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Question -

Classify the following function as injection, surjection or bijection:

(i) f: N → N given by f(x) = x2

(ii) f: Z → Z given by f(x) = x2

(iii) f: N → N given by f(x) = x3

(iv) f: Z → Z given by f(x) = x3

(v) f: R → R, defined by f(x) = |x|

(vi) f: Z → Z, defined by f(x) = x2 + x

(vii) f: Z → Z, defined by f(x) = x − 5

(viii) f: R → R, defined by f(x) = sin x

(ix) f: R → R, defined by f(x) = x3 +1

(x) f: R → R, defined by f(x) = x3 − x

(xi) f: R → R, defined by f(x) = sin2x +cos2x

(xii) f: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)

(xiii) f: Q → Q, defined by f(x) = x3 +1

(xiv) f: R → R, defined by f(x) = 5x3 +4

(xv) f: R → R, defined by f(x) = 5x3 +4

(xvi) f: R → R, defined by f(x) = 1 + x2

(xvii) f: R → R, defined by f(x) = x/(x+1)



Answer -

(i) Givenf: N → N, given by f(x) = x2

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

x= y2

x =y (We do not get ± because x and y are in Nthat is natural numbers)

So, f is aninjection.

Surjection condition:

Let y be anyelement in the co-domain (N), such that f(x) = y for someelement x in N (domain).

f(x) = y

x2= y

x =√y, which may not be in N.

For example, if y =3,

x =√3 is not in N.

So, f is nota surjection.

Also f is not abijection.

(ii) Givenf: Z → Z, given by f(x) = x2

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x= y2

x = ±y

So, f is notan injection.

Surjection test:

Let y be anyelement in the co-domain (Z), such that f(x) = y for someelement x in Z (domain).

f(x) = y

x= y

x = ±√y which may not be in Z.

For example, if y =3,

x = ± √ 3is not in Z.

So, f is nota surjection.

Also f is notbijection.

(iii) Givenf: N → N given by f(x) = x3

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

x3 = y3

x = y

So, f is aninjection 

Surjection condition:

Let y be anyelement in the co-domain (N), such that f(x) = y for someelement x in N (domain).

f(x) = y

x3= y

x = ywhich may not be in N.

For example, if y =3,

X = 3 is not in N.

So, f is nota surjection and f is not a bijection.

(iv) Givenf: Z → Z given by f(x) = x3

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (Z), such that f(x) = f(y)

f(x) = f(y)

x3 = y3

x = y

So, f is aninjection.

Surjection condition:

Let y be anyelement in the co-domain (Z), such that f(x) = y for someelement x in Z (domain).

f(x) = y

x3 = y

x = y which may not be in Z.

For example, if y =3,

x = 3 is not in Z.

So, f is nota surjection and f is not a bijection.

(v) Givenf: R → R, defined by f(x) = |x|

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y beany two elements in the domain (R), such that f(x) = f(y)

f(x) = f(y)

|x|=|y|

x = ±y

So, f is notan injection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y for someelement x in R (domain).

f(x) = y

|x|=y

x = ± y  Z

So, f is asurjection and f is not a bijection.

(vi) Givenf: Z → Z, defined by f(x) = x2 + x

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y beany two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x2+ x = y+ y

Here, we cannot say that x = y.

For example, x = 2 and y = – 3

Then,

x+ x= 2+ 2 = 6

y+ y= (−3)– 3 = 6

So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.

So, f is notan injection.

Surjection test:

Let y be anyelement in the co-domain (Z),

suchthat f(x) = y for some element x in Z (domain).

f(x) = y

x2 + x = y

Here, we cannot say x  Z.

For example, y =– 4.

x2 + x = − 4

x+ x + 4 = 0

x = (-1 ± √-5)/2= (-1 ± i √5)/2 which is not in Z.

So, f is nota surjection and f is not a bijection.

(vii) Givenf: Z → Z, defined by f(x) = x – 5

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (Z), suchthat f(x) = f(y).

f(x) = f(y)

x – 5 = y– 5

x = y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Z), suchthat f(x) = y for someelement x in Z (domain).

f(x) = y

x – 5 = y

x = y + 5, whichis in Z.

So, f is asurjection and f is a bijection

(viii) Givenf: R → R, defined by f(x) = sin x

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (R), suchthat f(x) = f(y).

f(x) = f(y)

Sin x = sin y

Here, x may not be equal to y because sin0 = sin π.

So, 0 and π have the same image 0.

So, f is notan injection.

Surjection test:

Range of f =[-1, 1]

Co-domainof f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) Givenf: R → R, defined by f(x) = x3 + 1

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (R), suchthat f(x) = f(y).

f(x) = f(y)

x3+1 = y3+ 1

x3= y3

x = y

So, f is aninjection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y for someelement x in R (domain).

f(x) = y

x3+1=y

x = (y – 1) R

So, f is asurjection.

So, f is abijection.

(x)  Givenf: R → R, defined by f(x) = x3 − x

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y beany two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

x– x= y− y

Here, we cannot say x= y.

For example, x= 1 and y = -1

x−x = 1 − 1 = 0

y– y= (−1)3− (−1) – 1 + 1 = 0

So, 1 and -1 have the same image 0.

So, f is notan injection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y for someelement x in R (domain).

f(x) = y

x3 − x = y

By observation we can say that there exist some x in R, such that x–x = y.

So, f is asurjection and f is not a bijection.

(xi) Givenf: R → R, defined by f(x) = sin2x + cos2x

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

f(x) = sin2x +cos2

We know that sin2x +cos2x = 1

So, f(x) = 1 forevery x in R.

So, for all elementsin the domain, the image is 1.

So, f is notan injection.

Surjection condition:

Range of f = {1}

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(xii) Givenf: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain(Q − {3}), such that f(x) = f(y).

f(x) = f(y)

(2x + 3)/(x – 3) = (2y+ 3)/(y – 3)

(2x + 3) (y − 3)= (2y + 3) (x − 3)

2xy − 6x+ 3y − 9 = 2xy − 6y + 3x − 9

9x = 9y 

x = y

So, f is aninjection.

Surjection test:

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

f(x) = y

(2x + 3)/(x – 3) = y

2x + 3 = x y − 3y

2x – x y = −3y − 3

x (2−y) =−3 (y + 1)

x = -3(y + 1)/(2 –y) which is not defined at y = 2.

So, f is nota surjection and f is not a bijection.

(xiii) Givenf: Q → Q, defined by f(x) = x3 + 1

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (Q), suchthat f(x) = f(y).

f(x) = f(y)

x+1 = y+ 1

x3 = y3

x = y

So, f is aninjection.

Surjection test:

Let y be anyelement in the co-domain (Q), such that f(x) = y for someelement x in Q (domain).

f(x) = y

x3+ 1 = y

x = (y-1),which may not be in Q.

For example, if y= 8,

x3+ 1 =  8

x3= 7

x = 7, which is not in Q.

So, f is nota surjection and f is not a bijection.

(xiv) Givenf: R → R, defined by f(x) = 5x3 + 4

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in the domain (R), suchthat f(x) = f(y).

f(x) = f(y)

5x+4 = 5y+ 4

5x3= 5y3

x= y3

x = y

So, f is aninjection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y for someelement x in R (domain).

f(x) = y

5x3+4 = y

x3 =(y – 4)/5 R

So, f is asurjection and f is a bijection.

(xv) Givenf: R → R, defined by f(x) = 5x3 + 4

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

5x+4 = 5y+ 4

5x= 5y3

x= y3

x = y

So, f is aninjection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y for someelement x in R (domain).

f(x) = y

5x+4 = y

x3 =(y – 4)/5 R

So, f is asurjection and f is a bijection.

(xvi) Givenf: R → R, defined by f(x) = 1 + x2

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

1 + x=1 + y2

x= y2

x = ± y

So, f is notan injection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y for someelement x in R (domain).

f(x) = y

1 + x= y

x= y − 1

x = ± √-1 = ± i` is not in R.

So, f is nota surjection and f is not a bijection.

(xvii) Givenf: R → R, defined by f(x) = x/(x+ 1)

Now we have to checkfor the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y beany two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

x /(x+1) = y /(y2 + 1)

x y2+ x = x2y + y

xy−x2y + x − y = 0

−x y (−y + x)+ 1 (x − y) = 0

(x − y) (1 – xy) = 0

x = y or x =1/y

So, f is notan injection.

Surjection test:

Let y be anyelement in the co-domain (R), such that f(x) = y forsome element x in R (domain).

f(x) = y

x /(x2 +1) = y

y x–x + y = 0

x = (-(-1) ± √ (1-4y2))/(2y) if y ≠ 0

= (1 ± √ (1-4y2))/(2y), which may not be in R

For example, if y=1, then

(1 ± √ (1-4)) / (2y) =(1 ± i √3)/2, which is not in R

So, f is not surjection and f is not bijection.

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