RD Chapter 2 Functions Ex 2.1 Solutions

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.

Answer - 1 : -

(i) Letf: Z → Z given by f(x) = 3x + 2

Let us check one-onecondition on f(x) = 3x + 2

Injectivity:

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

f (x)= f(y)

⇒ 3x + 2=3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x)= f(y)

⇒ x = y

So, f isone-one.

Surjectivity:

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

Let f(x) = y

⇒ 3x + 2 = y

⇒ 3x = y –2

⇒ x = (y –2)/3. It may not be in the domain (Z)

Because if we take y = 3,

x = (y – 2)/3 =(3-2)/3 = 1/3 ∉ domain Z.

So, for every elementin the co domain there need not be any element in the domain suchthat f(x) = y.

Thus, f is not onto.

(ii) Example for thefunction which is not one-one but onto

Letf: Z → N ∪{0} given by f(x) = |x|

Injectivity:

Let x and y be any two elements in the domain (Z),

Such that f(x)= f(y).

⇒ |x| = |y|

⇒ x = ± y

So, different elementsof domain f may give the same image.

So, f is not one-one.

Surjectivity:

Let y be any element in the co domain (Z), such that f(x)= y for some element x in Z (domain).

f(x) = y

⇒ |x| = y

⇒ x = ± y

Which is an elementin Z (domain).

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

Thus, f is onto.

(iii) Example for thefunction which is neither one-one nor onto.

Letf: Z → Z given by f(x) = 2x² + 1

Injectivity:

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

f(x) = f(y)

⇒ 2x²+1 = 2y²+1

⇒ 2x² = 2y²

⇒ x²= y²

⇒ x = ± y

So, different elementsof domain f may give the same image.

Thus, f is not one-one.

Surjectivity:

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

f (x) = y

⇒ 2x²+1=y

⇒ 2x²= y − 1

⇒ x² =(y-1)/2

⇒ x = √((y-1)/2) ∉ Z always.

For example, if we take, y = 4,

x = ± √ ((y-1)/2)

= ± √ ((4-1)/2)

= ± √ (3/2) ∉ Z

So, x may not be in Z (domain).

Thus, f isnot onto.

Question - 2 : -
Which of the following functions from A to B areone-one and onto?
(i) f₁ = {(1, 3), (2, 5), (3, 7)}; A = {1, 2,3}, B = {3, 5, 7}
(ii) f₂ = {(2, a), (3, b),(4, c)}; A = {2, 3, 4}, B = {a, b, c}
(iii) f₃ = {(a, x), (b, x), (c, z),(d, z)}; A = {a, b, c, d,}, B ={x, y, z}.

Answer - 2 : -

(i) Consider f₁ ={(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:

f₁ (1) = 3

f₁(2) = 5

f₁ (3) = 7

⇒ Every elementof A has different images in B.

So, f₁ is one-one.

Surjectivity:

Co-domain of f₁ = {3, 5, 7}

Range of f₁ =set of images = {3, 5, 7}

⇒ Co-domain = range

So, f₁ is onto.

(ii) Consider f₂ ={(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B ={a, b, c}

Injectivity:

f₂ (2) = a

f₂ (3) = b

f₂ (4) = c

⇒ Every elementof A has different images in B.

So, f₂ is one-one.

Surjectivity:

Co-domain of f₂ = {a, b, c}

Range of f₂ =set of images = {a, b, c}

⇒ Co-domain = range

So, f₂ isonto.

(iii) Consider f₃ ={(a, x), (b, x), (c, z), (d, z)} ; A ={a, b, c, d,}, B = {x, y, z}

Injectivity:

f₃ (a) = x

f₃ (b) = x

f₃ (c) = z

f₃ (d) = z

⇒a and b have the same image x.

Also c and d havethe same image z

So, f₃ is not one-one.

Surjectivity:

Co-domain of f₃ ={x, y, z}

Range of f₃ =set of images = {x, z}

So, the co-domain is not same as the range.

So, f₃ is not onto.

Question - 3 : - Prove that the function f: N → N, defined by f(x) = x² + x + 1, is one-one but not onto

Answer - 3 : -

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

Now we have to provethat given function is one-one

Injectivity:

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

⇒ x² +x + 1 = y² + y + 1

⇒ (x² –y²) + (x – y) = 0 `

⇒ (x + y) (x- y ) + (x– y ) = 0

⇒ (x – y) (x + y + 1) =0

⇒ x – y = 0 [x + y +1 cannot be zero because x and y are natural numbers

⇒ x = y

So, f isone-one.

Surjectivity:

When x = 1

x² + x+ 1 = 1 + 1 + 1 = 3

⇒ x² +x +1 ≥ 3, for every x in N.

⇒ f(x) will notassume the values 1 and 2.

So, f is not onto.

Question - 4 : -
Let A = {−1, 0, 1} and f = {(x, x²): x ∈ A}. Showthat f : A → A is neither one-one nor onto.

Answer - 4 : -

Given A = {−1, 0,1} and f = {(x, x²): x ∈ A}

Also given that, f(x) = x²

Now we have to provethat given function neither one-one or nor onto.

Injectivity:

Let x = 1

Therefore f(1) = 1²=1and

f(-1)=(-1)²=1

⇒ 1 and -1 have thesame images.

So, f is not one-one.

Surjectivity:

Co-domain of f = {-1, 0, 1}

f(1) = 1² =1,

f(-1) = (-1)² = 1 and

f(0) = 0

⇒ Rangeof f = {0, 1}

So, both are not same.

Hence, f is not onto

Question - 5 : -
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/(x²+1)

Answer - 5 : -

(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.

Question - 6 : - If f: A → B is an injection, such that range of f = {a}, determine the number of elements in A.

Answer - 6 : -

Givenf: A → B is an injection

And also given thatrange of f = {a}

So, the number of images of f = 1

Since, f is an injection, there will be exactly one image foreach element of f .

So, number of elements in A = 1.

Question - 7 : - Show that the function f: R − {3} → R − {2} given by f(x) = (x-2)/(x-3) is a bijection.

Answer - 7 : -

Given thatf: R − {3} → R − {2} given by f (x) = (x-2)/(x-3)

Now we have to showthat the given function is one-one and on-to

Injectivity:

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

f(x) = f(y)

⇒ (x – 2) /(x – 3) = (y– 2) /(y – 3)

⇒ (x – 2) (y – 3)= (y – 2) (x – 3)

⇒ x y – 3 x – 2 y+ 6 = x y – 3y – 2x + 6

⇒ x = y

So, f isone-one.

Surjectivity:

Let y be anyelement in the co-domain (R − {2}), such that f(x) = y for someelement x in R − {3} (domain).

f(x) = y

⇒ (x – 2) /(x – 3) = y

⇒ x – 2 = x y – 3y

⇒ x y – x = 3y – 2

⇒ x ( y – 1 ) = 3y – 2

⇒ x = (3y – 2)/ (y– 1), which is in R – {3}

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

⇒ f is onto.

Since, f isboth one-one and onto, it is a bijection.

Question - 8 : -
Let A = [-1, 1]. Then, discuss whether the followingfunction from A to itself is one-one, onto or bijective:
(i) f (x) = x/2
(ii) g (x) = |x|
(iii) h (x) = x²

Answer - 8 : -

(i) Given f: A →A, given by f (x) = x/2

Now we have to show thatthe given function is one-one and on-to

Injection test:

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

f(x) = f(y)

x/2 = y/2

x = y

So, f isone-one.

Surjection test:

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

f(x) = y

x/2 = y

x = 2y, which maynot be in A.

For example,if y = 1, then

x = 2, which is not in A.

So, f is not onto.

So, f is not bijective.

(ii) Given g: A →A, given by g (x) = |x|

Now we have to showthat the given function is one-one and on-to

Injection test:

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

g(x) = g(y)

|x| = |y|

x = ± y

So, f is notone-one.

Surjection test:

For y = -1,there is no value of x in A.

So, g is notonto.

So, g is not bijective.

(iii) Given h:A → A, given by h (x) = x²

Now we have to showthat the given function is one-one and on-to

Injection test:

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

h(x) = h(y)

x² = y²

x = ±y

So, f is notone-one.

Surjection test:

For y = – 1,there is no value of x in A.

So, h is not onto.

So, h is not bijective.

Question - 9 : -
Are the following set of ordered pair of a function? If so, examinewhether the mapping is injective or surjective:
(i) {(x, y): x is a person, y is the motherof x}
(ii) {(a, b): a is a person, b is an ancestorof a}

Answer - 9 : -

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

As, for eachelement x in domain set, there is a unique relatedelement y in co-domain set.

So, f is thefunction.

Injection test:

As, y can be mother of two or more persons

So, f is not injective.

Surjection test:

For everymother y defined by (x, y), there exists aperson x for whom y is mother.

So, f is surjective.

Therefore, f is surjective function.

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

Since, the ordered map (a, b) does not map ‘a’ – a person to a livingperson.

So, g is not a function.

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

Answer - 10 : -

Given A = {1, 2, 3}

Number of elements in A = 3

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

(i) {(1, 1), (2, 2),(3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

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RD Chapter 2 Functions Ex 2.1 Solutions

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