Functions

Definition:A relation R from a set A to a set B is called a function if each element of A has a unique image in B.

It is denoted by the symbol f:A→B which reads ‘f’ is a function from A to B ‘f’ maps A to B.

Let f:A→B,then the set A is known as the domain of f & the set B is known as co - domain of f .The set of images of all the elements of A is known as the range of f.

Thus, Domain of f = {a|a ∈ A,(a,f(a)) ∈ f )

Range of f = {f(a) | a∈ A ,f(a) ∈ B }

Example: The domain of y = sin x is all values of x i.e. R , since there are no restrictions on the values for x. The range of y is betweeen −1 and 1. We could write this as −1 ≤ y ≤ 1.

1)injective function

Definition: A function f: A → B is said to be a one - one function or injective mapping if different elements of A have different f images in B.

A function f is injective if and only if whenever f(x) = f(y), x = y.

Example: f(x) = x + 9 from the set of real number R to R is an injective function. When x = 3,then :f(x) = 12,when f(y) = 8,the value of y can only be 3,so x = y.

(ii) surjective function

Definition: If the function f:A→B is such that each element in B (co - domain) is the ‘f’ image of atleast one element in A , then we say that f is a function of A ‘onto’ B .Thus f: A→B is surjective if, for all b∈B, there are some a∈A such that f(a) = b.

Example: The function f(x) = 2x from the set of natural numbers N to the set of non negative even numbers is a surjective function.

(iii) bijective function

Definition: A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y.Alternatively, f is bijective if it is a one - to - one correspondence between those sets, in other words, both injective and surjective.

Example: If f(x) = x²,from the set of positive real numbers to positive real numbers is both injective and surjective.Thus it is a bijective function.

(iv)many - one function

Defintion : A function f: A→B is said to be a many one functions if two or more elements of A have the same f image in B.

trigonometric functions such as sinx are many - to - one since sinx = sin(2 + x) = sin(4 + x) and so one…

(v) into function

Definition: If f:A→B is such that there exists atleast one element in co - domain , which is not the image of any element in the domain , then f(x) is into.

Let f(x) = y = x – 1000

⇒ x = y + 1000 = g(y) (say)

Here g(y) is defined for each y∈ I , but g(y) ∉ N for y ≤ − 1000. Hence,f is into.

(i) one - one but not onto

f(x) = 6x

For One - One

f(x₁) = 6x₁

f(x₂) = 6x₂

put f(x₁) = f(x₂) we get

6x₁ = 6x₂

Hence, if f(x₁) = f(x₂) , x_{1 =} x₂

Function f is one - one

For Onto

f(x) = 6x

let f(x) = y ,such that y∈N

6x = y

⇒

If y = 1

x =

which is not possible as x∈N

Hence, f is not onto.

(ii) one - one and onto

f(x) = x⁵

⇒y = x⁵

Since the lines do not cut the curve in 2 equal valued points of y, therefore, the function f(x) is one - one.

The range of f(x) = ( - ∞,∞) = R(Codomain)

∴f(x) is onto

∴f(x) is one - one and onto.

(iii) neither one - one nor onto

f(x) = x²

for one one:

f(x₁) = (x₁)²

f(x₂) = (x₂)²

f(x₁) = f(x₂)

⇒(x₁)² = (x₂)²

⇒x₁ = x₂ or x₁ = - x₂

Since x₁ does not have a unique image it is not one - one

For onto

f(x) = y

such that y∈R

x² = y

⇒x =

If y is negative under root of a negative number is not real

Hence,f(x) is not onto.

∴f(x) is neither onto nor one - one

(iv) onto but not one - one.

Consider a function f:Z→N such that f(x) = |x|.

Since the Z maps to every single element in N twice, this function is onto but not one - one.

Z - integers

N - natural numbers.

i)f(2)

Since f(x) = x² - 2 , when x = 2

∴ f(2) = (2)² - 2 = 4 - 2 = 2

∴f(2) = 2

ii)f(4)

Since f(x) = 3x - 1 , when x = 4

∴f(4) = (3×4) - 1 = 12 - 1 = 11

∴f(4) = 11

iii)f( - 1)

Since f(x) = x² - 2 , when x = - 1

∴ f( - 1) = ( - 1)² - 2 = 1 - 2 = - 1

∴f( - 1) = - 1

iv)f( - 3)

Since f(x) = 2x + 3 , when x = - 3

∴f( - 3) = 2×( - 3) + 3 = - 6 + 3 = - 3

∴f( - 3) = - 3

To show: f: R → R : f(x) = 1 + x² is many - one into.

Proof:

f(x) = 1 + x²

⇒y = 1 + x²

Since the lines cut the curve in 2 equal valued points of y therefore the function f(x) is many one.

The range of f(x) = [1,∞)≠R(Codomain)

∴f(x) is not onto

⇒f(x) is into

Hence, showed that f: R → R : f(x) = 1 + x² is many - one into.

To show: f: R → R : f(x) = x⁴ is many - one into.

Proof:

f(x) = x⁴

⇒y = x⁴

Since the lines cut the curve in 2 equal valued points of y, therefore, the function f(x) is many ones.

The range of f(x) = [0,∞)≠R(Codomain)

∴f(x) is not onto

⇒f(x) is into

Hence, showed that f: R → R : f(x) = x⁴ is many - one into.

To show: f: R → R : : f(x) = x⁵ is one - one and onto.

Proof:

f(x) = x⁵

⇒y = x⁵

Since the lines do not cut the curve in 2 equal valued points of y, therefore, the function f(x) is one - one.

The range of f(x) = ( - ∞,∞) = R(Codomain)

∴f(x) is onto

Hence, showed f: R → R : f(x) = x⁵ is one - one and onto.

Here in this range, the lines do not cut the curve in 2 equal valued points of y, therefore, the function f(x) = sinx is one - one.

in this range, the lines do not cut the curve in 2 equal valued points of y, therefore, the function f(x) = cosx is also one - one.

(f + g):[0,] →R = sinx + cosx

in this range the lines cut the curve in 2 equal valued points of y, therefore, the function f(x) = cosx + sinx is not one - one.

Hence,showed that each one of f and g is one - one but (f + g) is not one - one.

(i) f : N → N : f(x) = x² is one - one into.

f(x) = x²

⇒y = x²

Since the function f(x) is monotonically increasing from the domain N → N

∴f(x) is one –one

Range of f(x) = (0,∞)≠N(codomain)

∴f(x) is into

∴f : N → N : f(x) = x² is one - one into.

(ii) f : Z → Z : f(x) = x² is many - one into

f(x) = x²

⇒y = x²

in this range the lines cut the curve in 2 equal valued points of y, therefore, the function f(x) = x² is many - one .

Range of f(x) = (0,∞)≠Z(codomain)

∴f(x) is into

∴ f : Z → Z : f(x) = x² is many - one into

(i) f : N → N : f(x) = x³ is one - one into.

f(x) = x³

Since the function f(x) is monotonically increasing from the domain N → N

∴f(x) is one –one

Range of f(x) = ( - ∞,∞)≠N(codomain)

∴f(x) is into

∴f : N → N : f(x) = x² is one - one into.

(ii) f : Z → Z : f(x) = x³ is one - one into

f(x) = x³

Since the function f(x) is monotonically increasing from the domain Z → Z

∴f(x) is one –one

Range of f(x) = ( - ∞,∞)≠Z(codomain)

∴f(x) is into

∴ f : Z → Z : f(x) = x³ is one - one into.

f(x) = sinx

y = sinx

Here in this range, the lines cut the curve in 2 equal valued points of y, therefore, the function f(x) = sinx is not one - one.

Range of f(x) = [ - 1,1]≠R(codomain)

∴f(x) is not onto.

Hence, showed that the function f : R → R : f(x) = sin x is neither one - one nor onto.

In the given range of N f(x) is monotonically increasing.

∴f(n) = n² + n + 1 is one one.

But Range of f(n) = [0.75,∞)≠N(codomain)

Hence,f(n) is not onto.

Hence, proved that the function f : N → N : f(n) = (n² + n + 1) is one - one but not onto.

f(1) = 0

f(2) = - 1

f(3) = 1

f(4) = - 2

f(5) = 2

f(6) = - 3

Since at no different values of x we get same value of y ∴f(n) is one –one

And range of f(n) = Z = Z(codomain)

∴ the function f: N → Z, defined by

is both one - one and onto.

Since the function f(x) can accept any values as per the given domain R, therefore, the domain of the function f(x) = x² + 1 is R.

The minimum value of f(x) = 1

⇒Range of f(x) = [ - 1,∞]

i.e range (f) = {y ∈ R : y ≥ 1}

Ans: dom (f) = R and range (f) = {y ∈ R : y ≥ 1}

For a relation to be a function each element of 1^st set should have different image in the second set(Range)

i) (i) f = {( - 1, 2), (1, 8), (2, 11), (3, 14)}

Here, each of the first set element has different image in second set.

∴f is a function whose domain = { - 1, 1, 2, 3} and range (f) = {2, 8, 11, 14}

(ii) g = {(1, 1), (1, - 1), (4, 2), (9, 3),
(16, 4)}

Here, some of the first set element has same image in second set.

∴ g is not a function.

(iii) h = {(a, b), (b, c), (c, b), (d, c)}

Here, each of the first set element has different image in second set.

∴h is a function whose domain = {a, b, c, d} and range (h) = {b, c}

(range is the intersection set of the elements of the second set elements.)

For domain (1 + x²)≠0

⇒x²≠ - 1

⇒dom(f) = R

For the range of x:

⇒

y_min = 0 (when x = 0)

y_max = 1 (when x = ∞)

∴range of f(x) = [0,1)

For many one the lines cut the curve in 2 equal valued points of y therefore the function f(x) is many - one.

Ans:

dom(f) = R

range(f) = [0,1)

function f(x) is many - one.

(i)

_{Here, x = 1/2,which is rational}

_{∴f(1/2) = 1}

(ii)

_{Here, x = √2,which is irrational}

_{∴f(√2) = - 1}

(iii)

_{Here, x = ∏, which is irrational}

_{= - 1}

(iv)

_{Here,x = 2 +√3, which is irrational}

_{∴f(2 +√3) = - 1}

Ans. (i) 1 (ii) - 1 (iii) - 1 (iv) - 1

(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1),

(3, 2), (4, 4)}

Solution: We have,

gof(1) = g(f(1)) = g(4) = 4

gof(2) = g(f(2)) = g(1) = 3

gof(3) = g(f(3)) = g(3) = 2

gof(4) = g(f(4)) = g(2) = 1

Ans) g o f = {(1, 4), (2, 3), (3, 2), (4, 1)}

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1),

(3, 2), (4, 4)}

Solution: We have,

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

fog(2) = f(g(2)) = f(1) = 4

fog(3) = f(g(3)) = f(2) = 1

fog(4) = f(g(4)) = f(4) = 2

Ans) f o g = {(1, 3), (2, 4), (3, 1), (4, 2)}

(iii) f o f

To find: f o f

Formula used: f o f = f(f(x))

Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)}

Solution: We have,

fof(1) = f(f(1)) = f(4) = 2

fof(2) = f(f(2)) = f(1) = 4

fof(3) = f(f(3)) = f(3) = 3

fof(4) = f(f(4)) = f(2) = 1

Ans) f o f = {(1, 2), (2, 4), (3, 3), (4, 1)}

(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

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

Solution: We have,

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

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

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

Ans) g o f = {(3, 3), (9, 3), (12, 9)}

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

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

Solution: We have,

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

Ans) f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}

To prove: (g o f) ≠ (f o g)

Formula used: (i) g o f = g(f(x))

(ii) f o g = f(g(x))

Given: (i) f : R → R : f(x) = x²

(ii) g : R → R : g(x) = (x + 1)

Proof: We have,

g o f = g(f(x)) = g(x²) = ( x² + 1 )

f o g = f(g(x)) = g(x+1) = [ (x+1)² + 1 ] = x² + 2x + 2

From the above two equation we can say that (g o f) ≠ (f o g)

Hence Proved

(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

Given: (i) f : R → R : f(x) = (2x + 1)

(ii) g : R → R : g(x) = (x² - 2)

Solution: We have,

g o f = g(f(x)) = g(2x + 1) = [ (2x + 1)² – 2 ]

⇒ 4x² + 4x + 1 – 2

⇒ 4x² + 4x – 1

Ans). g o f (x) = 4x² + 4x – 1

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

Given: (i) f : R → R : f(x) = (2x + 1)

(ii) g : R → R : g(x) = (x² - 2)

Solution: We have,

f o g = f(g(x)) = f(x² - 2) = [ 2(x² - 2) + 1 ]

⇒ 2x² - 4 + 1

⇒ 2x² – 3

Ans). f o g (x) = 2x² – 3

(iii) f o f

To find: f o f

Formula used: f o f = f(f(x))

Given: (i) f : R → R : f(x) = (2x + 1)

Solution: We have,

f o f = f(f(x)) = f(2x + 1) = [ 2(2x + 1) + 1 ]

⇒ 4x + 2 + 1

⇒ 4x + 3

Ans). f o f (x) = 4x+ 3

(iv) g o g

To find: g o g

Formula used: g o g = g(g(x))

Given: (i) g : R → R : g(x) = (x² - 2)

Solution: We have,

g o g = g(g(x)) = g(x² - 2) = [ (x² - 2)² – 2]

⇒ x⁴ -4x² + 4 - 2

⇒ x⁴ -4x² + 2

Ans). g o g (x) = x⁴ -4x² + 2

(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

Given: (i) f : R → R : f(x) = (x² + 3x + 1)

(ii) g: R → R : g(x) = (2x - 3)

Solution: We have,

g o f = g(f(x)) = g(x² + 3x + 1) = [ 2(x² + 3x + 1) – 3 ]

⇒ 2x² + 6x + 2 – 3

⇒ 2x² + 6x – 1

Ans). g o f (x) = 2x² + 6x – 1

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

Given: (i) f : R → R : f(x) = (x² + 3x + 1)

(ii) g: R → R : g(x) = (2x - 3)

Solution: We have,

f o g = f(g(x)) = f(2x - 3) = [ (2x - 3)² + 3(2x – 3) + 1 ]

⇒ 4x² - 12x + 9 + 6x – 9 + 1

⇒ 4x² - 6x + 1

Ans). f o g (x) = 4x² - 6x + 1

(iii) g o g

To find: g o g

Formula used: g o g = g(g(x))

Given: (i) g: R → R : g(x) = (2x - 3)

Solution: We have,

g o g = g(g(x)) = g(2x - 3) = [ 2(2x – 3) - 3 ]

⇒ 4x – 6 - 3

⇒ 4x - 9

Ans). g o g (x) = 4x – 9

To prove: f o f = f

Formula used: f o f = f(f(x))

Given: (i) f : R → R : f(x) = |x|

Solution: We have,

f o f = f(f(x)) = f(|x|) = = |x| = f(x)

Clearly f o f = f.

Hence Proved.

|

To find: formula for h o (g o f)

To prove:

Formula used: f o f = f(f(x))

Given: (i) f : R → R : f(x) = x²

(ii) g : R → R : g(x) = tan x

(iii) h : R → R : h(x) = log x

Solution: We have,

h o (g o f) = h o g(f(x)) = h o g(x²)

= h(g(x²)) = h (tan x²)

= log (tan x²)

h o (g o f) = log (tan x²)

For,

= 0

Hence Proved.

To prove: (f o g) = I_R = (g o f).

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = (2x - 3)

(ii)

Solution: We have,

f o g = f(g(x))

= x + 3 – 3

= x

= I_R

g o f = g(f(x))

=

= x

= I_R

Clearly we can see that (f o g) = I_R = (g o f) = x

Hence Proved.

To find: g : Z → Z : g o f = I_Z

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) g : Z → Z : g o f = I_Z

Solution: We have,

f(x) = 2x

Let f(x) = y

⇒ y = 2x

Where g: Z → Z

For g o f,

⇒ g(f(x))

⇒ g(2x)

⇒ x = I_Z

Clearly we can see that (g o f) = x = I_Z

To show: h o (g o f ) = (h o g) o f

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : N → N : f(x) = 2x

(ii) g : N → N : g(y) = 3y + 4

(iii) h : N → N : h(z) = sin z

Solution: We have,

LHS = h o (g o f )

⇒ h o (g(f(x))

⇒ h(g(2x))

⇒ h(3(2x) + 4)

⇒ h(6x +4)

⇒ sin(6x + 4)

RHS = (h o g) o f

⇒ (h(g(x))) o f

⇒ (h(3x + 4)) o f

⇒ sin(3x+4) o f

Now let sin(3x+4) be a function u

RHS = u o f

⇒ u(f(x))

⇒ u(2x)

⇒ sin(3(2x) + 4)

⇒ sin(6x + 4) = LHS

Hence Proved.

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f is a greatest integer function

(ii) g is an absolute value function

f(x) = [x] (greatest integer function)

g(x) = (absolute value function)

Substituting values from (i) and (ii)

⇒ [1.5] +

⇒ 1 + 1 = 2

Ans) 2

To find: f o g, g o f ,(f o g) (2) and (g o f) (-3)

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = x² + 2

f o g = f(g(x))

Ans) = 6

g o f = g(f(x))

⇒ g(x²+2)

To prove: function is one-one and onto

Given: f: R → R : f(x)= 2x

We have,

f(x) = 2x

For, f(x₁) = f(x₂)

⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

f(x) = 2x

Let f(x) = y such that

⇒ y = 2x

Since ,

⇒ x will also be a real number, which means that every value of y is associated with some x

∴ f(x) is onto

Hence Proved

To prove: function is one-one and into

Given: f: N → N : f(x)= 3x

We have,

f(x) = 3x

For, f(x₁) = f(x₂)

⇒ 3x₁ = 3x₂

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

f(x) = 3x

Let f(x) = y such that

⇒ y = 3x

If y = 1,

But as per question , hence x can not be

Hence f(x) is into

Hence Proved

To prove: function is neither one-one nor onto

Given: f : R → R : f (x) = x²

Solution: We have,

f(x) = x²

For, f(x₁) = f(x₂)

⇒ x₁² = x₂²

⇒ x₁ = x₂ or, x₁ = -x₂

Since x₁ doesn’t has unique image

∴ f(x) is not one-one

f(x) = x²

Let f(x) = y such that

⇒ y = x²

If y = -1, as

Then x will be undefined as we cannot place the negative value under the square root

Hence f(x) is not onto

Hence Proved

To prove: function is one-one and into

Given: f : N → N : f (x) = x²

Solution: We have,

f(x) = x²

For, f(x₁) = f(x₂)

⇒ x₁² = x₂²

⇒ x₁ = x₂

Here we can’t consider x₁ = -x₂ as , we can’t have negative values

∴ f(x) is one-one

f(x) = x²

Let f(x) = y such that

⇒ y = x²

If y = 2, as

Then we will get the irrational value of x, but

Hence f(x) is not into

Hence Proved

To prove: function is neither one-one nor onto

Given: f : R → R : f (x) = x⁴

We have,

f(x) = x⁴

For, f(x₁) = f(x₂)

⇒ x₁⁴ = x₂⁴

⇒ (x₁⁴ - x₂⁴) = 0

⇒(x₁² - x₂²) (x₁² + x₂²) = 0

⇒ (x₁ - x₂) (x₁ + x₂) (x₁² + x₂²) = 0

⇒ x₁ = x₂ or, x₁ = -x₂ or, x₁² = -x₂²

We are getting more than one value of x₁ (no unique image)

∴ f(x) is not one-one

f(x) = x⁴

Let f(x) = y such that

⇒ y = x⁴

If y = -2, as

Then x will be undefined as we can’t place the negative value under the square root

Hence f(x) is not onto

Hence Proved

To prove: function is one-one and into

Given: f : Z → Z : f (x) = x³

Solution: We have,

f(x) = x³

For, f(x₁) = f(x₂)

⇒ x₁³ = x₂³

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

f(x) = x³

Let f(x) = y such that

⇒ y = x³

If y = 2, as

Then we will get an irrational value of x, but

Hence f(x) is into

Hence Proved

To prove: function is one-one and onto

We have,

For, f(x₁) = f(x₂)

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

Let f(x) = y such that

Since ,

⇒ x will also , which means that every value of y is associated with some x

∴ f(x) is onto

Hence Proved

To prove: function is many-one into

Given: f : R → R : f(x) = 1 + x²

We have,

f(x) = 1 + x²

For, f(x₁) = f(x₂)

⇒ 1 + x₁² = 1 + x₂²

⇒ x₁² = x₂²

⇒ x₁² - x₂² = 0

⇒ (x₁ – x₂) (x₁ + x₂) = 0

⇒ x₁ = x₂ or, x₁ = –x₂

Clearly x₁ has more than one image

∴ f(x) is many-one

f(x) = 1 + x²

Let f(x) = y such that

⇒ y = 1 + x²

⇒ x² = y – 1

If y = 3, as

Then x will be undefined as we can’t place the negative value under the square root

Hence f(x) is into

Hence Proved

To find: f^-1

Given:

We have,

Let f(x) = y such that

⇒ 4y = 2x – 7

⇒ 4y + 7 = 2x

To find: f^-1

Given: f : R → R : f(x) = 10x + 3

We have,

f(x) = 10x + 3

Let f(x) = y such that

⇒ y = 10x + 3

⇒ y – 3 = 10x

To prove: function is many-one and into

We have,

f(x) = 1 when x is rational

It means that all rational numbers will have same image i.e. 1

⇒ f(2) = 1 = f (3) , As 2 and 3 are rational numbers

Therefore f(x) is many-one

The range of function is [{-1},{1}] but codomain is set of real numbers.

Therefore f(x) is into

To find: (f o g) (7)

Formula used: f o g = f(g(x))

Given: (i) f (x) = x + 7

(ii) g (x) = x – 7

We have,

f o g = f(g(x)) = f(x – 7) = [ (x – 7) + 7 ]

⇒ x

(f o g) (x) = x

(f o g) (7) = 7

Ans). (f o g) (7) = 7

To prove: g o f ≠ f o g

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = x²

(ii)

We have,

f o g = f(g(x)) = f(x + 7)

f o g = (x + 7)² = x² + 14x + 49

g o f = g(f(x)) = g(x²)

g o f = (x² + 1) = x² + 1

Clearly g o f ≠ f o g

Hence Proved

To find: f o f

Formula used: (i) f o f = f(f(x))

Given: (i) f : R → R : f(x) = (3 - x³)^1/3

We have,

f o f = f(f(x)) =

f o f =

=

=

=

= x

Ans) f o f (x) = x

To find: f{f(x)}

Formula used: (i) f o f = f(f(x))

Given: (i) f : R → R : f(x) = 3x + 2

We have,

f{f(x)} = f(f(x)) = f(3x + 2)

f o f =3(3x + 2) + 2

= 9x + 6 + 2

= 9x + 8

Ans) f{f(x)} = 9x + 8

To find: g o f

Formula used: g o f = g(f(x))

Given: (i) f = {(1, 2), (3, 5), (4, 1)}

(ii) g = {(1, 3), (2, 3), (5, 1)}

We have,

gof(1) = g(f(1)) = g(2) = 3

gof(3) = g(f(3)) = g(5) = 1

gof(4) = g(f(4)) = g(1) = 3

Ans) g o f = {(1, 3), (3, 1), (4, 3)}

To find: f o f

Formula used: f o f = f(f(x))

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

We have,

fof(1) = f(f(1)) = f(4) = 2

fof(2) = f(f(2)) = f(1) = 4

fof(3) = f(f(3)) = f(3) = 3

fof(4) = f(f(4)) = f(2) = 1

Ans) f o f = {(1, 2), (2, 4), (3, 3), (4, 1)}

To find: g o f and f o g

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f(x) = 8x³

(ii) g(x) = x^1/3

We have,

g o f = g(f(x)) = g(8x³)

g o f = = 2x

f o g = f(g(x)) = f(x^1/3)

f o g = = 8x

Ans) g o f = 2x and f o g = 8x

To find: the function g : R → R : g o f = f o g = I_g

Formula used: (i) g o f = g(f(x))

(ii) f o g = f(g(x))

Given: f : R → R : f(x) = 10x + 7

We have,

f(x) = 10x + 7

Let f(x) = y

⇒ y = 10x + 7

⇒ y – 7 = 10x

Let where g: R → R

g o f = g(f(x)) = g(10x + 7)

= x

= I_g

f o g = f(g(x)) =

=

= x – 7 + 7

= x

Clearly g o f = f o g = I_g

To state: Whether f is one-one

Given: f = {(1, 4), (2,5), (3, 6)}

Here the function is defined from A → B

For a function to be one-one if the images of distinct elements of A under f are distinct

i.e. 1,2 and 3 must have a distinct image.

From f = {(1, 4), (2, 5), (3, 6)} we can see that 1, 2 and 3 have distinct image.

Therefore f is one-one

Ans) f is one-one

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x_l, x₂ A & f(x₁), f(x₂) B, f(x1) = f(x2) x₁= x₂ or x₁ x₂f(x1) f(x2)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

As we see that inthe above figure (2 is mapped with 7), (3 is mapped with 9), (4 is mapped with 11),

(5 is mapped with 13)

So it is one-one functions.

Now elements of B are known as co-domain. Also, a range of a function is also the elements of B(by definition)

So it is onto functions.

Hence Proved that f is invertible.

Now, We know that if f : A B then f^-1 : B A (if it is invertible)

So,

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

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ R and f(x) = 2x+3.So f(x₁) = f(x₂) 2x₁+3 = 2x₂+3 x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is R.

Let y = f(x) = 2x+3 , So x = [Range of f(x) = Domain of y]

So Domain of y is R(real no.) = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) =

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = 3x-4.So f(x₁) = f(x₂) 3x₁ - 4 = 3x₂ - 4 x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is Q.

Let y = f(x) = 3x- 4 , So x = [Range of f(x) = Domain of y]

So Domain of y is Q = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = Q

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) =

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = .So f(x₁) = f(x₂) = x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is R.

Let y = f(x) =, So x = [Range of f(x) = Domain of y]

So Domain of y is R = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) =

To Show: that f o f (x) = x

Finding (f o f) (x) == ==.

To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = .So f(x₁) = f(x₂) = on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is R except 3x-2=0.

Let y = f(x) = So x = [Range of f(x) = Domain of y]

So Domain of y is R (except 3x-2=0) = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R except 3x-2=0

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = .

To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = .So f(x₁) = f(x₂) = on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is R except 3x+4=0.

Let y = f(x) = So x = [Range of f(x) = Domain of y]

So Domain of y is R = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R except 3x+4=0

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) =

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ R and f(x) = (9x² + 6x – 5).So f(x₁) = f(x₂) (9² + 6 – 5) = (9² + 6 – 5) on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is [-5, ∞]

Let y = f(x) = (9x² + 6x – 5), So x = [Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = [-5, ∞]

Hence, Range of f(x) = co-domain of f(x) =[-5, ∞]

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) =.

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ R and f(x) = 4x² + 12x + 15 So f(x₁) = f(x₂) (4² + 12 +1 5) = (4² + 12 +1 5), on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is Range(f).

Let y = f(x) = 4x² + 12x + 15, So x = [Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = [6, ∞]

Hence, Range of f(x) = co-domain of f(x) = [6, ∞]

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) =.

To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ A & f(x₁), f(x₂) B, f(x₁) = f(x₂) x₁= x₂ or x₁ x₂f(x₁) f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = .So f(x₁) = f(x₂) = , on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is R – {1}

Let y = f(x) =, So x = [Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = R – {1}

Hence, Range of f(x) = co-domain of f(x) = R – {1}.

So, f(x) is onto function

As it is a bijective function. So it is invertible

Invers of f(x) is f^-1(y) =

To Find: Inverse of f o g and g o f.

Given: f(x) = |x| + x and g(x) = |x| - x for all x ∈ R

f o g (x) = f(g(x)) = |g(x)| + g(x) = ||x| - x | + |x| - x

Case 1) when x0

f(g(x)) = 0 (i.e. |x| - x)

Case 2) when x 0

f(g(x)) = -4x

g o f (x) = g(f(x)) = |f(x)| - f(x) = ||x| + x | - |x| - x

Case 1) when x0

g(f(x)) = 0 (i.e. |x| - x)

Case 2) when x 0

g(f(x)) = 0

f(x) = 2x

For One - One

f(x₁) = 2x₁

f(x₂) = 2x₂

put f(x₁) = f(x₂) we get

2x₁ = 2x₂

Hence, if f(x₁) = f(x₂), x_{1 =} x₂

Function f is one - one

For Onto

f(x) = 2x

let f(x) = y, such that y∈N

2x = y

⇒

If y = 1

x =

which is not possible as x∈N

Hence, f is not onto., f is into

Hence, option b is correct

In the given range of N f(x) is monotonically increasing.

∴f(x) = x² + x + 1 is one one.

But Range of f(n) = [0.75,∞)≠N(codomain)

Hence, f(x) is not onto.

Hence, the function f : N → N : f(x) = (x² + x + 1) is one - one but not onto. i.e. into

f(x) = x²

⇒y = x²

in this range the lines cut the curve in 2 equal valued points of y, therefore, the function f(x) = x² is many - one .

Range of f(x) = (0,∞)≠R(codomain)

∴f(x) is into

∴ f : R → R: f(x) = x² is many - one into

f(x) = x³

Since the function f(x) is monotonically increasing from the domain R → R

∴f(x) is one –one

Range of f(x) = ( - ∞,∞)≠R(codomain)

∴f(x) is into

∴ f : R → R: f(x) = x³ is one - one into.

f(x) = e^x

Since the function f(x) is monotonically increasing from the domain R ⁺ → R ⁺

∴f(x) is one –one

Range of f(x) = (1,∞) = R ⁺ (codomain)

∴f(x) is onto

∴ f : R ⁺ → R ⁺ : f(x) = e^x is one - one onto.

_{Here in this range, the function is NOT repeating its value,}

Therefore it is one - one.

Range = Codomain

∴Function is onto

Hence, option B is the correct choice.

f(x) = cosx

y = cosx

Here in this range the lines cut the curve in many equal valued points of y therefore the function f(x) = cosx is not one - one.

⇒f(x) = many one

Range of f(x) = [ - 1,1]≠R(codomain)

∴f(x) is not onto.

⇒f(x) = into

Hence,f(x) = cosx is many one and into

Ans: (c) many - one and into

Here in this range the lines cut the curve in 2 equal valued points of y therefore the function f(z) = |z| is not one - one

⇒f(z) = many one.

Range of f(z) = [0,∞)≠R(codomain)

∴f(z) is not onto.

⇒f(z) = into

Hence, f(z) = |z| is many one and into

_{In this function}

_{x = 3 and y = 1 are the asymptotes of this curve and these are not included in the functions of the domain and range respectively therefore the function f(x) is one one sice there are no different values of x which has same value of y .}

_{and the function has no value at y = 1 here range = codomain}

_{∴f(x) is onto}

SINCE, f(a, b) = (b, a).There is no same value of y at different values of x ∴function is one one

∴Range(A×B)≠Codomain(B × A)

⇒function is into

f(x) = 2x + 3

⇒y = 2x + 3

x⟺y

⇒x = 2y + 3

⇒x - 3 = 2y

⇒

x⟺y

⇒

f(x) =

⇒y =

x⟺y

⇒x =

⇒3yx + 4x = 4y

⇒y(3x - 4) = - 4x

⇒y =

x⟺y

⇒x =

f(x) = (x² – 1)

g(x) = (2x + 3)

∴(g o f) (x) = g(f(x))

f(x) = x² – 3x + 2

⇒f(x) = x² - 2x - x + 2 = x(x - 2) - 1(x - 2)

⇒f(x) = (x - 2)(x - 1)

⇒f(x) = (x - 2)(x - 1)

⇒f(f(x)) = ( f(x) - 2)( f(x) - 1)

⇒f(f(x)) = ((x - 2)(x - 1) - 2) ((x - 2)(x - 1) - 1)

⇒f(f(x)) = (x² – 3x + 2 - 2) (x² – 3x + 2 - 1)

⇒f(f(x)) = (x² – 3x) (x² – 3x + 1)

⇒f(f(x)) = x⁴ - 3x³ + x² - 3x³ + 9x² - 3x

⇒f(f(x)) = x⁴ - 6x³ + 10x² - 3x

f(x) = 8x³

g(x) = x^1/3

f(x) = x², g(x) = tan x and h(x) = log x

g = {(2, 3), (5, 1), (1, 3)}

(g o f) = {(dom(f), 3), (dom(f), 1), (dom(f), 3)}

⇒(g o f) = {(1, 3), (3, 1), (4, 3)}

F(x) =

∴dom(f) = [ - 3, 3]

And x≠4

And

Taking the intersection we get

Dom(f) = (1, ∞)

X² - 1≠0

x≠(1, - 1)

∴ Dom(f) = R - { - 1,1}

Given:

From f(x), x ≠ 0

Now, domain of sin^-1x is [-1, 1] as the values of sin^-1x lies between -1 and 1.

We can see that from this graph:

Domain of f(x) = [-1, 1] – 0

Hence, B is the correct answer.

f(x) = cos ^{- 1} 2x.

domain of cos ^{- 1}x = [ - 1,1]

on multiplying by an integer the domain decreases by same number

domain of cos ^{- 1}2x = [ - 1/2,1/2]

f(x) = cos ^{- 1} (3x - 1).

domain of cos ^{- 1}x = [ - 1,1]

on multiplying by an integer the domain decreases by same number

domain of cos ^{- 1}3x = [ - 1/3,1/3]

domain of cos ^{- 1} (3x - 1) = [1/3 - 1/3,1/3 + 1/3] = [0,2/3]

As per the diagram

We can imply that domain of √cosx

is

f(x) = √log (2x – x²).

Dom(f) = ( - ∞, 1)

According to sketched graph of x²

Domain of f(x) = R

And Range of f(x) = R ⁺

According to sketched graph of x³

Domain of f(x) = R

And Range of f(x) = R

Since x³ is a, monotonically increasing function

log (1 – x) + √(x² – 1)

1 - x>0

X² - 1≥0

X²1

Taking intersection of the ranges we get

Dom (f) = (b) ( - ∞, - 1]

∴ range (f) = [1, ∞)

⇒y≥0

And

1 - y>0

⇒y<1

Taking intersection we get

range (f) = [0, 1)

For this type

Range is

f(x) = a^x

when x<0

0<a^x<1

When x≥0

a^x>0

Therefore range of f(x) = a^x = (0, ∞)

f(1) = 1

f(2) = 1

f(3) = 2

f(4) = 2

f(5) = 3

f(6) = 3

Since at different values of x we get same value of y ∴f(n) is many –one

And range of f(n) = N = N(codomain)

∴ the function f: N → Z, defined by

is both many - one and onto.

f(x) = 4x² + 12x + 15

⇒y = 4x² + 12x + 15

⇒y = (2x + 3)² + 6

⇒√(y - 6) = 2x + 3

⇒

f ^{- 1}(y) =