RD Sharma Class 12 Ex 1.1 Solutions Chapter 2 Function

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter2
Exercise1.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 1.1 Solutions Chapter 2 Function

Question 1. Give an example of a function

(i) Which is one-one but not onto.

Solution:

Let f: R → R given by f(x) = 3x + 2

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

Injectivity: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f (x) = f(y)

⇒ 3x + 2 =3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x) = f(y)

⇒ x = y

So, f is 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).

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 element in the co domain there need not be any element in the domain such that f(x) = y. Thus, f is not onto.

(ii) Which is not one-one but onto.

Solution:

Example for the function which is not one-one but onto

Let f: 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 elements of 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 element in Z (domain).

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

Thus, f is onto.

(iii) Which is neither one-one nor onto.

Solution:

Example for the function which is neither one-one nor onto.

Let f: Z → Z given by f(x) = 2x² + 1

Injectivity:

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

f(x) = f(y)

⇒ 2x²+1 = 2y²+1

⇒ 2x² = 2y²

⇒ x² = y²

⇒ x = ± y

So, different elements of 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 is not onto.

Question 2. Which of the following functions from A to B are one-one and onto? 

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

Solution:

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

Solution:

(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 element of 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 is onto.

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

Solution:

(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 have the 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

Solution:

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

Now we have to prove that given function is one-one

Injectivity: Let x and y be any two elements in the domain (N), such that 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 is one-one.

Surjectivity:

When x = 1

x² + x + 1 = 1 + 1 + 1 = 3

⇒ x² + x +1 ≥ 3, for every x in N.

⇒ f(x) will not assume the values 1 and 2.

So, f is not onto.

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

Solution:

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

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

Injectivity: Let x = 1

Therefore f(1) = 1²=1 and f(-1)=(-1)²=1

⇒ 1 and -1 have the same 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 ⇒ Range of 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

Solution:

Given f: N → N, given by f(x) = x²

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

Injection condition:

Let x and y be any 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 N that is natural numbers)

So, f is an injection.

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element 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 not a surjection.

Also f is not a bijection.

(ii) f: Z → Z given by f(x) = x²

Solution:

Given f: Z → Z, given by f(x) = x²

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

Injection condition:

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

f(x) = f(y)

x² = y²

x = ±y

So, f is not an injection.

Surjection test:

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

x² = y

x = ± √y which may not be in Z.

For example, if y = 3, x = ± √ 3 is not in Z.

So, f is not a surjection.

Also f is not bijection.

(iii) f: N → N given by f(x) = x³

Solution:

Given f: N → N given by f(x) = x³

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

Injection condition:

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

f(x) = f(y)

x³ = y³

x = y

So, f is an injection

Surjection condition:

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

f(x) = y

x³= y

x = ∛y which may not be in N.

For example, if y = 3,

X = ∛3 is not in N.

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

(iv) f: Z → Z given by f(x) = x³

Solution:

Given f: Z → Z given by f(x) = x³

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

Injection condition:

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

f(x) = f(y)

x³ = y³

x = y

So, f is an injection.

Surjection condition:

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

x³ = y

x = ∛y which may not be in Z.

For example, if y = 3,

x = ∛3 is not in Z.

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

(v) f: R → R, defined by f(x) = |x|

Solution:

Given f: R → R, defined by f(x) = |x|

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

Injection test:

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

f(x) = f(y)

|x|=|y|

x = ±y

So, f is not an injection. Surjection test:

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

f(x) = y

|x|=y

x = ± y ∈ Z

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

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

Solution:

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

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

Injection test:

Let x and y be any 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 not an injection. Surjection test:

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

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 not a surjection and f is not a bijection.

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

Solution:

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

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

Injection test: Let x and y be any two elements in the domain (Z), such that 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), such that f(x) = y for some element x in Z (domain).

f(x) = y x – 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

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

Solution:

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

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

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y)

Sin x = sin y

Here, x may not be equal to y because sin 0 = sin π.

So, 0 and π have the same image 0.

So, f is not an injection. Surjection test:

Range of f = [-1, 1]

Co-domain of f = R Both are not same. So, f is not a surjection and f is not a bijection.

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

Solution:

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

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

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

x³+1 = y³+ 1

x³= y³

x = y

So, f is an injection.

Surjection test:

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

f(x) = y

x³+1=y

x = ∛ (y – 1) ∈ R

So, f is a surjection.

So, f is a bijection.

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

Solution:

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

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

Injection test:

Let x and y be any 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 not an injection. Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element 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 a surjection and f is not a bijection.

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

Solution:

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

Now we have to check for 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 for every x in R.

So, for all elements in the domain, the image is 1.

So, f is not an 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) f: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)

Solution:

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

Now we have to check for 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 an injection.

Surjection test:

Let y be any element 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 not a surjection and f is not a bijection.

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

Solution:

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

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

Injection test: Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x) = f(y)

x³ + 1 = y³ + 1

x³ = y³

x = y

So, f is an injection. Surjection test:

Let y be any element in the co-domain (Q), such that f(x) = y for some element 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 not a surjection and f is not a bijection

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

Solution:

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

Now we have to check for the given function is injection, surjection and bijection

condition.

Injection test: Let x and y be any 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 an injection.

Surjection test:

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

f(x) = y

5x³+ 4 = y

x³ = (y – 4)/5 ∈ R

So, f is a surjection and f is a bijection.

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

Solution:

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

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

Injection condition:

Let x and y be any 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 an injection.

Surjection test:

Let y be any element in the co-domain (R), 

such that f(x) = y for some element x in R (domain).

f(x) = y

5x³ + 4 = y

x³ = (y – 4)/5 ∈ R

So, f is a surjection and f is a bijection.

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

Solution:

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

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

Injection condition:

Let x and y be any 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 not an injection. Surjection test:

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

f(x) = y

1 + x² = y

x² = y − 1

x = ± √-1 = ± i` is not in R.

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

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

Solution:

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

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

Injection condition:

Let x and y be any 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 – x y) = 0

x = y or x = 1/y

So, f is not an injection. Surjection test:

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

Solution:

Given f: A → B is an injection

And also given that range of f = {a} 

So, the number of images of f = 1 Since, f is an injection,

 there will be exactly one image for each 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.

Solution:

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

Now we have to show that 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 is one-one.

Surjectivity:

Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element 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 element in the co-domain, there exists some pre-image in the domain.

⇒ f is onto.

Since, f is both one-one and onto, 

it is a bijection.

Question 8. Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective:

(i) f (x) = x/2

Solution:

Given f: A → A, given by f (x) = x/2

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

Injection test:

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

f(x) = f(y)

x/2 = y/2

x = y

So, f is one-one. Surjection test:

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

f(x) = y

x/2 = y

x = 2y, which may not 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) g (x) = |x|

Solution:

Given g: A → A, given by g (x) = |x|

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

Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).

g(x) = g(y)

|x| = |y|

x = ± y

So, f is not one-one. Surjection test:

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

So, g is not onto. 

So, g is not bijective.

(iii) h (x) = x²

Solution:

Given h: A → A, given by h (x) = x²

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

Injection test:

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

h(x) = h(y)

 x² = y²

x = ±y

So, f is not one-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, examine whether the mapping is injective or surjective:

(i) {(x, y): x is a person, y is the mother of x}

Solution:

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

As, for each element x in domain set, there is a unique related element y in co-domain set.

So, f is the function.

Injection test: As, y can be mother of two or more persons So, f is not injective.

Surjection test:

For every mother y defined by (x, y), there exists a person x for whom y is mother. 

So, f is surjective. 

Therefore, f is surjective function.

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

Solution:

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 living person. 

So, g is not a function.

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

Solution:

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 )}

Question 11. If f: R → R be the function defined by f(x) = 4×3 + 7, show that f is a bijection.

Solution:

Given f: R → R is a function defined by f(x) = 4×3 + 7 Injectivity:

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

⇒ 4×3 + 7 = 4y3 + 7

⇒ 4×3 = 4y3

⇒ x3 = y3

⇒ x = y

So, f is one-one.

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

f(x) = y

⇒ 4×3 + 7 = y

⇒ 4×3 = y − 7

⇒ x3 = (y – 7)/4

⇒ x = ∛(y-7)/4 in R

So, for every element in the co-domain, there exists some pre-image in the domain. f is onto.

Since, f is both one-to-one and onto,

 it is a bijection.

Question 12: Show that the exponential function f: R → R, given by f(x) = ex is one one but not onto. What happens if the codomain is replaced by Ro+.

Solution:

We have f: R → R, given by f(x) = ex.

Let x,y ϵ R, such that

=> f(x) = f(y)

=> ex = ey

=> e(x-y) = 1 = e0

=> x – y = 0

=> x = y

Hence,

f is one-one.

Clearly range of f = (0, INFINITY) not equals to R.

Hence, f is not onto.

When co-domain is replaced by Ro+ i.e. (0, INFINITY) then f becomes an onto function.

Question 13: Show that the logarithmic function f: Ro+→ R given by f(x) = log(a)x, a>0 is a bijection.

Solution:

We have f: Ro+→ R given by f(x) = log(a)x, a>0.

let x,y ϵ Ro+ such that,

f(x) = f(y)

=> log(a)x = log(a)y

=> log(a)x (x/y) = 0 [log(a)x = 0]

=> x/y = 1

=> x = y

Hence, f is not one-one.

Now, let y ϵ R be arbitrary, then f(x) = y

=> log (a)x = y

=>x = ay ϵ Ro+

Thus for all yϵR there exist x = ay such that f(x) = y.

Hence,

f is onto.

f is one-one.

=> f is bijective.

Question 14: If A = {1,2,3}, show that a one-one function f: A → A must be onto.

Solution:

Since f is one-one, three elements of {1,2,3} must be taken to the 3 different elements of the co-domain {1,2,3} under f. Hence f has to be onto.

Question 15: If A={1, 2, 3}, show that an onto function f: A→A must be one-one.

Solution:

A={1,2,3}

Possible onto functions from A to A can be the following:

(0){(1,1),(2,2).(3,3)

(ii) (1,1),(2,3),(3,2)

(ii){(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)

Here, in each function, different elements of the domain have different images.

Therefore,

All the functions are one-one

Question 16: Find the number of all onto functions from the set A = { 1, 2, 3, ….n} to itself.

Solution:

We know that every onto function from A to itself is one-one.

Therefore,

The number of one-one functions=number of bijections =n!

Question 17: Give examples of two one-one functions f1 and f2. From R to R such that f1 + f2: R→ R defined by (f1 + f2)(x) = f1(x) + f2(x) is not one-one.

Solution:

We know that f1: R→ R, given by f1 (x)= x, and f2 (x)=-x are one-one.

Proving f1, is one-one:

Let f1(x)= f1(y)

Implies that x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let f2(x) = f2(y)

Implies that – x=-y

Implies that x = y

Therefore,

f2 is one-one.

Proving (f1 + f2) is not one-one:

Given:(f1+f2)(x)= f1(x)+ f2 (x)= x+(-x) = 0

Therefore,

For every real number x,(f1 + f2)(x)=0

Therefore,

The image of ever number in the domain is same as 0.

Thus, (f1 + f2) is not one-one.

Question 18: Given an example of two surjective functions f1 and f2 from Z to Z such that f1 + f2 is not surjective.

Solution:

We know that f1: R → R, given by f1(x) = x, and f2(x)=-x are surjective functions.

Proving f1 is surjective:

Let y be an element in the co-domain (R), such that f1(x)= y.

f1(x)= y

Implies that x = y, which is in R.

Therefore,

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

Therefore,

f1 is surjective

Proving f2 is surjective:

Let f2 (x)= y

x = y, which is in R.

Therefore,

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

Therefore,

f2 is surjective.

Proving (f1+f2) is not surjective:

Given:(f1 + f2)(x) = f1(x)+ f2(x)=x+(-x)=0

Therefore, for every real number x, (f1 + f2)(x) = 0

Therefore, the image of every number in the domain is same as 0.

Implies that Range = {0}

Co-domain = R

Therefore, both are not same.

Therefore, f1+f2 is not surjective.

Question 19: Show that if f1 and f2 are one-one maps, from R to R then the product f1 X f2: R→R defined by (f1 X f2)(x) = f1(x)f2(x) need not be one-one.

Solution:

We know that f: R→ R, given by f1(x) = x, and f2(x) = x are one-one.

Proving f1 is one-one:

Let x and y be two elements in the domain R, such that f1(x) = f1(y)

f1(x) = f1(y)

x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let x and y be two elements in the domain R, such that f2 (x) = f2(y)

f2(x)= f2(y)

Implies that x = y

Therefore,

f2 is one-one.

Proving f1 X f2 is not one-one:

Given:

(f1 X f2)(x) = f1(x) X f2(x) = x * x = x²

Let x and y be two elements in the domain R, such that

(f1 X f2)(x) = (f1 X f2)(y)

Implies that x² = y²

Implies that x = (+-)y

Therefore,

(f1 X f2) is not one-one.

Question 20: Suppose f1 and f2 are non-zero one-one functions from R to R. Is (f1/f2) necessarily one-one? Justify.

Solution:

We know that f1: R→R given by f1(x) = x³ and f2(x)= x are one-one.

Injectivity of f1:

Consider x and y be two elements in the domain R, such that

f1(x)= f1(y)

Implies that x³ = y

x=3√y belongs to R

Therefore,

f1 is one-one.

Injectivity of f2:

Consider x and y be two elements in the domain R, such that

f2(x)= f2(y)

Implies that x = y

x belongs to R

Therefore,

f2 is one-one.

Providing (f1 / f2) is not one-one:

Given that (f1/f2)(x)= = f1(x)/f2(x) = (x³ / x) = x²

Consider x and y be two elements in the domain R, such that

(f1/f2)(x) = (f1/f2)(y)

f2 f2

x² = y²

x= (+-)y

Therefore,

(f1/f2) is not one-one.

Question 21: Given A = {2, 3, 4}, B = {2,5,6,7}. Construct an example of each of the following.

(i) An injective map from A to B.

(ii) A mapping from A to B which is not injective.

(iii)A mapping from A to B.

Solution:

Given A={1,2,3,4}, B = {2,5,6,7}

Let f: A → Bf: A → B be a mapping from A to B f = {(2,5)(3,6)(4,7)}

f is an injective mapping.

Since for every element a € A there is an unique element b € B

Let us define a mapping: A→B given by g = {(2,2)(2,5)(3,6)(4,7)}

g is not an injective mapping.

since the element 2 € A is not uniquely mapped

Since (2,2) and (2,5) both belong to the mapping g, g is not injective

Let us define a mapping h: A→B

h: A → B given by h = {(2,2),(5,3),(7,4)}

h is a mapping from A to B

B to A since the every ordered puts {2,5,7} € B to elements in {2,3,4} € A

Question 22: Show that f: R → R, Given by f(x) = x – [x] is neither one-one nor onto.

Solution:

f:R → R, given by f (x)= x-[x]

Injectivity:

f(x)=0 for all x belongs to Z,

Therefore,

f is not one-one.

Surjectivity:

Range of f = (0,1) not equals to R.

Co-domain of f = R

Both are not same.

Therefore,

f is not onto.

Question 23: Let f:N→N defined by

f(n) = n + 1, if n is odd.

f(n) = n – 1, if n is even. Show that if f is a bijection.

Solution:

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Let both x and y be even and

Let x, y belongs to N such that f (x) = f(y)

As, f (x) = f(y)

Implies that x – 1= x-1

Implies that x = y

Case-2: Let both x and y be odd and

Let x, y belongs to N such that f (x)= f (y)

As, f (x)= f (y)

Implies that x +1 = y +1

Implies that x = y

Case-3:Let x be even and y be odd then, x y.

Then,

x+1 is odd and y-1 is even.

Implies that x+1≠ y-1

Implies that f(x) ≠ f(y)

Therefore,

x ≠ y

= f(x) ≠ f(y)

In all the 3 cases,

Therefore,

f is one-one.

Surjectivity:

Co-domain off = {1, 2,3,4,…}

Range of f = {1+1,2 – 1,3+1,4 – 1,…} = {2,1,4, 3,…}={1, 2, 3, 4,…}

Both are same.

Implies that f is onto.

Therefore,

f is a bijection.

Question 12: Show that the exponential function f: R → R, given by f(x) = ex is one one but not onto. What happens if the codomain is replaced by Ro+.

Solution:

We have f: R → R, given by f(x) = ex.

Let x,y ϵ R, such that

=> f(x) = f(y)

=> ex = ey

=> e(x-y) = 1 = e0

=> x – y = 0

=> x = y

Hence,

f is one-one.

Clearly range of f = (0, INFINITY) not equals to R.

Hence, f is not onto.

When co-domain is replaced by Ro+ i.e. (0, INFINITY) then f becomes an onto function.

Question 13: Show that the logarithmic function f: Ro+→ R given by f(x) = log(a)x, a>0 is a bijection.

Solution:

We have f: Ro+→ R given by f(x) = log(a)x, a>0.

let x,y ϵ Ro+ such that,

f(x) = f(y)

=> log(a)x = log(a)y

=> log(a)x (x/y) = 0 [log(a)x = 0]

=> x/y = 1

=> x = y

Hence, f is not one-one.

Now, let y ϵ R be arbitrary, then f(x) = y

=> log (a)x = y

=>x = ay ϵ Ro+

Thus for all yϵR there exist x = ay such that f(x) = y.

Hence,

f is onto.

f is one-one.

=> f is bijective.

Question 14: If A = {1,2,3}, show that a one-one function f: A → A must be onto.

Solution:

Since f is one-one, three elements of {1,2,3} must be taken to the 3 different elements of the co-domain {1,2,3} under f. Hence f has to be onto.

Question 15: If A={1, 2, 3}, show that an onto function f: A→A must be one-one.

Solution:

A={1,2,3}

Possible onto functions from A to A can be the following:

(0){(1,1),(2,2).(3,3)

(ii) (1,1),(2,3),(3,2)

(ii){(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)

Here, in each function, different elements of the domain have different images.

Therefore,

All the functions are one-one

Question 16: Find the number of all onto functions from the set A = { 1, 2, 3, ….n} to itself.

Solution:

We know that every onto function from A to itself is one-one.

Therefore,

The number of one-one functions=number of bijections =n!

Question 17: Give examples of two one-one functions f1 and f2. From R to R such that f1 + f2: R→ R defined by (f1 + f2)(x) = f1(x) + f2(x) is not one-one.

Solution:

We know that f1: R→ R, given by f1 (x)= x, and f2 (x)=-x are one-one.

Proving f1, is one-one:

Let f1(x)= f1(y)

Implies that x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let f2(x) = f2(y)

Implies that – x=-y

Implies that x = y

Therefore,

f2 is one-one.

Proving (f1 + f2) is not one-one:

Given:(f1+f2)(x)= f1(x)+ f2 (x)= x+(-x) = 0

Therefore,

For every real number x,(f1 + f2)(x)=0

Therefore,

The image of ever number in the domain is same as 0.

Thus, (f1 + f2) is not one-one.

Question 18: Given an example of two surjective functions f1 and f2 from Z to Z such that f1 + f2 is not surjective.

Solution:

We know that f1: R → R, given by f1(x) = x, and f2(x)=-x are surjective functions.

Proving f1 is surjective:

Let y be an element in the co-domain (R), such that f1(x)= y.

f1(x)= y

Implies that x = y, which is in R.

Therefore,

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

Therefore,

f1 is surjective

Proving f2 is surjective:

Let f2 (x)= y

x = y, which is in R.

Therefore,

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

Therefore,

f2 is surjective.

Proving (f1+f2) is not surjective:

Given:(f1 + f2)(x) = f1(x)+ f2(x)=x+(-x)=0

Therefore, for every real number x, (f1 + f2)(x) = 0

Therefore, the image of every number in the domain is same as 0.

Implies that Range = {0}

Co-domain = R

Therefore, both are not same.

Therefore, f1+f2 is not surjective.

Question 19: Show that if f1 and f2 are one-one maps, from R to R then the product f1 X f2: R→R defined by (f1 X f2)(x) = f1(x)f2(x) need not be one-one.

Solution:

We know that f: R→ R, given by f1(x) = x, and f2(x) = x are one-one.

Proving f1 is one-one:

Let x and y be two elements in the domain R, such that f1(x) = f1(y)

f1(x) = f1(y)

x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let x and y be two elements in the domain R, such that f2 (x) = f2(y)

f2(x)= f2(y)

Implies that x = y

Therefore,

f2 is one-one.

Proving f1 X f2 is not one-one:

Given:

(f1 X f2)(x) = f1(x) X f2(x) = x * x = x²

Let x and y be two elements in the domain R, such that

(f1 X f2)(x) = (f1 X f2)(y)

Implies that x² = y²

Implies that x = (+-)y

Therefore,

(f1 X f2) is not one-one.

Question 20: Suppose f1 and f2 are non-zero one-one functions from R to R. Is (f1/f2) necessarily one-one? Justify.

Solution:

We know that f1: R→R given by f1(x) = x³ and f2(x)= x are one-one.

Injectivity of f1:

Consider x and y be two elements in the domain R, such that

f1(x)= f1(y)

Implies that x³ = y

x=3√y belongs to R

Therefore,

f1 is one-one.

Injectivity of f2:

Consider x and y be two elements in the domain R, such that

f2(x)= f2(y)

Implies that x = y

x belongs to R

Therefore,

f2 is one-one.

Providing (f1 / f2) is not one-one:

Given that (f1/f2)(x)= = f1(x)/f2(x) = (x³ / x) = x²

Consider x and y be two elements in the domain R, such that

(f1/f2)(x) = (f1/f2)(y)

f2 f2

x² = y²

x= (+-)y

Therefore,

(f1/f2) is not one-one.

Question 21: Given A = {2, 3, 4}, B = {2,5,6,7}. Construct an example of each of the following.

(i) An injective map from A to B.

(ii) A mapping from A to B which is not injective.

(iii)A mapping from A to B.

Solution:

Given A={1,2,3,4}, B = {2,5,6,7}

Let f: A → Bf: A → B be a mapping from A to B f = {(2,5)(3,6)(4,7)}

f is an injective mapping.

Since for every element a € A there is an unique element b € B

Let us define a mapping: A→B given by g = {(2,2)(2,5)(3,6)(4,7)}

g is not an injective mapping.

since the element 2 € A is not uniquely mapped

Since (2,2) and (2,5) both belong to the mapping g, g is not injective

Let us define a mapping h: A→B

h: A → B given by h = {(2,2),(5,3),(7,4)}

h is a mapping from A to B

B to A since the every ordered puts {2,5,7} € B to elements in {2,3,4} € A

Question 22: Show that f: R → R, Given by f(x) = x – [x] is neither one-one nor onto.

Solution:

f:R → R, given by f (x)= x-[x]

Injectivity:

f(x)=0 for all x belongs to Z,

Therefore,

f is not one-one.

Surjectivity:

Range of f = (0,1) not equals to R.

Co-domain of f = R

Both are not same.

Therefore,

f is not onto.

Question 23: Let f:N→N defined by

f(n) = n + 1, if n is odd.

f(n) = n – 1, if n is even. Show that if f is a bijection.

Solution:

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Let both x and y be even and

Let x, y belongs to N such that f (x) = f(y)

As, f (x) = f(y)

Implies that x – 1= x-1

Implies that x = y

Case-2: Let both x and y be odd and

Let x, y belongs to N such that f (x)= f (y)

As, f (x)= f (y)

Implies that x +1 = y +1

Implies that x = y

Case-3:Let x be even and y be odd then, x y.

Then,

x+1 is odd and y-1 is even.

Implies that x+1≠ y-1

Implies that f(x) ≠ f(y)

Therefore,

x ≠ y

= f(x) ≠ f(y)

In all the 3 cases,

Therefore,

f is one-one.

Surjectivity:

Co-domain off = {1, 2,3,4,…}

Range of f = {1+1,2 – 1,3+1,4 – 1,…} = {2,1,4, 3,…}={1, 2, 3, 4,…}

Both are same.

Implies that f is onto.

Therefore,

f is a bijection.

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