Question -
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)
x2 = 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)
x2 = 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
x2 = 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 = y2 + y
Here, we cannot say that x = y.
For example, x = 2 and y = – 3
Then,
x2 + x= 22 + 2 = 6
y2 + y= (−3)2 – 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
x2 + 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)
x3 – x= y3 − y
Here, we cannot say x= y.
For example, x= 1 and y = -1
x3 −x = 1 − 1 = 0
y3 – 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 x3 –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 +cos2x
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)
x3 +1 = y3 + 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)
5x3 +4 = 5y3 + 4
5x3= 5y3
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
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)
5x3 +4 = 5y3 + 4
5x3 = 5y3
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
5x3 +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 + x2 =1 + y2
x2 = 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 + x2 = y
x2 = 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/(x2 + 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 /(x2 +1) = y /(y2 + 1)
x y2+ x = x2y + y
xy2 −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 x2 –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.