RD Sharma Class 12 Solutions Chapter 2 Functions

<str

(i) Let f: Z → Z 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) 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) 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

<str

(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) 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) 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

<str

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

<str

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

<str

(i) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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) 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

<str

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

<str

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

<str

(i) 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) 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) 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

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) 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

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 )}

11. If f: R → R be the function defined by f(x) = 4x³ + 7, show that f is a bijection.

Solution:

Given f: R → R is a function defined by f(x) = 4x³ + 7

Injectivity:

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

⇒ 4x³ + 7 = 4y³ + 7

⇒ 4x³ = 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

⇒ 4x³ + 7 = y

⇒ 4x³ = 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.

Solution:

(i) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

Also given that f(x) = 2x + 3 and g(x) = x² + 5

Now, (gof) (x) = g (f (x))

= g (2x +3)

= (2x + 3)² + 5

= 4x²+ 9 + 12x +5

=4x²+ 12x + 14

Now, (fog) (x) = f (g (x))

= f (x² + 5)

= 2 (x² + 5) +3

= 2 x²+ 10 + 3

= 2x² + 13

(ii) Given, f: R → R and g: R → R

so, gof: R → R and fog: R → R

f(x) = 2x + x² and g(x) = x³

(gof) (x)= g (f (x))

= g (2x+x²)

= (2x+x²)³

Now, (fog) (x) = f (g (x))

= f (x³)

= 2 (x³) + (x³)²

= 2x³ + x⁶

(iii) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = x² + 8 and g(x) = 3x³ + 1

(gof) (x) = g (f (x))

= g (x² + 8)

= 3 (x²+8)³ + 1

Now, (fog) (x) = f (g (x))

= f (3x³ + 1)

= (3x³+1)² + 8

= 9x⁶ + 6x³ + 1 + 8

= 9x⁶ + 6x³ + 9

(iv) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = x and g(x) = |x|

(gof) (x) = g (f (x))

= g (x)

= |x|

Now (fog) (x) = f (g (x))

= f (|x|)

= |x|

(v) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = x² + 2x − 3 and g(x) = 3x − 4

(gof) (x) = g (f(x))

= g (x² + 2x − 3)

= 3 (x² + 2x − 3) − 4

= 3x² + 6x − 9 − 4

= 3x² + 6x − 13

Now, (fog) (x) = f (g (x))

= f (3x − 4)

= (3x − 4)² + 2 (3x − 4) −3

= 9x² + 16 − 24x + 6x – 8 − 3

= 9x² − 18x + 5

(vi) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = 8x³ and g(x) = x^1/3

(gof) (x) = g (f (x))

= g (8x³)

= (8x³)^1/3

= [(2x)³]^1/3

= 2x

Now, (fog) (x) = f (g (x))

= f (x^1/3)

= 8 (x^1/3)³

= 8x

Solution:

Given f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}

f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}

Co-domain of f is a subset of the domain of g.

So, gof exists and gof: {3, 9, 12} → {3, 9}

(gof) (3) = g (f (3)) = g (1) = 3

(gof) (9) = g (f (9)) = g (3) = 3

(gof) (12) = g (f (12)) = g (4) = 9

⇒ gof = {(3, 3), (9, 3), (12, 9)}

Co-domain of g is a subset of the domain of f.

So, fog exists and fog: {1, 3, 4, 5} → {3, 9, 12}

(fog) (1) = f (g (1)) = f (3) = 1

(fog) (3) = f (g (3)) = f (3) = 1

(fog) (4) = f (g (4)) = f (9) = 3

(fog) (5) = f (g (5)) = f (9) = 3

⇒ fog = {(1, 1), (3, 1), (4, 3), (5, 3)}

Solution:

Given f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}

f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6, 8}

Co-domain of f = domain of g

So, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}

(gof) (1) = g (f (1)) = g (−1) = −2

(gof) (4) = g (f (4)) = g (−2) = −4

(gof) (9) = g (f (9)) = g (−3) = −6

(gof) (16) = g (f (16)) = g (4) = 8

So, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}

But the co-domain of g is not same as the domain of f.

So, fog does not exist.

Solution:

Given f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.

Also given that A = {a, b, c}, B = {u, v, w}

Now we have to show f and g both are bijective.

Consider f = {(a, v), (b, u), (c, w)} and f: A → B

Injectivity of f: No two elements of A have the same image in B.

So, f is one-one.

Surjectivity of f: Co-domain of f = {u, v, w}

Range of f = {u, v, w}

Both are same.

So, f is onto.

Hence, f is a bijection.

Now consider g = {(u, b), (v, a), (w, c)} and g: B → A

Injectivity of g: No two elements of B have the same image in A.

So, g is one-one.

Surjectivity of g: Co-domain of g = {a, b, c}

Range of g = {a, b, c}

Both are the same.

So, g is onto.

Hence, g is a bijection.

Now we have to find fog,

we know that Co-domain of g is same as the domain of f.

So, fog exists and fog: {u v, w} → {u, v, w}

(fog) (u) = f (g (u)) = f (b) = u

(fog) (v) = f (g (v)) = f (a) = v

(fog) (w) = f (g (w)) = f (c) = w

So, fog = {(u, u), (v, v), (w, w)}

Now we have to find gof,

Co-domain of f is same as the domain of g.

So, fog exists and gof: {a, b, c} → {a, b, c}

(gof) (a) = g (f (a)) = g (v) = a

(gof) (b) = g (f (b)) = g (u) = b

(gof) (c) = g (f (c)) = g (w) = c

So, gof = {(a, a), (b, b), (c, c)}

Solution:

Given f: R → R; f(x) = x² + 8 and g: R → R; g(x) = 3x³ + 1.

Consider (fog) (2) = f (g (2))

= f (3 × 2³ + 1)

= f(3 × 8 + 1)

= f (25)

= 25² + 8

= 633

(gof) (1) = g (f (1))

= g (1² + 8)

= g (9)

= 3 × 9³ + 1

= 2188

Solution:

Given f: R⁺ → R⁺ and g: R⁺ → R⁺

So, fog: R⁺ → R⁺ and gof: R⁺ → R⁺

Domains of fog and gof are the same.

Now we have to find fog and gof also we have to check whether they are equal or not,

Consider (fog) (x) = f (g (x))

= f (√x)

= √x²

= x

Now consider (gof) (x) = g (f (x))

= g (x²)

= √x²

= x

So, (fog) (x) = (gof) (x), ∀x ∈ R⁺

Hence, fog = gof

Solution:

Given f: R → R and g: R → R.

So, the domains of f and g are the same.

Consider (fog) (x) = f (g (x))

= f (x + 1) = (x + 1)²

= x² + 1 + 2x

Again consider (gof) (x) = g (f (x))

= g (x²) = x² + 1

So, fog ≠ gof

Solution:

(i) Given f (x) = e^x, g(x) = log_e x

Let f: R → (0, ∞); and g: (0, ∞) → R

Now we have to calculate fog,

Clearly, the range of g is a subset of the domain of f.

fog: ( 0, ∞) → R

(fog) (x) = f (g (x))

= f (log_e x)

= log_e e^x

= x

Now we have to calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof) (x) = g (f (x))

= g (e^x)

= log_e e^x

= x

(ii) f (x) = x², g(x) = cos x

f: R→ [0, ∞) ; g: R→[−1, 1]

Now we have to calculate fog,

Clearly, the range of g is not a subset of the domain of f.

⇒ Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}

⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}

⇒ Domain of (fog) = R

(fog): R→ R

(fog) (x) = f (g (x))

= f (cos x)

= cos² x

Now we have to calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→R

(gof) (x) = g (f (x))

= g (x²)

= cos x²

(iii) Given f (x) = |x|, g(x) = sin x

f: R → (0, ∞) ; g : R→[−1, 1]

Now we have to calculate fog,

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R→R

(fog) (x) = f (g (x))

= f (sin x)

= |sin x|

Now we have to calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog : R→ R

(gof) (x) = g (f (x))

= g (|x|)

= sin |x|

(iv) Given f (x) = x + 1, g(x) = e^x

f: R→R ; g: R → [ 1, ∞)

Now we have calculate fog:

Clearly, range of g is a subset of domain of f.

⇒ fog: R→R

(fog) (x) = f (g (x))

= f (e^x)

= e^x + 1

Now we have to compute gof,

Clearly, range of f is a subset of domain of g.

⇒ fog: R→R

(gof) (x) = g (f (x))

= g (x+1)

= e^x+1

(v) Given f (x) = sin ⁻¹ x, g(x) = x²

f: [−1,1]→ [(-π)/2 ,π/2]; g : R → [0, ∞)

Now we have to compute fog:

Clearly, the range of g is not a subset of the domain of f.

Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}

Domain (fog) = {x: x ∈ R and x² ∈ [−1, 1]}

Domain (fog) = {x: x ∈ R and x ∈ [−1, 1]}

Domain of (fog) = [−1, 1]

fog: [−1,1] → R

(fog) (x) = f (g (x))

= f (x²)

= sin⁻¹ (x²)

Now we have to compute gof:

Clearly, the range of f is a subset of the domain of g.

fog: [−1, 1] → R

(gof) (x) = g (f (x))

= g (sin⁻¹ x)

= (sin⁻¹ x)²

(vi) Given f(x) = x+1, g(x) = sin x

f: R→R ; g: R→[−1, 1]

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

Set of the domain of f.

⇒ fog: R→ R

(fog) (x) = f (g (x))

= f (sin x)

= sin x + 1

Now we have to compute gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R → R

(gof) (x) = g (f (x))

= g (x+1)

= sin (x+1)

(vii) Given f (x) = x+1, g (x) = 2x + 3

f: R→R ; g: R → R

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R→ R

(fog) (x) = f (g (x))

= f (2x+3)

= 2x + 3 + 1

= 2x + 4

Now we have to compute gof

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R → R

(gof) (x) = g (f (x))

= g (x+1)

= 2 (x + 1) + 3

= 2x + 5

(viii) Given f (x) = c, g (x) = sin x²

f: R → {c} ; g: R→ [ 0, 1 ]

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

fog: R→R

(fog) (x) = f (g (x))

= f (sin x²)

= c

Now we have to compute gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof) (x) = g (f (x))

= g (c)

= sin c²

(ix) Given f (x) = x²+ 2 and g (x) = 1 – 1 / (1 – x)

f: R → [ 2, ∞ )

For domain of g: 1− x ≠ 0

⇒ x ≠ 1

⇒ Domain of g = R − {1}

g (x )= 1 – [1/(1 – x)] = (1 – x – 1)/ (1 – x) = -x/(1 – x)

For range of g

y = (- x)/ (1 – x)

⇒ y – x y = − x

⇒ y = x y − x

⇒ y = x (y−1)

⇒ x = y/(y – 1)

Range of g = R − {1}

So, g: R − {1} → R − {1}

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R − {1} → R

(fog) (x) = f (g (x))

= f (-x/ (1 – x))

= ((-x)/ (1 – x))² + 2

= (x² + 2x² + 2 – 4x) / (1 – x)²

= (3x² – 4x + 2)/ (1 – x)²

Now we have to compute gof

Clearly, the range of f is a subset of the domain of g.

⇒ gof: R→R

(gof) (x) = g (f (x))

= g (x² + 2)

= 1 – 1 / (1 – (x² + 2))

= – 1/ (1 – (x² + 2))

= (x² + 2)/ (x² + 1)

Solution:

Given f(x) = x² + x + 1 and g(x) = sin x

Now we have to prove fog ≠ gof

(fog) (x) = f (g (x))

= f (sin x)

= sin² x + sin x + 1

And (gof) (x) = g (f (x))

= g (x²+ x + 1)

= sin (x²+ x + 1)

So, fog ≠ gof.

Solution:

Given f(x) = |x|,

Now we have to prove that fof = f.

Consider (fof) (x) = f (f (x))

= f (|x|)

= ||x||

= |x|

= f (x)

So,

(fof) (x) = f (x), ∀x ∈ R

Hence, fof = f

Solution:

f(x) and g(x) are polynomials.

⇒ f: R → R and g: R → R.

So, fog: R → R and gof: R → R.

(i) (fog) (x) = f (g (x))

= f (x² + 1)

= 2 (x² + 1) + 5

=2x² + 2 + 5

= 2x² +7

(ii) (gof) (x) = g (f (x))

= g (2x +5)

= (2x + 5)² + 1

= 4x² + 20x + 26

(iii) (fof) (x) = f (f (x))

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

(iv) f² (x) = f (x) x f (x)

= (2x + 5) (2x + 5)

= (2x + 5)²

= 4x² + 20x +25

Hence, from (iii) and (iv) clearly fof ≠ f²

Solution:

Given f(x) = sin x and g(x) = 2x

We know that

f: R→ [−1, 1] and g: R→ R

Clearly, the range of f is a subset of the domain of g.

gof: R→ R

(gof) (x) = g (f (x))

= g (sin x)

= 2 sin x

Clearly, the range of g is a subset of the domain of f.

fog: R → R

So, (fog) (x) = f (g (x))

= f (2x)

= sin (2x)

Clearly, fog ≠ gof

Hence they are not equal functions.

Solution:

Given that f(x) = sin x, g (x) = 2x and h (x) = cos x

We know that f: R→ [−1, 1] and g: R→ R

Clearly, the range of g is a subset of the domain of f.

fog: R → R

Now, (f h) (x) = f (x) h (x) = (sin x) (cos x) = ½ sin (2x)

Domain of f h is R.

Since range of sin x is [-1, 1], −1 ≤ sin 2x ≤ 1

⇒ -1/2 ≤ sin x/2 ≤ 1/2

Range of f h = [-1/2, 1/2]

So, (f h): R → [(-1)/2, 1/2]

Clearly, range of f h is a subset of g.

⇒ go (f h): R → R

⇒ Domains of fog and go (f h) are the same.

So, (fog) (x) = f (g (x))

= f (2x)

= sin (2x)

And (go (f h)) (x) = g ((f(x). h(x))

= g (sin x cos x)

= 2sin x cos x

= sin (2x)

⇒ (fog) (x) = (go (f h)) (x), ∀x ∈ R

Hence, fog = go (f h)

Solution:

(i) Given f: {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

We have:

f (1) = f (2) = f (3) = f (4) = 10

⇒ f is not one-one.

⇒ f is not a bijection.

So, f does not have an inverse.

(ii) Given g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

from the question it is clear that g (5) = g (7) = 4

⇒ f is not one-one.

⇒ f is not a bijection.

So, f does not have an inverse.

(iii) Given h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Here, different elements of the domain have different images in the co-domain.

⇒ h is one-one.

Also, each element in the co-domain has a pre-image in the domain.

⇒ h is onto.

⇒ h is a bijection.

Therefore h inverse exists.

⇒ h has an inverse and it is given by

h^-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

Solution:

(i) Given A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Here, different elements of the domain have different images in the co-domain.

Clearly, this is one-one.

Range of f = Range of f = B

so, f is a bijection and,

Thus, f ^-1 exists.

Hence, f ^-1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²

So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Here, different elements of the domain have different images in the co-domain.

Clearly, f is one-one.

But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

⇒ f is not a bijection.

So, f ^-1does not exist.

Solution:

Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.

So, f and g are invertible.

Now,

f ^-1 = {(a ,1) , (b , 2) , (3,c)} and g^-1 = {(apple, a), (ball , b), (cat , c)}

So, f^-1 o g^-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)

f: {1,2,3,} → {a, b, c} and g: {a, b, c} → {apple, ball, cat}

So, gof: {1, 2, 3} → {apple, ball, cat}

⇒ (gof) (1) = g (f (1)) = g (a) = apple

(gof) (2) = g (f (2))

= g (b)

= ball,

And (gof) (3) = g (f (3))

= g (c)

= cat

∴ gof = {(1, apple), (2, ball), (3, cat)}

Clearly, gof is a bijection.

So, gof is invertible.

(gof)^-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)

Form (1) and (2), we get

(gof)^-1 = f^-1 o g ^-1

Solution:

Given that f (x) = 2x + 1

⇒ f= {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}

= {(1, 3), (2, 5), (3, 7), (4, 9)}

Also given that g(x) = x²−2

⇒ g = {(3, 3²−2), (5, 5²−2), (7, 7²−2), (9, 9²−2)}

= {(3, 7), (5, 23), (7, 47), (9, 79)}

Clearly f and g are bijections and, hence, f⁻¹: B→ A and g⁻¹: C→ B exist.

So, f⁻¹= {(3, 1), (5, 2), (7, 3), (9, 4)}

And g⁻¹= {(7, 3), (23, 5), (47, 7), (79, 9)}

Now, (f⁻¹ o g⁻¹): C→ A

f⁻¹ o g⁻¹ = {(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)

Also, f: A→B and g: B → C,

⇒ gof: A → C, (gof) ⁻¹ : C→ A

So, f⁻¹ o g⁻¹and (gof)⁻¹ have same domains.

(gof) (x) = g (f (x))

=g (2x + 1)

=(2x +1 )²− 2

⇒ (gof) (x) = 4x² + 4x + 1 − 2

⇒ (gof) (x) = 4x²+ 4x −1

Then, (gof) (1) = g (f (1))

= 4 + 4 − 1

=7,

(gof) (2) = g (f (2))

= 4(2)² + 4(2) – 1 = 23,

(gof) (3) = g (f (3))

= 4(3)² + 4(3) – 1 = 47 and

(gof) (4) = g (f (4))

= 4(4)² + 4(4) − 1 = 79

So, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

⇒ (gof)^{– 1} = {(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)

From (1) and (2), we get:

(gof)⁻¹ = f⁻¹ o g⁻¹

Solution:

Given function f: Q → Q, defined by f(x) = 3x + 5

Now we have to show that the given function is invertible.

Injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ 3x + 5 = 3y + 5

⇒ 3x = 3y

⇒ x = y

so, f is one-one.

Surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ 3x +5 = y

⇒ 3x = y – 5

⇒ x = (y -5)/3 belongs to Q domain

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f^-1:

Let f^-1(x) = y…… (1)

⇒ x = f(y)

⇒ x = 3y + 5

⇒ x −5 = 3y

⇒ y = (x – 5)/3

Now substituting this value in (1) we get

So, f^-1(x) = (x – 5)/3

Solution:

Given f: R → R given by f(x) = 4x + 3

Now we have to show that the given function is invertible.

Consider injection of f:

Let x and y be two elements of domain (R),

Such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

So, f is one-one.

Now surjection of f:

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

Such that f(x) = y.

⇒ 4x + 3 = y

⇒ 4x = y -3

⇒ x = (y-3)/4 in R (domain)

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f ^-1

Let f^-1(x) = y……. (1)

⇒ x = f (y)

⇒ x = 4y + 3

⇒ x − 3 = 4y

⇒ y = (x -3)/4

Now substituting this value in (1) we get

So, f^-1(x) = (x-3)/4

Solution:

Given f: R → R⁺ → [4, ∞) given by f(x) = x² + 4.

Now we have to show that f is invertible,

Consider injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ x² + 4 = y² + 4

⇒ x² = y²

⇒ x = y (as co-domain as R+)

So, f is one-one

Now surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ x² + 4 = y

⇒ x² = y – 4

⇒ x = √ (y-4) in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f^-1:

Let f⁻¹ (x) = y…… (1)

⇒ x = f (y)

⇒ x = y² + 4

⇒ x − 4 = y²

⇒ y = √ (x-4)

So, f^-1(x) = √ (x-4)

Now substituting this value in (1) we get,

So, f^-1(x) = √ (x-4)

Solution:

It is given that f(x) = (4x + 3)/ (6x – 4), x ≠ 2/3

Now we have to show fof(x) = x

(fof)(x) = f (f(x))

= f ((4x+ 3)/ (6x – 4))

= (4((4x + 3)/ (6x -4)) + 3)/ (6 ((4x +3)/ (6x – 4)) – 4)

= (16x + 12 + 18x – 12)/ (24x + 18 – 24x + 16)

= (34x)/ (34)

= x

Therefore, fof(x) = x for all x ≠ 2/3

=> fof = 1

Hence, the given function f is invertible and the inverse of f is f itself.

Solution:

Given f: R⁺ → [−5, ∞) given by f(x) = 9x² + 6x – 5

We have to show that f is invertible.

Injectivity of f:

Let x and y be two elements of domain (R⁺),

Such that f(x) = f(y)

⇒ 9x² + 6x – 5 = 9y² + 6y − 5

⇒ 9x² + 6x = 9y² + 6y

⇒ x = y (As, x, y ∈ R⁺)

So, f is one-one.

Surjectivity of f:

Let y is in the co domain (Q)

Such that f(x) = y

⇒ 9x² + 6x – 5 = y

⇒ 9x² + 6x = y + 5

⇒ 9x² + 6x +1 = y + 6 (By adding 1 on both sides)

⇒ (3x + 1)² = y + 6

⇒ 3x + 1 = √(y + 6)

⇒ 3x = √ (y + 6) – 1

⇒ x = (√ (y + 6)-1)/3 in R⁺ (domain)

f is onto.

So, f is a bijection and hence, it is invertible.

Now we have to find f^-1

Let f⁻¹(x) = y….. (1)

⇒ x = f (y)

⇒ x = 9y² + 6y − 5

⇒ x + 5 = 9y² + 6y

⇒ x + 6 = 9y²+ 6y + 1 (adding 1 on both sides)

⇒ x + 6 = (3y + 1)²

⇒ 3y + 1 = √ (x + 6)

⇒ 3y =√(x +6) -1

⇒ y = (√ (x+6)-1)/3

Now substituting this value in (1) we get,

So, f^-1(x) = (√ (x+6)-1)/3

Solution:

Given f: R → R be defined by f(x) = x³ −3

Now we have to prove that f⁻¹ exists

Injectivity of f:

Let x and y be two elements in domain (R),

Such that, x³ − 3 = y³ − 3

⇒ x³ = y³

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R)

Such that f(x) = y

⇒ x³ – 3 = y

⇒ x³ = y + 3

⇒ x = ∛(y+3) in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f ^-1:

Let f^-1(x) = y…….. (1)

⇒ x= f(y)

⇒ x = y³ − 3

⇒ x + 3 = y³

⇒ y = ∛(x + 3) = f^-1(x) [from (1)]

So, f^-1(x) = ∛(x + 3)

Now, f^-1(24) = ∛ (24 + 3)

= ∛27

= ∛3³

= 3

And f^-1(5) =∛ (5 + 3)

= ∛8

= ∛2³

= 2

Solution:

Given that f: R → R is defined as f(x) = x³ + 4

Injectivity of f:

Let x and y be two elements of domain (R),

Such that f (x) = f (y)

⇒ x³ + 4 = y³ + 4

⇒ x³ = y³

⇒ x = y

So, f is one-one.

Surjectivity of f:

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

Such that f(x) = y.

⇒ x³ + 4 = y

⇒ x³ = y – 4

⇒ x = ∛ (y – 4) in R (domain)

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f^-1:

Let f⁻¹ (x) = y…… (1)

⇒ x = f (y)

⇒ x = y³ + 4

⇒ x − 4 = y³

⇒ y =∛ (x-4)

So, f^-1(x) =∛ (x-4) [from (1)]

f^-1 (3) = ∛(3 – 4)

= ∛-1

= -1