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

Answer:

\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. 

Answer:

\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.

Answer:

(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.

Answer:

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.

Answer:

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.

Answer:

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. }

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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 ?

Answer:

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 .

Answer:

\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.

Answer:

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.

Answer:

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]

Answer:

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. 

Answer:

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. }

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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

Answer:

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. }

Answer:

\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]

Answer:

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. }

Answer:

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.

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

Answer:

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.

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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]

Answer:

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.

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\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. } 

Answer:

\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.

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\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.

Answer:

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.