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Given,

X = {–1, 0, 3, 9, 7} 

f : X → R and 

f(x) = x³ + 1 

When x = –1, we have 

f(–1) = (–1)³ + 1 

⇒ f(–1) = –1 + 1 

∴ f(–1) = 0 

When x = 0, we have 

f(0) = 0³ + 1 

⇒ f(0) = 0 + 1 

∴ f(0) = 1 

When x = 3, we have 

f(3) = 3³ + 1 

⇒ f(3) = 27 + 1 

∴ f(3) = 28 

When x = 9, we have 

f(9) = 9³ + 1 

⇒ f(9) = 729 + 1 

∴ f(9) = 730 

When x = 7, we have 

f(7) = 7³ + 1 

⇒ f(7) = 343 + 1 

∴ f(7) = 344 

Thus, 

f = {(–1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}

Given,

f : R → R and f(x) = x². 

i. range of f 

Domain of f = R 

(set of real numbers) 

We know that the square of a real number is always positive or equal to zero. 

Hence, 

The range of f is the set of all non-negative real numbers. 

Thus, 

Range of f = R +∪ {0} 

ii. {x: f(x) = 4} 

Given,

f(x) = 4 

⇒ x² = 4 

⇒ x² – 4 = 0 

⇒ (x – 2)(x + 2) = 0 

∴ x = ±2 

Thus, 

{x: f(x) = 4} = {–2, 2} 

iii. {y: f(y) = –1} 

Given,

f(y) = –1 

⇒ y² = –1 

However, 

The domain of f is R, and for every real number y, the value of y² is non-negative. 

Hence, 

There exists no real y for which y² = –1. 

Thus,

{y: f(y) = –1} = ∅

Given as

f : R → R and f(x) = x²

(i) The domain of f = R (the set of real numbers)

As we know that the square of a real number is always positive or equal to zero.

∴ range of f = R⁺∪ {0}

(ii) Given as

f(x) = 4

As we know, x² = 4

x² – 4 = 0

(x – 2)(x + 2) = 0

∴ x = ± 2

∴ {x: f(x) = 4} = {–2, 2}

(iii) Given as

f(y) = –1

y² = –1

However, the domain of f is R, and for every real number y, the value of y² is non-negative.

Thus, there exists no real y for which y² = –1.

∴ {y: f(y) = –1} = ∅

Given : 

A = {–2, –1, 0, 2} 

f : A → Z: f(x) = x² – 2x – 3 

Finding f(x) for each value of x, 

f(-2) = (-2)² – 2(-2) – 3 = 4 + 4 -3 = 5 

f(-1) = (-1)² – 2(-1) – 3 = 1 + 2 -3 = 0 

f(0) = (0)² – 2(0) – 3 = 0 + 0 -3 = -3 

f(2) = (2)² – 2(2) – 3 = 4 - 4 -3 = -3 

f in ordered pair is represented as 

f = {(-2,5),(-1,0),(0,-3),(2,-3)}

Given, X = {–1, 0, 2, 5} 

f : X → R Z: f(x) = x³ + 1 

Finding f(x) for each value of x, 

(1) f(-1) = (-1)³ + 1 = -1 + 1 = 0 

(2) f(0) = (0)³ + 1 = 0 + 1 = 1 

(3) f(2) = (2)³ + 1 = 8 + 1 = 9 

(4) f(5) = (5)³ + 1 = 125 + 1 = 126 

f in ordered pair is represented as 

f = {(-1,0),(0,1),(2,9),(5,126)}

f(x) = x + 3, g(x) = 3x² – 1 

To find:- Set of values of x for which f(x) = g(x) 

Consider, 

f(x) = g(x) 

x + 3 = 3x² – 1 

3x² - x – 4 = 0 

3x² - 4x + 3x -4 = 0 

x(3x-4) +(3x-4) = 0 

(3x - 4)(x + 1) = 0 

x = 4/3 or x= -1 

The set values for which f(x) and g(x) have same value is { 4/3 , -1}.

f(x) = 1 – 3x, g(x) = 2x² – 1 

To find:- Set of values of x for which f(x) = g(x) 

Consider, 

f(x) = g(x) 

1 – 3x = 2x² – 1 

2x² + 3x -2 = 0 

2x² + 4x – x – 2 = 0 

2x ( x + 2) –( x + 2 ) = 0 

(x+2)(2x-1) = 0 

x = -2 or x = 1/2

The set values for which f(x) and g(x) have same value is { -2 , 1/2 }.

One – One Function: – A function f : A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f : A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Now, f : A → B, denotes a mapping such that

f = {(2,2),(5,3),(6,4),(7,4)}

Here it is clear that every first component is from B and second component is from A

Hence this is mapping from B to A

One – One Function: – A function f : A → B, is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f : A → B, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Now, f : A → B, denotes a mapping such that

f = {(2,2),(3,5),(4,5)}

Hence this is not injective mapping

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

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

Therefore, f is the function.

Injection test:

As, y can be mother of two or more persons

Therefore, 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}

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

Therefore, g is not a function.

(i) domain

f(x) = \(\frac{ax+b}{bx-a}\)

As f(x) is a polynomial function whose domain is R except for the points where the denominator becomes 0. 

Hence x ≠ a/b

Domain is R a/{-b} 

(ii) Range

Let y = \(\frac{ax+b}{bx-a}\)

Y(bx-a) = ax +b 

byx -ay = ax + b 

byx -ax= ay +b 

x(by -a) = ay + b

x = \(\frac{ay+b}{by-a}\)

x is not defined when denominator is zero.

by – a ≠ 0 

y ≠ a/b 

Range is R-{a/b}.

Given,

f(x) = 1/√(x-|x|)

For f(x) to be defined,

x-|x|>0 

But x-|x|≤0 

So, 

f(x) does not exist.. 

Therefore,

D(f) = R(f) = ϕ

Given,

f(x) = √(x-[x])

For f(x) to be defined,

x-[x]≥0 

We know that,

{x} + {x} = x

Where {x} is fractional part function and [x] is greatest integer function.

{x}≥0

Also,

0 ≤ {x} < 1

Therefore, 

D(f) = R and range = [0, 1)

Given,

f(x) = √([x]-x)

For function to be defined, 

[x]-x≥0 -{x}≥ 

Therefore, 

Domain of f(x) is integers. 

D(f)∈I Range = {0}.

f(x) = 3x² -1;

g(x) = 3+x 

For f(x) = g(x)

3x² -1 = 3+x 

3x² -x-4 = 0 

(3x-4)(x+1) = 0 

3x-4 = 0 or x+1 = 0

x = \(\frac{4}{3}\), 1

For each value of set A, 

We can have q functions as each value of A pair up with all the values of B. 

So, 

Total number of functions from A to B = q× q× q…..{p times} 

= q^p

Option : (C)

A function is said to be defined from A to B if each element in set A has an unique image in set B. 

Not all the elements in set B are the images of any element of set A. 

Therefore, 

Option C is correct.

Given,

f = {(2, 4), (5, 6), (8, -1), (10, -3)} and 

g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, -5)}.

D(f) = {2, 5, 8, 10} 

D(g) = {2, 7, 8, 10, 11} 

Therefore, 

D(f+g) = {2, 8, 10}

Given,

f = {(0, 1), (2, 0), (3, -4), (4, 2), (5, 1)} and 

g = {(1, 0), (2, 2), (3, -1), (4, 4), (5, 3)}

D(f) = {0, 2, 3, 4, 5} 

D(g) = {1, 2, 3, 4, 5} 

So,

D(fg) = {2, 3, 4, 5}

|x| is defined as 

|x| = x; x > = 0 

-x; x<0 

\(\frac{1}{|x|}\)= \(\frac{1}{x}\); x>0 

= \(\frac{-1}{x}\) x<0 

\(\frac{x}{|x|}\)= 1; x > 0

= -1; x < 0

Hence f(x) gives output values 1 and -1 only. 

Range is {1,-1}.

Given: f(x) = \(\frac{x-5}{5-x}\)

 (i) dom(f) 

Here f(x) is a polynomial function whose domain is R except for points at which denominator becomes zero. 

Hence x ≠ 5 

The domain is (−∞,∞) – {5} 

(ii) range(f)

Let y = \(\frac{x-5}{5-x}\)

For the specified domain 

y = - 1 

Range is {-1}.

Given: f(x) = \(\frac{x^2-9}{x-3}\)

Need to find: Where the functions are defined. 

To find the domain of the function f(x) we need to equate the denominator of the function to 0. 

Therefore, 

x – 3 = 0 

⇒ x = 3 

It means that the denominator is zero when x = 3 

So, the domain of the function is the set of all the real numbers except 3. 

The domain of the function, D_f(x) = (- ∞, 3) ∪ (3, ∞). 

Now if we put any value of x from the domain set the output value will be either (-ve) or (+ve), but the value will never be 6 

So, the range of the function is the set of all the real numbers except 6. 

The range of the function, R_f(x) = (-∞, 6) ∪(6, ∞).

Given: f(x) = \(\frac{1}{\sqrt{2x-3}}\)

Need to find: Where the functions are defined.

2x - 3 > 0

⇒ x > \(\frac{3}2\)

So, the domain of the function is the set of all the real numbers greater than \(\frac{3}2\).

The domain of the function, D_f(x) = (\(\frac{3}2\),∞).

Now putting any value of x within the domain set we get the value of the function always a fraction whose denominator is not equals to 0. 

The range of the function, R_f(x) = (0, 1).

Given: f(x) = \(\frac{x^2-16}{x-4}\)

Need to find: Where the functions are defined. 

To find the domain of the function f(x) we need to equate the denominator of the function to 0. 

Therefore, 

x – 4 = 0 

⇒ x = 4 

It means that the denominator is zero when x = 4 

So, the domain of the function is the set of all the real numbers except 4. 

The domain of the function, D_f(x) = (- ∞, 4) ∪ (4, ∞). 

Now if we put any value of x from the domain set the output value will be either (-ve) or (+ve), but the value will never be 8

So, the range of the function is the set of all the real numbers except 8. 

The range of the function, R_f(x) = (-∞, 8) ∪(8, ∞).

Given: f(x) = \(\frac{3x-2}{x+2}\)

Need to find: Where the functions are defined.

Let, f(x) = \(\frac{3x-2}{x+2}\) = y....(1)

To find the domain of the function f(x) we need to equate the denominator of the function to 0. 

Therefore, 

x + 2 = 0 

⇒ x = -2 

It means that the denominator is zero when x = -2 

So, the domain of the function is the set of all the real numbers except -2. 

The domain of the function, D_f(x) = (- ∞, -2) ∪ (-2, ∞). 

Now, to find the range of the function we need to interchange x and y in the equation no. (1) 

So the equation becomes,

\(\frac{3y-2}{2+y}\) = x

⇒ 3y - 2 = 2x + xy

⇒ 3y - xy = 2x + 2

⇒ y = \(\frac{2x+2}{3-x}\) = f(x₁)

To find the range of the function f(x₁) we need to equate the denominator of the function to 0. 

Therefore, 

3 – x = 0 

⇒ x = 3 

It means that the denominator is zero when x = 3 

So, the range of the function is the set of all the real numbers except 3. 

The range of the function, R_f(x) = (- ∞, 3) ∪ (3, ∞).

Given,

f(x) = {(3x-2,x < 0)(1,x = 0)(4x +1,x > 0)

We need to find,

f(1), f(–1), f(0) and f(2). 

When x > 0, f(x) = 4x + 1 

Substituting x = 1 in the above equation, we get 

f(1) = 4(1) + 1 

⇒ f(1) = 4 + 1 

∴ f(1) = 5 

When x < 0, f(x) = 3x – 2 

Substituting x = –1 in the above equation, we get 

f(–1) = 3(–1) – 2 

⇒ f(–1) = –3 – 2 

∴ f(–1) = –5 

When x = 0, f(x) = 1 

∴ f(0) = 1 

When x > 0, f(x) = 4x + 1 

Substituting x = 2 in the above equation, we get 

f(2) = 4(2) + 1 

⇒ f(2) = 8 + 1

∴ f(2) = 9 

Thus, 

f(1) = 5, 

f(–1) = –5, 

f(0) = 1 and f(2) = 9.

Given as

Let us find the f(1), f(–1), f(0) and f(2).

If x > 0, f (x) = 4x + 1

By substituting x = 1 in the above equation, we get

f (1) = 4(1) + 1

= 4 + 1

= 5

If x < 0, f(x) = 3x – 2

By substituting x = –1 in the above equation, we get

f (–1) = 3(–1) – 2

= –3 – 2

= –5

If x = 0, f(x) = 1

By, substituting x = 0 in the above equation, we get

f(0) = 1

If x > 0, f(x) = 4x + 1

By substituting x = 2 in the above equation, we get

f(2) = 4(2) + 1

= 8 + 1

= 9

Thus f(1) = 5, f(–1) = –5, f(0) = 1 and f(2) = 9.

Given: f(x) = \(\frac{1}{2-sin3x}\)

Need to find: Where the functions are defined. 

The maximum value of an angle is 2π 

So, the maximum value of x = 2π/3. 

Whereas, the minimum value of x is 0 

Therefore, the domain of the function, D_f(x) = (0, 2π/3). 

Now, the minimum value of sinθ = 0 and the maximum value of sinθ = 1. So, the minimum value of the denominator is 1, and the maximum value of the denominator is 2. 

Therefore, the range of the function, R_f(x) = (1/2, 1).

Given: f(x) = \(\frac{1}x\)

Need to find: Where the functions are defined.

Let, f(x) = \(\frac{1}x\) = y.......(1)

To find the domain of the function f(x) we need to equate the denominator of the function to 0. 

Therefore, 

x = 0 

It means that the denominator is zero when x = 0 

So, the domain of the function is the set of all the real numbers except 0. 

The domain of the function, D_f(x) = (- ∞, 0) ∪ (0, ∞). 

Now, to find the range of the function we need to interchange x and y in the equation no. (1) 

So the equation becomes,

\(\frac{1}y\) = x

⇒ y = \(\frac{1}x\) = f(x₁)

To find the range of the function f(x1) we need to equate the denominator of the function to 0. 

Therefore, 

x = 0 

It means that the denominator is zero when x = 0 

So, the range of the function is the set of all the real numbers except 0. 

The range of the function, R_f(x) = (- ∞, 0) ∪ (0, ∞).

Given: 

f : R → R : f(x) = x² + 1 

To find inverse of f(x) 

Let y = f(x) 

y = x² + 1 

y -1 = x²

x = \(\sqrt{y-1}\)

f^-1 (x) = \(\sqrt{x-1}\)

Substituting x = 10

f^–1 (10) = \(\sqrt{10-1}\) = √9 = 3

Given as f: R⁺→ R and f(x) = log_e x.

(i) The domain of f = R⁺ (the set of positive real numbers)

As we know the value of logarithm to the base e (natural logarithm) can take all possible real values.

∴ The image set of f = R

(ii) Given as f(x) = –2

log_e x = –2

∴ x = e^-2 [since, log_b a = c ⇒ a = b^c]

∴ {x: f(x) = –2} = {e^–2}

(iii) Here, we have f (x) = log_e x ⇒ f (y) = log_e y

Then, let us consider the f(xy)

F(xy) = log_e (xy)

f(xy) = log_e (x × y) [since, log_b (a×c) = log_b a + log_b c]

f(xy) = log_e x + log_e y

f(xy) = f (x) + f (y)

Thus the equation f(xy) = f(x) + f(y) holds.

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

∴ f(2) = 3, f(4) = 6, f(5) = 2

g ={(2, 4), (3, 4), (6, 2)}

∴ g(2) = 4, g(3) = 4, g(6) = 2

gof: {2, 4, 5} → {2, 4}

(gof) (2) = g(f(2)) = g(3) = 4

(gof) (4) = g(f(4)) = g(6) = 2

(gof) (5) = g(f(5)) = g(2) = 4

∴ gof = {(2, 4), (4, 2), (5, 4)}

Given as

f(x) = 2x + 5 and g(x) = x² + x

Here, both f(x) and g(x) are defined for all x ∈ R.

Therefore, domain of f = domain of g = R

(i) f + g

As we know, (f + g)(x) = f(x) + g(x)

(f + g)(x) = 2x + 5 + x² + x

= x² + 3x + 5

Now, (f + g)(x) is defined for all real numbers x.

∴ The domain of (f + g) is R

(ii) f – g

As we know, (f – g)(x) = f(x) – g(x)

(f – g)(x) = 2x + 5 – (x² + x)

= 2x + 5 – x² – x

= 5 + x – x²

(f – g)(x) is defined for all real numbers x.

∴ The domain of (f – g) is R

(iii) fg

As we know, (fg)(x) = f(x)g(x)

(fg)(x) = (2x + 5)(x² + x)

= 2x(x² + x) + 5(x² + x)

= 2x³ + 2x² + 5x² + 5x

= 2x³ + 7x² + 5x

(fg)(x) is defined for all real numbers x.

∴ The domain of fg is R

(iv) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = (2x+5)/(x²+x)
Here, (f/g) (x) is defined for all real values of x, except for the case when x² + x = 0.

x² + x = 0

x(x + 1) = 0

x = 0 or x + 1 = 0

x = 0 or –1

When x = 0 or –1, (f/g) (x) will be undefined as the division result will be indeterminate.

Thus, the domain of f/g = R – {–1, 0}

f(x) = 3x + 5, g (x) = 6x – 1

(i) (f + g) (x) = f (x) + g (x)

= 3x + 5 + 6x – 1

= 9x + 4

(ii) (f – g) (2) = f(2) – g(2)

= [3(2) + 5] – [6(2) – 1]

= 6 + 5 – 12 + 1

= 0

(iii) (fg) (3) = f (3) g(3)

= [3(3) + 5] [6(3) – 1]

= (14) (17)

= 238

(iv) \((\frac fg)x=\frac{f(x)}{g(x)}=\frac{3x+5}{6x-1}, x\neq \frac16\)

Domain = R - \(\{\frac16\}\) 

Given as

f(x) = √(x + 1) and g(x) = √(9 - x²)

As we know the square of a real number is never negative.

Therefore, f(x) takes real values only when x + 1 ≥ 0

x ≥ –1, x ∈ [–1, ∞)

The domain of f = [–1, ∞)

Similarly, g(x) takes real values only when 9 – x² ≥ 0

9 ≥ x²

x² ≤ 9

x² – 9 ≤ 0

x² – 3² ≤ 0

(x + 3)(x – 3) ≤ 0

x ≥ –3 and x ≤ 3

∴ x ∈ [–3, 3]

The domain of g = [–3, 3]

(i) f + g

As we know, (f + g)(x) = f(x) + g(x)

(f + g) (x) = √(x + 1) + √(9 - x²)

The domain of f + g = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

∴ f + g: [–1, 3] → R is given by (f + g) (x) = f(x) + g(x) = √(x+1) + √(9-x²) 

(ii) g – f

As we know, (g – f)(x) = g(x) – f(x)

(g – f) (x) = √(9 - x²) – √(x + 1)
The domain of g – f = Domain of g ∩ Domain of f

= [–3, 3] ∩ [–1, ∞) 

= [–1, 3]

∴ g – f: [–1, 3] → R is given by (g – f) (x) = g(x) – f(x) = √(9 - x²) – √(x + 1)

(iii) fg

As we know, (fg) (x) = f(x)g(x)

(fg) (x) = √(x + 1)√(9 - x²)
= √[(x + 1) (9 - x²)]

= √[x(9 - x²) + (9 - x²)]

= √(9x - x³ + 9 - x²)

= √(9 + 9x - x² - x³)

The domain of fg = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

∴ fg: [–1, 3] → R is given by (fg) (x) = f(x) g(x) = √(x + 1)√(9 - x²) = √(9 + 9x - x² - x³)

(iv) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = √(x + 1)/√(9 - x²)

= √[(x + 1)/(9 - x²)]

The domain of f/g = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

However, (f/g)(x) is defined for all real values of x ∈ [–1, 3], except for the case when 9 – x² = 0 or x = ± 3

When x = ±3, (f/g) (x) will be undefined as the division result will be indeterminate.

The domain of f/g = [–1, 3] – {–3, 3}

The domain of f/g = [–1, 3)

∴ f/g: [–1, 3) → R is given by (f/g) (x) = f(x)/g(x) = √(x + 1) / √(9 - x²)

(v) g/f

As we know, (g/f) (x) = g(x)/f(x)  

(g/f) (x) = √(9 - x²)/√(x + 1)

= √[(9 - x²)/(x + 1)]

The domain of g/f = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]
However, (g/f) (x) is defined for all real values of x ∈ [–1, 3], except for the case when x + 1 = 0 or x = –1

If x = –1, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = [–1, 3] – {–1}

Domain of g/f = (–1, 3]

∴ g/f: (–1, 3] → R is given by (g/f) (x) = g(x)/f(x) = √(9 - x²)/√(x + 1)

(vi) 2f – √5g   

As we know, (2f – √5g) (x) = 2f(x) – √5g(x)

(2f – √5g) (x) = 2f (x) – √5g (x)

= 2√(x + 1) – √5√(9 - x²)

= 2√(x + 1) – √(45 - 5x²)

The domain of 2f – √5g = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

∴ 2f – √5g: [–1, 3] → R is given by (2f – √5g) (x) = 2f (x) – √5g (x) = 2√(x + 1) – √(45 - 5x²)

(vii) f² + 7f

As we know, (f² + 7f) (x) = f²(x) + (7f)(x)

(f² + 7f) (x) = f(x) f(x) + 7f(x)

= √(x + 1)√(x + 1) + 7√(x + 1)

= x + 1 + 7√(x + 1)

Domain of f² + 7f is same as domain of f.

Domain of f² + 7f = [–1, ∞)

∴ f² + 7f: [–1, ∞) → R is given by (f² + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7√(x + 1)

(viii) 5/g 

As we know, (5/g) (x) = 5/g(x)

(5/g) (x) = 5/√(9 - x²)

The domain of 5/g = Domain of g = [–3, 3]

However, (5/g) (x) is defined for all real values of x ∈ [–3, 3], except for the case when 9 – x² = 0 or x = ± 3

If x = ±3, (5/g) (x) will be undefined as the division result will be indeterminate.

The domain of 5/g = [–3, 3] – {–3, 3}

= (–3, 3)

∴ 5/g: (–3, 3) → R is given by (5/g) (x) = 5/g(x) = 5/√(9 - x²) 

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 \(y\in R\)

⇒ y = x⁴

^{\(\Rightarrow x=\sqrt[4]{y}\)}

If y = -2, as \(y\in R\)

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

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

It is given that

f(x) = x⁴

By substituting the values

f(1) = 1⁴ = 1

f(-1) = -1⁴ = 1

Hence, f is many-one.

Consider -1 in the co-domain R with no x ∈ R such that f(x) = x⁴ = – 1

f is into.

One – One Function: – A function f: A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Now, We have, A = {–1, 0, 1} and f = {(x, x²) : x ∈ A}.

To Prove: – f : A → A is neither One – One nor onto function

Check for Injectivity:

We can clearly see that

f(1) = 1

and f( – 1) = 1

Therefore

f(1) = f( – 1)

⇒ Every element of A does not have different image from A

Hence f is not One – One function

Check for Surjectivity:

Since, y = – 1 be element belongs to A

i.e -1 ∈ A in co – domain does not have any pre image in domain A.

Hence, f is not Onto function.

(i) Domain 

|x | is defined for all real values. 

Hence -|x| is also defined for all real values. 

The domain is R.

(ii) Range 

Range for |x| is [0, ∞) 

Therefore, range for - |x| is ( -∞,0].

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 \(y\in N\)

⇒ y = 3x

⇒ \(x=\frac{y}{3}\)

If y = 1,

⇒ \(x=\frac{1}{3}\)

But as per question \(x\in E\), hence x can not be \(\frac{1}{3}\)

Hence f(x) is into

Hence Proved

log x function has domain R⁺. 

When x is replaced by |x|, the function f shows value as 

f(x) = log(x); x>0 

= log(-x); x<0 

Hence in the function x cannot be zero as log function is not defined for x = 0. 

Domain of f(x) is R – {0}

\(f(x)=\begin{cases}\frac{1}{2}(n-1)\text{ when n is odd}\\-\frac{1}{2}n,\text{ when n is even}\end{cases}\)

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

\(f(x)=\begin{cases}\frac{1}{2}(n-1)\text{ when n is odd}\\-\frac{1}{2}n,\text{ when n is even}\end{cases}\)

is both one - one and onto.