Question 1: Find whether the following functions are one-one or not:

$\text{(i) } f : R \rightarrow R \text{ given by } f(x)= x^3 + 2 \text{ for all } x \in R$

$\text{(ii) } f : Z \rightarrow Z \text{ given by } f(x)= x^2 + 1 \text{ for all } x \in Z$

$\text{(i) Let } x, y \text{ be two arbitrary elements of } R (\text{domain of } f) \\ \text{ such that } f (x) = f (y). \text{ Then, }$

$f(x) = f( y) \Rightarrow x^3+ 2 = y^3 + 2 \Rightarrow x^3 = y^3 \Rightarrow x = y$

Hence, $f$ is a one-one function from $R$ to itself.

$\text{(ii) Let } x, y \text{ be two arbitrary elements of } Z \text{ such that } f (x) = f(y). \text{ Then, }$

$f(x) = f( y) \Rightarrow x^2 + 1 = y^2 + 1 \Rightarrow x^2 = y^2 \Rightarrow x = \pm y$

Here, $f(x) = f(y)$ does not provide the unique solution $x =y$ but it provides $x =\pm y .$ Hence, $f$ is not a one-one function.

$\text{Infact, } f (2) = 2^2 + 1= 5 \text{ and } f(- 2) = (- 2)^2 + 1=5. \text{ Hence, } 2 \text{ and, } 1 \\ \text{ are two distinct elements having the same image. }$

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Question 2: Show that the function $f : Z \rightarrow Z$ defined by $f(x)= x^2 + x$ for all $x \in Z$ is a many-one function.

$\text{Let } x , y \in Z. \text{Then, } f(x)=f(y)$

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

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

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

$\Rightarrow x = y \text{ or } y = - x - 1$

$\text{Since, } f(x) = f(y) \text{ does not provide the unique solution } x=y \\ \\ \text{but it also provides } y= - x -1. \\ \\ \text{This means that } x \neq y \text{ but } f(x)=f(y) \text{ when } y =-x-1.$

$\text{For example, if we put } x =1 \text{ in } y = - x - 1 \text{ we obtain } y = - 2.$

$\text{This shows that } 1 \text{ and } -2 \text{ have the same image under } f. \\ \\ \text{Hence, } f \text{ is a many-one function. }$

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Question 3:  Discuss the surjectivity of the following functions:

$\text{(i) } f : R \rightarrow R \text{ given by } f(x) = x^3 + 2 \text{ for all } x \in R.$

$\text{(ii) } f :R \rightarrow R \text{ given by } f(x) : x^2 + 2 \text{ for all } x \in R.$

$\text{(iii) } f : Z \rightarrow Z \text{ given by } f(x) = 3x + 2 \text{ for all } x \in Z.$

(i) Let $y$ be an arbitrary element of $R$. Then,

$\displaystyle f(x) = y \Rightarrow x^3 + 2 = y \Rightarrow x = ( y-2)^{\frac{1}{3}}$

$\displaystyle \text{Clearly, for all } y \in R, ( y-2)^{\frac{1}{3}} \text{ for all } x \in R.$

Hence, negative real numbers in R(co-domain) do not have their pre-images in R(domain). Hence, f is not an onto function.

$\displaystyle \text{(ii) Clearly, } f(x) = x^2 + 2 \geq 2 \text{ for all } x \in R.$

Hence, negative real numbers in $R$(co-domain) do not have their pre-images in $R$(domain). Hence, $f$ is not an onto function.

(iii) Let y be an arbitrary element of $Z$(co-domain). Then,

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

$\displaystyle \text{Clearly, if } y = 0 , \text{ then } x = \frac{-2}{3} \notin Z.$

Thus, $y =0 \in Z$ does not have its pre-image in Z(domain). Hence, $f$ is not an onto function.

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Question 4: Show that the function $f:N \rightarrow N \text{ given by } f(1)=f(2)=1$ and $f (x)=x-1$ for every $x \geq 2$, is onto but not one-one.

It is given that

$\displaystyle f(x) = \Bigg\{ \begin{array}{ll} 1, & x = 1, 2 \\ \\ x-1, & x \geq 2\end{array}$

$\text{Clearly, } f (1) = f (2)=1 \text{ i.e. } 1 \text{ and } 2 \text{ have the same image. }$

$\text{Hence, } f : N \rightarrow N \text{is a many-one function }$

Let $y$ be an arbitrary element in $N$(co-domain). Then,

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

$\text{Clearly, } y +1 \in N \text{(domain) } \text{ for all } y \in N \text{(Co-domain). }$

$\text{Thus, for each } y \in N \text{(co-domain) there exists } y + 1 \in N \\ \text{(domain) } \text{ such that } f (y +1) =y + 1 - 1 =y.$

$\text{Hence, } f : N \rightarrow N \text{ is an onto function. }$

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Question 5: Show that the Signum function $f : R \rightarrow R$, given by

$\displaystyle f(x) = \Bigg\{ \begin{array}{ll} 1, & \text{ if } x > 0 \\ 0, & \text{ if } x = 0 \\ -1, & \text{ if } x < 0 \end{array}$

is neither one-one nor onto.

Clearly, all positive real numbers have the same image equal to 1.

Hence, $f$ is a many-one function.

We observe that the range of $f \text{ is } \{-1,0,1 \}$ which is not equal to the co-domain of $f$ .

Hence,  $f$ is not onto.

Hence, $f$ is neither one-one nor onto

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Question 6: Prove that the function f : Q \rightarrow Q given by f (x) = 2x – 3 for all x \in Q is a bijection.

We observe the following properties of $f$.

Injectivity:  Let $x, y$ be two arbitrary elements in $Q$. Then,

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

$\text{Thus, } f(x) = f(y)\Rightarrow x = y \text{ for all } x,y \in Q.$

Hence, $f$ is an injective map.

Surjectivity:  Let $y$ be an arbitrary element of $Q$. Then,

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

$\displaystyle \text{Clearly, for all } y \in Q, x = \frac{y+3}{2} \in Q.$

Thus, for all $y \in Q$(co-domain) there exists $x \in Q$(domain).

$\displaystyle \text{Given by } x = \frac{y+3}{2} \text{ such that } f(x) = f (\frac{y+3}{2}) = 2(\frac{y+3}{2}) - 3 = y$

That is every element in the co-domain has its pre-image in $x$.

Hence, $f$ is a surjection.

$\text{Hence } f : Q \rightarrow Q \text{ is a bijection. }$

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$\text{Question 7: Show that the function } f : R \rightarrow R \\ \\ \text{ defined by } f (x) : 3x^3 + 5 \text{ for all } x \in R \text{ is a bijection. }$

We observe the following properties of $f.$

Injectivity: Let $x, y$ be any two elements of $R$(domain). Then,

$f(x) = f(y)= 3x^3 +5 = 3y^3 +5 \Rightarrow x^3 = y^3 \Rightarrow x = x$

$\text{Thus, } f(x) = f(y) \Rightarrow x = y \text{ for all } x,y \in R.$

Hence, $f$ is an injective map.

Surjectivity: Let $y$ be an arbitrary element of $R$(co-domain). Then,

$\displaystyle f(x) = y \Rightarrow 3x^3 + 5 = y \Rightarrow x^3 = \frac{y-5}{3} \Rightarrow x = \Big( \frac{y-5}{3} \Big)^{1/3}$

$\displaystyle \text{Thus, we find that for all } y \in R \text{(co-domain) there exists } \\ \\ x = \Big( \frac{y-5}{3} \Big)^{1/3} \in R \text{(domain) such that }$

$\displaystyle f(x) = f \Bigg( \Big( \frac{y-5}{3} \Big)^{1/3} \Bigg) = 3\Bigg( \Big( \frac{y-5}{3} \Big)^{1/3} \Bigg)^3 + 5 = y - 5+5 = y$

This shows that every element in the co-domain has its pre-image in the domain. Hence, $f$ is a surjection.

Hence, $f$ is a bijection.

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Question 8: Let $A =\{ x \in R : -1 \leq x \leq 1 \} = B$. Show that $f : A \rightarrow B$ given by $f(x) = x |x|$ is a  bijection.

We observe the following properties of $f$.

Injectivity: Let $x, y$ be any two elements in $A$. Then,

$x \neq y \Rightarrow x |x| \neq y|y| \Rightarrow f(x) \neq f(y)$

Hence, $f : A \rightarrow B$ is an injective map.

Surjectivity: We have

$\displaystyle f(x) = x |x| = \Bigg\{ \begin{array}{rr} x^2, & \text{ if } x \geq 0 \\ \\ -x^2, & \text{ if } x < 0 \end{array}$

If $0 \leq x \leq 1, \text{ then } f (x)=x^2$ takes all values between 0 and 1 including these two points.

Also, if $-1 \leq x < 0, \text{ then } f(x)=-x^2$ takes all values between -1 and 0 including -1. Therefore, $f(x)$ takes every value between – 1 and 1 including – 1 and 1. Hence, range of $f$ is same as its co-domain.

$\text{Hence, } f: A \rightarrow B \text{ is an onto function. }$

$\text{Thus, } f : A \rightarrow B \text{ is both one-one and onto. }$

Hence, it is a bijection.

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Question 9: Let A be the set of all 50 students of class XII in a central school. Let $f : A\rightarrow N$ be a function defined by $f (x) =$ Roll number of student $x$. Show that $f$ is one-one but not onto.

Here, $f$ associates each students to his (her) roll number. Since no two different students of the class can have the same roll number.

Therefore, $f$ is one-one.

We observe that $f (A)= \text{ Range of } f =\{1,2,3,...,50)\} \neq N$ i.e. range of $f$ is not same as its co-domain. Hence, $f$ is not onto.

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Question 10: Prove that $f : R \rightarrow R, \text{ given by } f (x) = 2x,$ is one-one and onto

We observe the following properties of $f$.

$\text{Injectivity: Let } x_1, x_2 \in R \text{ such that } f(x_1)=f(x_2). \text{ Then, }$

$f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_ 2$

Hence, $f: R \rightarrow R$ is one-one.

Surjectivity: Let $y$ be any real number in $R$(co-domain). Then,

$\displaystyle f(x) = y \Rightarrow 2x = y \Rightarrow x = \frac{y}{2}$

$\displaystyle \text{Clearly, } \frac{y}{2} \in R \text{ for any } y \in R \text{ such that } f \Big(\frac{y}{2}\Big) = 2\Big(\frac{y}{2}\Big) = y.$

Thus, for each $y \in R$(co-domain)  has its pre-image in domain.

This means that each element in co-domain has its pre-image in domain.

Hence, $f: R \rightarrow R$ is onto.

Hence, $f : R \rightarrow R$ is a bijection.

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Question 11: Show that the function $f :R \rightarrow R, \text{ defined as } f (x)=x^2$, is neither one-one nor onto.

We observe that $1 \text{ and } -1 \in R \text{ such that } f (- 1) = f (1)$ i.e. there are two distinct elements in $R$ which have the same image. Hence, f is not one-one.

Since $f (x )$ assumes only non-negative values. Hence, no negative real number in $R$(co-domain) has its pre-image in domain of $f \text{ i.e. } R$. Consequently $f$ is not onto.

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Question 12: Show that $f : R \rightarrow R, \text{ defined as } f (x)=- x^3$, is a bijection.

We observe the following properties of $f$.

Injectivity: Let $x,y \in R$ such that $f(x) = f (y).$ Then,

$f(x)=f(y) \Rightarrow x^3 =y^3 \Rightarrow x=y$

Hence, $f: R \rightarrow R$ is one-one.

Surjectivity: Let $y \in R$(co-domain). Then,

$f(x)=y \Rightarrow x^3=y \Rightarrow x - y^{1/3}$

Clearly, $y^{1/3} \in R$(domain) for all $y \in R$(co-domain).

Thus, for each $y \in R$(co-domain) there exists $x=y^{1/3} \in R$(domain) such that $f(x)=x^3 =y.$

Hence, $f: R \rightarrow R$ is onto.

Hence, $f :R \rightarrow R$ is a bijection.

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Question 13: Show that the function $f : R_0 \rightarrow R_0, \text{ defined as } f (x) = \frac{1}{x},$  is one-one onto, where $R_0$ is the set of all non-zero real numbers. ls the result true, if the domain $R_0$ is replaced by $N$ with co-domain being same as $R_0$?

We observe the following properties of $f$.

Injectivity: Let $x, y \in R_0$ such that $f (x) = f (y$). Then,

$\displaystyle f(x) = f(y) \Rightarrow \frac{1}{x} = \frac{1}{y} \Rightarrow x = y$

Hence, $f: R_0 \rightarrow R_0$ is one-one

Surjectivity: Let $y$ be an arbitrary element of $R_0$(co-domain) such that $f(x) = y$. Then,

$\displaystyle f(x) = y \Rightarrow \frac{1}{x} = y \Rightarrow x = \frac{1}{y}$

$\displaystyle \text{Clearly, } x = \frac{1}{y} \in R_0 \text{(domain) for all } y \in R_0 \text{(co-domain). }$

$\displaystyle \text{Thus, for each } y \in R_0 \text{(co-domain) there exists } x = \frac{1}{y} \in R_0 \\ \text{ (domain) such that } f(x) \frac{1}{x} = y.$

Hence, $f: R_0 \rightarrow R_0$ is onto.

Hence, $f:R_0 \rightarrow R_0$ is one-one onto.

$\text{Let us now consider } f:N \rightarrow R_0 \text{ given by } f(x) = \frac{1}{x}$

For any $x, y \in N$, we find that

$\displaystyle f(x) = f(y) \Rightarrow \frac{1}{x} = \frac{1}{y} \Rightarrow x = y$

Hence, $f: N \rightarrow R_0$ is one-one.

$\displaystyle \text{We find that } \frac{2}{3}, \frac{3}{5} \text{ etc. in co-domain } R_0 \text{ do not have their pre-image in domain } N.$

$\text{Hence, } f:N \rightarrow R_0 \text{ is not onto. }$

$\text{Thus, } f:N \rightarrow R_0 \text{ is one-one but not onto. }$

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Question 14: Prove that the greatest integer function $f : R \rightarrow R$, given by $f (x) =[x]$ , is neither one-one nor, where $[x]$ denotes the greatest integer less than or equal to $x$.

$\text{We observe that } f (x) =0 \text{ for all } x \in [0, 1) \text{ Hence, } f :R \rightarrow R \text{ is not one-one }.$

Also, $f: R \rightarrow R$ does not attain non-integral values. Therefore, non-integer points in $R$(co-domain) do not have their pre-images in the domain. Hence, $f : R \rightarrow R$ is not onto.

Hence, $f: R \rightarrow R$ is neither one-one nor onto.

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Question 15: show that the modulus function $f : R \rightarrow R, \text{ given by } f (x) =|x|$ is neither one-one nor onto.

We observe that $f (- 2) = f (2)$. Hence, f is not one-one.

Also, $f(x) = |x|$ assumes only non-negative values. Hence, negative real numbers in $R$(co-domain) do not have their pre-images in $R$(domain).

Hence, $f$ is neither one-one nor onto

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Question 16: Let $C$ and $R$ denote the set of all complex numbers and all real numbers respectively. Then show that $f : C \rightarrow R$ given by $f(z) = |z|$ for all $z \in C$ is neither one-one nor onto.

Injectivity: We find that $z_1=1 -i$ and $z_2 = 1 + i$ are two distinct complex numbers in $C$ such that $|z_1| = |z_2| \text{ i.e. } z_1 \neq z_2 \text{ but } f (z_1) = f (z_2).$

This shows that different elements in $C$ may have the same image. Hence, $f$ is not an injection.

Surjectivity: $f$ is not a surjection, because negative real numbers in $R$ do not have their pre-images in $C$. In other words, for every negative real number a there is no complex number $z \in C$ such that $f (z) =|z| = a.$

Hence, $f$ is not a surjection.

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Question 17: Show that the function $f:R \rightarrow R \text{ given by } f(x)=ax+b, \text{ where } a,b \in R, a \neq 0$ is a bijection.      [CBSE 2010]

Injectivity: Let $x, y$ be any two real numbers. Then,

$f(x) = f(y) \Rightarrow ax+b = ay+b \Rightarrow ax = ay \Rightarrow x = y$

$\text{Thus, } f(x) = f(y) \Rightarrow x = y \text{ for all. } x,y \in R \text{(domain). }$

Hence, $f$ is an injection.

Surjectivity:  Let $y$ be an arbitrary element of $R$(co-domain). Then,

$\displaystyle f(x) = y \Rightarrow ax+b = y \Rightarrow x = \frac{y-b}{a}$

$\displaystyle \text{Clearly, } x = \frac{y-b}{a} \in R \text{(domain) for all } y \in R \text{(co-domain). }$

$\displaystyle \text{Thus, for all } y \in R \text{(co-domain) there exists } x = \frac{y-b}{a} \in R \text{(domain) such that}.$

$\displaystyle f(x) = f (\frac{y-b}{a}) = a(\frac{y-b}{a}) + b = y$

This shows that every element in co-domain has its pre-image in domain. Hence, f is a surjection.

Hence, $f$ is a bijection.

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Question 18: Show that the function $f : R \rightarrow R \text{ given by } f(x) = \cos x \text{ for all } x \in R$, is neither one-one nor onto.

Injectivity: We know that $f(0) = \cos 0 = 1, \text{ and } f (2\pi) = \cos 2\pi =1.$

hence, different elements in $R$ may have the same image. Hence, $f$ is not an injection.

Surjectivity: Since the values of $cos x$ lie between $-1$ and $1$, it follows that the range of $f(x)$ is not equal to its co-domain. Hence, f is not a surjection.

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$\displaystyle \text{Question 19: Let } A = R- \{ 2 \} \text{ and } B = R - \{ 1 \}. \text{ If } f :A \rightarrow B \text{ is a mapping} \\ \\ \text{defined by } f(x) = \frac{x-1}{x-2} , \text{ show that f is bijective. }$

Injectivity: Let $x, y$ be any two elements of $A$. Then,

$f(x) = f(y)$

$\displaystyle \Rightarrow \frac{x-1}{x-2} = \frac{y-1}{y-2} \\ \\ \Rightarrow (x-1)(y-2) = (x-2)(y-1) \\ \\ \Rightarrow xy - y - 2x + 2 = xy - x - 2y + 2 \\ \\ \Rightarrow x = y$

$\text{Thus, } f(x) = f(y) \Rightarrow x = y \text{ for all } x , y \in A.$

Hence, $f$ is an injective.

Surjectivity: Let $y$ be an arbitrary element of $B$. Then,

$\displaystyle f(x) = y \Rightarrow \frac{x-1}{x-2} = y \Rightarrow (x-1) = y(x-2) \Rightarrow x = \frac{1-2y}{1-y}$

$\displaystyle \text{Clearly, } x = \frac{1-2y}{1-y} \text{ is a real number for all } y \neq 1.$

$\displaystyle \text{Also, } \frac{1-2y}{1-y} \neq 2 \text{ for any } y , \text{ for, it we take } \frac{1-2y}{1-y} = 2,\text{ then we get } \\ \\ 1 = 2 \text{ which is not possible. }$

$\displaystyle \text{Thus, every element } y \text{ in } B \text{ has its pre-image } x \text{ in } A \text{ given by } x = \frac{1-2y}{1-y} .$

Hence, $f$ is a surjective.

Hence, $f$ is a bijective.

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Question 20: Let $A$ and $B$ be two sets. Show that $f :A \times B \rightarrow B \times A$ defined by $f (a,b)=f(b,a)$ is a bijection.

Injectivity: Let $(a_1, b_1)$ and $(a_2, b_2) \in A \times B$ such that

$f (a_1, b_1) = f (a_2,b_2)$

$\Rightarrow (b_1, a_1) = (b_2, a_2)$

$\Rightarrow b_1 = b_2 \text{ and } a_1 = a_2$

$\Rightarrow (a_1, b_1) = (a_2, b_2)$

Thus, $f(a_1, b_1) = f( a_2, b_2) \Rightarrow (a_1, b_1) = (a_2, b_2) \text{ for all } ( a_1, b_1), (a_2, b_2) \in A \times B.$

Hence, $f$ is an injective map.

Surjectivity: Let $(b, a)$ be an arbitrary element of $B \times A$. Then,

$b \in B \text{ and } a \in A\Rightarrow (a,b) \in A \times B.$

Thus, for all, $(b, a) \in B \times A$ there exists $(a,b) \in A \times B$ such that $f (a, b)=(b, a).$

Hence, $f : A \times B \rightarrow B \times A$ is an onto function.

Hence, $f$ is a bijection.

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Question 21: Let $A$ be any non-empty set. The, prove that the identity function on set $A$ is a bijection.

The identity function $I_A \rightarrow A$ is defined as

$I_A(x) = x \text{ for all } x \in A.$

Injectivity: Let $x, y$ be any two elements of $A$. Then,

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

Hence, $I_A$ is an injective map.

Surjectivity: Let $y \in A$. Then, there exists $x =y \in A$ such that

$I_A(x) = x = y$

Hence, $I_A$ is a surjective map.

Hence, $I_A: A \rightarrow A$ is a bijection.

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Question 22: Let $A = \{1, 2 \}$. Find all one-to-one functions from $A$ to $A$.

Let $f : A \rightarrow A$ be a one-one function.

Then, $f (1 )$ has two choices, namely, 1 or 2.

$\text{Hence, } f(1) :1 \text{ or } f (1) = 2.$

Case 1: When $f(1) = 1:$

As $f : A \rightarrow A$ is one-one. Therefore, $f (2) =2.$

Thus, we have $f(1) =1 \text{ and } f (2) =2.$

Case 2: When $f (1) =2:$

Since $f : A \rightarrow A$ is one-one. Therefore, $f (2) =1.$

Thus, in this case, we have $f (1)=2 \text{ and } f (2)=1$

Hence, there are two one-one functions say $f \text{ and } g$ from $A \text{ to } A$ given by

$f (1)=1, f (2) =2 \text{ and, } g (1) = 2, g (2) =1.$

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Question 23: Consider the identity function $I_N: N \rightarrow N$ defined as, $I_N(x) =x$ for all $x \in N$. Show that although $I_N$ is onto but $I_N+I_N : N \rightarrow N$ defined as $(I_N +I_N)(x) =I_N(x) + I_N(x) = x + x = 2x$

We know that the identity function on a given set is always a bijection. Therefore, $I_N : N \rightarrow N$ is onto. We have,

$(I_N +I_N)(x) = 2x \text{ for all } x \in N$

This means that under $I_N+I_N,$ images of natural numbers are even natural numbers.

Hence, odd natural numbers in $N$ (co-domain) do not have their pre-images in domain $N$.

For example,1,3,5 etc. do not have their pre-images. Hence, $I_N + I_N : N \rightarrow N$ is not onto.

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$\displaystyle \text{Question 24: Consider the function } f : \Big[ 0, \frac{\pi}{2} \Big] \rightarrow R \text{ given by } f(x) = \sin x \\ \text{ and } g : \Big[0, \frac{\pi}{2} \Big] \rightarrow R \text{ given by } g(x) = \cos x . \text{ Show that } f \text{ and } g \\ \text{ are one-one, but } f + g \text{ is not one-one. }$

$\displaystyle \text{We observe that for any two distinct elements } x_1 \text{ and } x_2 \text{ in } \Big[ 0, \frac{\pi}{2} \Big]$

$\sin x_1 \neq \sin x_2 \text{ and } \cos x_1 \neq \cos x_2$

$\Rightarrow f(x_1) \neq f(x_2) \text{ and } g(x_1) \neq g(x_2)$

$\Rightarrow f \text{ and } g \text{ are one-one. }$

$\text{We have } (f+g) (x) = f(x) + g( x) = \sin x + \cos x$

$\displaystyle \Rightarrow (f+g) (0) = \sin 0 + \cos 0 = 1 \text{ and } (f+g) \Big(\frac{\pi}{2}\Big) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1$

$\displaystyle \text{Thus, } 0 \neq \frac{\pi}{2} \text{ but } (f+g) (0) = (f+g) \Big(\frac{\pi}{2}\Big).$

Hence, $f+g$ is not one-one.

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Question 25: Let $f:X \rightarrow Y$ be a function. Define a relation $R$ on $X$ given by $R= \{ (a, b) :f (a) = f(b) \}$.

Show that $R$ is an equivalence relation on $X.$                          [CBSE 2010]

We observe the following properties of relation R.

Reflexivity: For any $a \in X$, we have

$f(a)=f(a) \Rightarrow (a,a) \in R \Rightarrow R \text{ is reflexive.}$

Symmetry: Let $a, b \in X$ be such that $(a, b) \in R.$ Then,

$(a, b) \in R \Rightarrow f (a)=f (b) \Rightarrow f (b)=f (a) \Rightarrow (b,a) \in R$

Hence, $R$ is symmetric.

Transitivity: Let $a, b, c \in X$ be such that $(a, b) \in R \text{ and } (b, c) \in R.$ Then,

$(a, b) \in R \text{ and } (b, c) \in R$

$\Rightarrow f (a) = f (b) \text{ and } f (b)=f (c)$

$\Rightarrow f (a) = f (c)$

$\Rightarrow (a, c) \in R$

Hence, $R$ is transitive.

Hence, $R$ is an equivalence relation.

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$\displaystyle \text{Question 26: Show that the function } f:R \rightarrow \{ x \in R : -1 < x < 1 \} \\ \\ \text{ defined by } f(x) = \frac{x}{1+|x|}, x \in R \text{ is one-one onto function. }$

We have,

$\displaystyle f(x) = \frac{x}{1+|x|} = \Bigg\{ \begin{array}{rr} \frac{x}{1+x}, & \text{ if } x \geq 0 \\ \\ \frac{x}{1-x}, & \text{ if } x < 0 \end{array}$

$\displaystyle \text{Case 1: When } x \geq 0$

$\displaystyle \text{In this case we have } f(x) = \frac{x}{1+x}$

$\displaystyle \text{Injectivity: Let } x, y \in R \text{ such that } x \geq 0, y \geq 0. \text{ Then, }$

$\displaystyle f(x) = f(y) \Rightarrow \frac{x}{1+x} = \frac{y}{1+y} \Rightarrow x + xy = y + xy \Rightarrow x = y$

Hence, $f$ is an injective map.

Surjectivity: When $x \geq 0$, we have

$\displaystyle \text{In this case we have } f(x) = \frac{x}{1+x} \geq 0 \text{ and } f(x) < 1$

Let $y \in [0, 1)$ be any real number. Then,

$\displaystyle f(x) = y \Rightarrow \frac{x}{1+x} = y \Rightarrow x = \frac{y}{1-y}$

$\displaystyle \text{Clearly, } x \geq 0 \text{ for all } y \in [0,1). \text{ Thus, for each } y \in [0, 1) \text{ there exists } \\ \\ x = \frac{y}{1-y} \geq \text{ such that } f(x) = y.$

Hence, $f$ is an onto function from $[0, 1) \text{ to } [0, 1)$

$\displaystyle \text{Case 2: When } x < 0$

$\displaystyle \text{In this case we have } f(x) = \frac{x}{1-x}$

$\displaystyle \text{Injectivity: Let } x, y \in R \text{ such that } x < 0, y < 0. \text{ Then, }$

$\displaystyle f(x) = f(y) \Rightarrow \frac{x}{1-x} = \frac{y}{1-y} \Rightarrow x - xy = y - xy \Rightarrow x = y$

Hence, $f$ is an injective map.

Surjectivity: When $x < 0$, we have

$\displaystyle \text{In this case we have } f(x) = \frac{x}{1-x} < 0$

$\displaystyle \text{Also, } f(x) = \frac{x}{1-x} = -1 + \frac{1}{1-x} > - 1$

$\displaystyle \therefore -1 < f(x) < 0$

Let $y \in (- 1, 0)$ be an arbitrary real number such that $f (x) = y$. Then,

$\displaystyle f(x) = y \Rightarrow \frac{x}{1-x} = y \Rightarrow x = \frac{y}{1+y}$

$\displaystyle \text{Clearly, } x < 0 \text{ for all } y \in (-1,0). \text{ Thus, for each } y \in (-1,0) \text{ there exists } \\ \\ x = \frac{y}{1+y} < 0 \text{ such that } f(x) = y.$

Hence, $f$ is an onto function from $(-1, 1) \text{ to } (-1, 1)$

Hence, $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ is a one-one onto function.

$\\$

Question 27: Show that the function $f :R \rightarrow R \text{ given by } f (x): x^3 + x$ is a bijection.

Injectivity: Let $x, y \in R$ such that

$f(x) = f(y)$

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

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

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

$\because x^2 + xy + y^2 \geq 0 \text{ for all } x, y \in R \\ \\ \therefore x^2 + xy + y^2+ 1 \geq 1 \text{ for all } x, y \in R$

$\Rightarrow x-y = 0$

$\Rightarrow x = y$

$\text{Thus, } f(x) =f(y) \Rightarrow x=y \text{ for all } x, y \in R$

Hence, $f$ is an injective map.

Surjective: Let $y$ be an arbitrary element of $R$. Then,

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

We know w that an odd degree equation has at least one real root. Therefore, for every real value of $y$  the equation $x^3 + x - y = 0$ has a real root $\alpha$ such that

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

Thus, for every $y \in R$ there exists $\alpha \in R$ such that $f(\alpha) =y$. Hence, $f$ is a surjective map.

Hence, $f:R \rightarrow R$ is a bijection.

$\\$

Question 28: Show that $f:N \rightarrow N$ defined by

$\displaystyle f(n) = \Bigg\{ \begin{array}{rr} \frac{n+1}{2}, & \text{ if } n \text{ is odd} \\ \\ \frac{n}{2}, & \text{ if } n \text{ is even} \end{array}$

is many-one onto function.                                              [CBSE 2009]

We observe that

$\displaystyle f(1) = \frac{1+1}{2} = 1 \text{ and } f(2) = \frac{1}{2} = 1$

Thus $1, 2 \in N \text{ such that } 1 \neq 2 \text{ but } f (1)=f (2)$. Hence, f is a many-one function.

Surjectivity: Let $n$ be an arbitrary element of $N$.

If n is an odd natural number, then $2n-1$ is also an odd natural number such that

$\displaystyle f(2n-1) = \frac{2n-1+1}{2} = n$

If n is an even natural number, then $2n$ is also an even natural number such that

$\displaystyle f(2n) = \frac{2n}{2} = n$

Thus, for every $n \in N$ (whether even or odd) there exists its pre-image in $N$. Hence, f is a surjection.

Hence, $f$ is a many-one onto function.

$\\$

$\text{Question 29: Show that the function } f : N \rightarrow N \text{ given by, } \\ \\ f (n) = n - (- 1)^n \text{ for all } n \in N \text{ is a bijection. }$

$\text{We have, } f (n) = n-(-1)^n \text{ for all } n \in N$

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

lnjectivity: Let $n, m$ be any two even natural numbers. Then,

$f (n) = f (m) \Rightarrow n-1=m-1 \Rightarrow n = m$

If $n, m$ are any two odd natural numbers. Then,

$f (n) = f (m) \Rightarrow n+1=m+1 \Rightarrow n = m.$

Thus in both the cases, $f(n) = f (m) \Rightarrow n = m.$

$\text{If } n \text{ is even and } m \text{ is odd, then } n \neq m . \text{ Also } f (n) \text{ is odd and } f (m) \text{ is even.} \\ \\ \text{Hence, } f (n) \neq f (m).$

$\text{Thus, } n \neq m \Rightarrow f (n) \neq f (m).$

Hence, $f$ is an injective map.

Surjectivity: Let $N$ he an arbitrary natural number.

If $n$ is an odd natural number, then there exists an even natural number $n + 1$ such that

$f(n+1) = n+1-1 = n$

If $n$ is an even natural number, then there exists an odd natural number $(n - 1)$ such that

$f (n-1) = n-1+1 = n$

Thus, every $m \in N$ has its pre-image in $N.$ Hence, $f: N \rightarrow N$ is a surjection.

Hence, $f: N \rightarrow N$ is a bijection.

$\\$

$\text{Question 30: Let } f:N \cup \{ 0 \} \rightarrow N \cup \{ 0 \} \text{ be defined by }$

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

Show that $f$ is a bijection.

$f$ is an injection : Let $n, m \in N \cup \{ 0 \}$

If $n$ and $m$ are even, then

$f (n) = f (m) \Rightarrow n+1=m+1\Rightarrow n = m$

If $n$ and $m$ are odd, then

$f (n) = f (m) \Rightarrow n-1=m-1 \Rightarrow n = m$

Thus, in both case, we have

$f(n)=f(m) \Rightarrow n=m.$

If $n$ is odd and $m$ is even then $f (n) = n-1$ is even and $f (m) = m + 1$ is odd. Therefore,

$n \neq m \Rightarrow f(n) \neq f(m).$

Similarly, if $n$ is even and $m$ is odd, then

$n \neq m \Rightarrow f(n) \neq f(m).$

Hence, $f$ is an injection.

$f$ is a surjection: Let $n$ be an arbitrary element of $N \cup \{ 0 \}$

If $n$ is an odd natural number, there exist an even natural number $n - 1 \in N \cup \{ 0 \}$ (domain) such that $f (n-1) =n-1+1 =n.$

If $n$ is an even natural number, then there exists an odd natural number $n + 1 \in N \cup \{ 0 \}$ (domain) such that $f(n + 1) = n + 1 -1 =n.$

Also, $f (1) = 0.$

Thus, every element of $N \cup \{ 0 \}$ (co-domain) has its pre-image in $N \cup \{ 0 \}$ (domain). Hence, $f$ is an onto function.