarrow R \text{ is a function defined by } f(x) = 4x^3 + 7$

Injectivity:

$\displaystyle \text{ Let } x \text{ and } y \text{ be any two elements in domain } (R), \text{ such that } f(x) = f(y)$

$\displaystyle \Rightarrow 4x^3 + 7 = 4y^3 + 7$

$\displaystyle \Rightarrow 4x^3 = 4y^3$

$\displaystyle \Rightarrow x^3 = y^3$

$\displaystyle \Rightarrow x = y$

$\displaystyle \text{ Therefore, } f \text{ is one-one. }$

Surjectivity:

$\displaystyle \text{ Let } y \text{ be any element in co-domain } (R), \text{ such that }f(x) = y \text{ for some element } x \\ \\ \text{ in } R \text{ (domain) } f(x) = y$

$\displaystyle \Rightarrow 4x^3 + 7 = y$

$\displaystyle \Rightarrow 4x^3 = y - 7$

$\displaystyle \Rightarrow x^3 = \frac{y - 7}{4}$

$\displaystyle \Rightarrow x = \sqrt[3]{\frac{y-7}{4}} \text{ in } R$

Therefore, for every element in the co-domain, there exists some pre-image in the domain. $\displaystyle f$ is onto.

Since, $\displaystyle f$ is both one-to-one and onto, it is a bijection.

$\\$

Question 12: Show that the exponential function $f: R \rightarrow R,$ given by $f(x) = e^x,$ is one-one but not onto. What happens if the co-domain is replaced by $R_{0}^{+}$  (set of all positive real numbers).

$f: R \rightarrow R, \text{ given by } f(x) = e^x,$

$\text{Let } x_1 \text{ and } x_2 \text{ be any two elements in the domain } (R), \text{ such that } f(x_1) = f(x_2)$

$\text{Calculate } f(x_1) : \Rightarrow f(x_1) = e^{x_1}$

$\text{Calculate } f(x_2) : \Rightarrow f(x_2) = e^{x_2}$

$\text{Now, } f(x_1) = f(x_2)$

$\Rightarrow e^{x_1} = e^{x_2}$

$\Rightarrow x_1 = x_2$

Therefore $f$ is one-one function.

We know that range of $e^x \text{ is } (0,\infty)=R^+$

$\Rightarrow \text{ Co-domain } = R$

Both are not same.

Therefore $f$ is not onto function.

If the co-domain is replaced by $R^+$, then the co-domain and range become the same and in that case, $f$ is onto and hence, it is a bijection.

$\\$

Question 13: Show that the logarithmic function $f : R_{0}^{+} \rightarrow R$ given by $f (x) = \log_a x , a >0$ is a bijection.

$f: R \rightarrow R, \text{ given by } f(x) = \log_a x, a > 0,$

$\text{Let } x_1 \text{ and } x_2 \text{ be any two elements in the domain } (R), \text{ such that } f(x_1) = f(x_2)$

$\text{Calculate } f(x_1) : \Rightarrow f(x_1) = \log_a x_1$

$\text{Calculate } f(x_2) : \Rightarrow f(x_2) = \log_a x_2$

$\text{Now, } f(x_1) = f(x_2)$

$\Rightarrow \log_a x_1 = \log_a x_2$

$\Rightarrow x_1 = x_2$

Therefore $f$ is one-one function.

Let $f(x) = y$

$\Rightarrow \log_a x = y$

$\Rightarrow x = a^y$

$\Rightarrow x = a^y \in R^+$

So for every element in the co-domain there exists some pre-image in the domain.

Therefore $f$ is onto.

Therefore f is one-one and onto, therefore it is bijection.

$\\$

Question 14: . If $A = \{1, 2, 3 \},$ show that a one-one function $f :A \rightarrow A$ must be onto.

A function $f : A \rightarrow B$ is called an onto function if the range of $f$ is $B.$

In other words, if each  $b \in B$ there exists at least one $a \in A$ such that $f ( a ) = b ,$ then $f$ is an on-to function.

Here, $f : \{ 1 , 2 , 3 \} \rightarrow \{ 1 , 2 , 3 \}$

Here, $A = \{ 1 , 2 , 3 \} , B=\{1,2,3\}$

Here, as given function is one-one and for each value of $B,$ there exists a value in $A,$ given function is onto function.

$\\$

Question 15: If $A=\{1, 2, 3 \},$ show that an onto function $f:A \rightarrow A$ must be one-one.

Suppose $f$ is not one-one.

Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.

Also, the image of 3 under f can be only one element.

Therefore, the range set can have at most two elements of the co-domain $\{1, 2, 3 \}$

i.e $f$ is not an onto function,  a contradiction.

Hence, $f$ must be one-one.

$\\$

Question 16: Find the number of all onto functions from the set $A = \{ 1, 2, 3, \ldots , n \}$ to itself.

Taking set $\{ 1, 2, 3 \}$

Since f is onto, all elements of $\{ 1, 2, 3 \}$ have unique pre-image.

Total number of one-one function $= 3 \times 2 \times 1 = 6$

Since f is onto, all elements of $\{ 1, 2, 3 \ldots n \}$ have unique pre-image.

Total number of one-one function $= n \times (n-1) \times (n-2) \times \ldots 3 \times 2 \times 1 = n!$

$\\$

$\text{Question 17: Give examples of two one-one functions } f_1 \text{ and } f_2 \text{ from } R \text{ to } R \\ \\ \text{ such that } f_1+ f_2 : R \rightarrow R, \text{ defined by } (f_1+ f_2) (x) = f_1(x) + f_2 (x) \\ \\ \text{ is not one-one. }$

$\text{ Let } f_1 : R \rightarrow R \text{ and } f_2 : R \rightarrow R \text{ be two functions given by: }$

$f_1(x) = x$

$f_2(x) = - x$

$\text{We can easily verify that } f_1 \text{ and } f_2 \text{ are one-one functions. }$

$\text{ Now, } (f_1+f_2)(x) = f_1(x) + f_2(x) = x -x = 0$

$\therefore f_1 + f_2 : R \rightarrow R \text{ is a function given by } (f_1+f_2) (x) = 0$

Since $f_1 + f_2$ is a constant function, it is not one-one.

$\\$

$\text{Question 18: Give examples of two surjective function } f_1 \text{and } f_2 \text{ from } Z \text{ to } Z \\ \\ \text{ such that } f_1 + f_2 \text{ is not surjective. }$

$\text{ Let } f_1 : Z \rightarrow Z \text{ be defined by } f_1(x) = x \text{ and }$

$f_2 : Z \rightarrow Z \text{ be defined by } f_2(x) = -x$

$\text{ Then, } f_1 \text{ and } f_2 \text{ are surjective functions. }$

$\text{ Now } f_1 + f_2 : Z \rightarrow Z \text{ is given by }$

$(f_1+f_2)(x) = f_1(x) + f_2(x) = x -x = 0$

Since $f_1 + f_2$ is a constant function, it is not surjective.

$\\$

Question 19: Show that if $f_1$ and $f_2$ are one-one maps from $R$ to $R,$ then the product $f_1 \times f_2 : R \rightarrow R \text{ defined by } (f_1 \times f_2) (x) = f_1(x) f_x(x))$ need not be one-one.

$\text{ Let } f_1 : R \rightarrow R \text{ be defined by } f_1(x) = x \text{ and }$

$f_2 : R \rightarrow R \text{ be defined by } f_2(x) = x$

$\text{ Clearly, } f_1 \text{ and } f_2 \text{ are one-one functions. }$

$\text{ Now, } F = f_1 \times f_2 : R \rightarrow R \text{ is defined by }$

$F(x) = (f_1 \times f_2) (x) = f_1(x) \times f_2(x) = x^2$

$\text{ Clearly } F(-1) = 1 = F(1)$

$\text{ Therefore } F \text{ is not one-one. }$

$\text{ Hence } f_1 \times f_2 : R \rightarrow R \text{ is not one-one. }$

$\\$

Question 20: Suppose $f_1$ and $f_2$ are non-zero one-one functions from $R$ to $R.$

$\displaystyle \text{Is } \frac{f_1}{f_2} \text{ necessarily one-one? } \text{Justify your answer. Here, } \frac{f_1}{f_2} : R \rightarrow R \text{ is given by } \\ \\ \Big( \frac{f_1}{f_2} \Big) (x) = \frac{f_1(x)}{f_2(x)} \text{ for all } x \in R.$

$\displaystyle \text{ Let } f_1 : R \rightarrow R \text{ and } f_2 : R \rightarrow R \text{ be two functions given by: }$

$\displaystyle f_1(x) = x^3 \text{ and } f_2(x) = -x$

$\displaystyle \text{ Clearly, } f_1 \text{ and } f_2 \text{ are one-one functions. }$

$\displaystyle \text{ Now, } \frac{f_1}{f_2} : R \rightarrow R \text{ given by }$

$\displaystyle \Big( \frac{f_1}{f_2} \Big) (x) = \frac{f_1(x)}{f_2(x)} = x^2 \text{ for all } x \in R$

$\displaystyle \text{ Let } \frac{f_1}{f_2} = f$

$\displaystyle \therefore F : R \rightarrow R \text{ defined by } f(x) = x^2$

$\displaystyle \text{ Now, } F(1) = 1 = F(-1)$

$\displaystyle \text{ Therefore } F \text{ is not one-one }$

$\displaystyle \therefore \frac{f_1}{f_2} = R \rightarrow R \text{ is not one-one }$

$\\$

Question 21: Given $A = \{ 2, 3, 4 \} , B = \{ 2, 5, 6, 7 \}.$ Construct an example of each of the following:
(i) an injective map from $A$ to $B$
(ii) a mapping from $A$ to $B$ which is not injective
(iii) a mapping from $A$ to $B.$

Given $A = \{ 2, 3, 4 \} , B = \{ 2, 5, 6, 7 \}.$

$\text{ (i) Let } f:A \rightarrow B \text{ denote a mapping such that } f = \{ (x, y) : y = x+3 \} \\ \\ \text{ or } f = \{ (2, 5), (3, 6), (4, 7) \} \text{ which is injective mapping. }$

$\text{ (ii) Let } g:A \rightarrow B \text{ denote a mapping such that } g = \{ (2, 2), (3, 2), (4, 5) \} \\ \\ \text{ which is not an injective mapping. }$

$\text{ (iii) Let } h:B \rightarrow A \text{ denote a mapping such that } h = \{ (2, 2), (5, 3), (6, 4), (7, 4) \} \\ \\ \text{ which is one of the mappings from } B \text{ and } A.$

$\\$

Question 22: Show that $f : R \rightarrow R, \text{ given by } f (x) = x - [x],$ is neither one-one nor onto.

$\text{ We have } f: R \rightarrow R \text{ given by } f(x) = x - [x]$

Now, check for Injection:

$\because f(x) = x - [x] \Rightarrow f(x) = 0 \text{ for } x \in Z$

$\therefore f \text{ is not one-one where as many-one. }$

Now, check for Surjection:

$\text{ Again Range of } f = [0, 1) \neq R$

$\therefore f \text{ is an into function. }$

$\text{ Therefore } f \text{ is neither one-one nor onto. }$

$\\$

Question 23: Let $f: N \rightarrow N$ be defined by

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

Show that $f$ is a bijection.

We have,

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

Injection Test:

Case 1: If $n$ is odd

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

$As f(x) = f(y)$

$\Rightarrow x+1 = y+ 1$

$\Rightarrow x = y$

Case 2: If $n$ is even

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

$\text{ As } f(x) = f(y)$

$\Rightarrow x-1 = y- 1$

$\Rightarrow x = y$

Hence, $f$ is invective.

Surjection Test:

$\text{ As for every } n \in N, \text{ there exists } y = n+1 \text{ in } N \text{ such that } \\ \\ f(y) = f(n+1) = n+1 -1 = n$

Hence, $f$ is surjective.

Therefore $f$ is a bijection.