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Given as

A = {–2, –1, 0, 1, 2}

f : A → Z such that f(x) = x² – 2x – 3

(i) A is the domain of the function f. Thus, range is the set of elements f(x) for all x ∈ A.

By substituting x = –2 in f(x), we get

f(–2) = (–2)² – 2(–2) – 3

= 4 + 4 – 3

= 5

By substituting x = –1 in f(x), we get

f(–1) = (–1)² – 2(–1) – 3

= 1 + 2 – 3

= 0

By substituting x = 0 in f(x), we get

f(0) = (0)² – 2(0) – 3

= 0 – 0 – 3

= – 3

By substituting x = 1 in f(x), we get

f(1) = 1² – 2(1) – 3

= 1 – 2 – 3

= – 4

By substituting x = 2 in f(x), we get

f(2) = 2² – 2(2) – 3

= 4 – 4 – 3

= –3

Hence, the range of f is {-4, -3, 0, 5}.

(ii) pre-images of 6, –3 and 5

Suppose x be the pre-image of 6 ⇒ f(x) = 6

 x² – 2x – 3 = 6

x² – 2x – 9 = 0

x = [-(-2) ± √ ((-2)² – 4(1) (-9))] / 2(1)

= [2 ± √ (4+36)] / 2

= [2 ± √40] / 2

= 1 ± √10

However, 1 ± √10 ∉ A

Hence, there exists no pre-image of 6.

Then, suppose x be the pre-image of –3 ⇒ f(x) = –3

x² – 2x – 3 = –3

x² – 2x = 0

x(x – 2) = 0

x = 0 or 2

It is clear, both 0 and 2 are elements of A.

Hence, 0 and 2 are the pre-images of –3.

Then, suppose x be the pre-image of 5 ⇒ f(x) = 5

x² – 2x – 3 = 5

x² – 2x – 8= 0

x² – 4x + 2x – 8= 0

x(x – 4) + 2(x – 4) = 0

(x + 2)(x – 4) = 0

x = –2 or 4

However, 4 ∉ A but –2 ∈ A

Hence, –2 is the pre-images of 5.

Thus, Ø, {0, 2}, -2 are the pre-images of 6, -3, 5

A function from is defined by a set of ordered pairs such that any two ordered pairs should not have the same first component and the different second component. 

This means that each element of a set, say X is assigned exactly to one element of another set, say Y. 

The set X containing the first components of a function is called the domain of the function. 

The set Y containing the second components of a function is called the range of the function. 

For example,

f = {(a, 1), (b, 2), (c, 3)} is a function. 

Domain of f = {a, b, c} 

Range of f = {1, 2, 3}

A function from a set X to a set Y is defined as a correspondence between sets X and Y such that for each element of X, there is only one corresponding element in Y. 

The set X is called the domain of the function. 

The set Y is called the range of the function. 

For example, 

X = {a, b, c}, Y = {1, 2, 3, 4, 5} and f be a correspondence which assigns the position of a letter in the set of alphabets. 

Therefore, 

f(a) = 1, f(b) = 2 and f(c) = 3. 

As there is only one element of Y for each element of X, f is a function with domain X and range Y.

Random numbers are used for fames, simulations, testing, security, and privacy applicatons. Python includes following functions that are commonly used.

Given f(x) = x³ - \(\frac{1}{x^3}\)

Need to prove: f(x) + f\((\frac{1}{x})\) = 0

Replacing x by \(\frac{1}{x}\) we get,

f\((\frac{1}{x})\) = \(\frac{1}{x^3}\) - \(\frac{1}{\frac{1}{x^3}}\) = \(\frac{1}{x^3}\) - x³

Now according to the problem,

f(x) + f\((\frac{1}{x})\) = x³ - \(\frac{1}{x^3}\) + \(\frac{1}{x^3}\) - x³

⇒ f(x) + f\((\frac{1}{x})\) = 0 

[proved]

Given,

f(x) = 2x + 5 and g(x) = x² + x 

Clearly,

Both f(x) and g(x) are defined for all x ∈ R. 

Hence, 

Domain of f = domain of g = R 

i. f + g 

We know,

(f + g)(x) = f(x) + g(x) 

⇒ (f + g)(x) = 2x + 5 + x² + x 

∴ (f + g)(x) = x² + 3x + 5 

Clearly, 

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

∴ The domain of (f + g) is R 

ii. f – g 

We know,

(f – g)(x) = f(x) – g(x) 

⇒ (f – g)(x) = 2x + 5 – (x² + x) 

⇒ (f – g)(x) = 2x + 5 – x² – x 

∴ (f – g)(x) = 5 + x – x² 

Clearly, 

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

∴ The domain of (f – g) is R 

iii. fg 

We know,

(fg)(x) = f(x)g(x) 

⇒ (fg)(x) = (2x + 5)(x² + x) 

⇒ (fg)(x) = 2x(x² + x) + 5(x² + x) 

⇒ (fg)(x) = 2x³ + 2x² + 5x² + 5x 

∴ (fg)(x) = 2x³ + 7x² + 5x 

Clearly, 

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

∴ The domain of fg is R

 iv. \(\frac{f}{g}\) 

We know,

(\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\)

∴ (\(\frac{f}{g}\))(x) = \(\frac{2x+5}{x^2+x}\)

Clearly,

(\(\frac{f}{g}\))(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,(\(\frac{f}{g}\))(x) will be undefined as the division result will be indeterminate.

Thus, 

Domain of \(\frac{f}{g}\) = R – {–1, 0}

Suppose ‘f’ be the function and R be the relation defined from the set X to set Y.

A domain of a relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is because each element of the domain of a function must have an element associated with it, whereas this is not necessary for a relation.

In the relation, one element of X might be associated with the one or more elements of Y, while it must be associated with the only one element of Y in a function.

Hence, not every relation is a function. However, every function is necessarily a relation.

|x² – x – 6| = x + 2 …..(i) 

R.H.S. must be non-negative 

∴ x ≥ -2 …..(ii) 

|(x – 3) (x + 2)| = x + 2 

∴ (x + 2) |x – 3| = x + 2 as x + 2 ≥ 0 

∴ |x – 3| = 1 if x ≠ -2 

∴ x – 3 = ±1 

∴ x = 4 or 2 

∴ x = -2 also satisfies the equation 

∴ Solution set = {-2, 2, 4}

1 < |x – 1| < 4 

∴ -4 < x – 1 < -1 or 1 < x – 1 < 4 

∴ -3 < x < 0 or 2 < x < 5 

∴ Solution set = (-3, 0) ∪ (2, 5)

(i) {x} > 4

This is a meaningless statement as 0 ≤ {x} < 1

∴ The solution set = { } or Φ

(ii) {x} = 0

∴ x is an integer

∴ The solution set is Z.

(iii) {x} = 0.5

∴ x = ….., -2.5, -1.5, -0.5, 0.5, 1.5, …..

∴ The solution set = {x : x = n + 0.5, n ∈ Z}

(iv) 2{x} = x + [x]

= [x] + {x} + [x] ……[x = [x] + {x}]

∴ {x} = 2[x]

R.H.S. is an integer

∴ L.H.S. is an integer

∴ {x} = 0

∴ [x] = 0

∴ x = 0

g(x) = x³ – 2x² – 5x + 6

= ( x- 1) (x² – x – 6)

= (x – 1) (x + 2) (x – 3)

g(x) = 0

∴ (x – 1) (x + 2) (x – 3) = 0

∴ x – 1 = 0 or x + 2 = 0

or x – 3 = 0

∴ x = 1, -2, 3

f(x) = x⁴ + 2x² , g(x) = 11x²

f(x) = g(x)

∴ x⁴ + 2x² = 11x²

∴ x⁴ – 9x² = 0

∴ x² (x²– 9) = 0

∴ x² = 0 or x² – 9 = 0

∴ x = 0 or x² = 9

∴ x = 0, ±3

f(x) = 1/1+√x = y, (say)

∴ √x y + y = 1

∴ √x = 1-y/y ≥ 0

∴= y-1/y  ≤ 0

∴ o < y ≤ 1 

∴ Range of f = (0, 1]

f(x) = x/9+x² = y (say)

∴ x² y – x + 9y = 0 

For real x, Discriminant > 0 

∴ 1 – 4(y)(9y) ≥ 0

∴ y² ≤ 1/36

∴ -1/6 ≤ y ≤ 1/6

∴ Range of f = [-1/6, 1/6]

 f(x) = \(\sqrt{x-x^2} + \sqrt {5-x}\)

For f to be defined, 

log (x² – 6x + 6) ≥ 0 

∴ x² – 6x + 6 ≥ 1 

∴ x² – 6x + 5 ≥ 0

∴ (x – 5)(x – 1) ≥ 0 

∴ x ≤ 1 or x ≥ 5 …..(i) 

[∵ The solution of (x – a) (x – b) ≥ 0 is x ≤ a or

x ≥ b, for a < b]

and x² – 6x + 6 > 0 

∴ (x – 3)² > -6 + 9 

∴ (x – 3)² > 3 

∴ x < 3 – √3 0r x > 3 + √3 ……..(ii) 

From (i) and (ii), we get 

x ≤ 1 or x ≥ 5 

Solution set = (-∞, 1] ∪ [5, ∞) 

∴ Domain of f = (-∞, 1] ∪ [5, ∞)

Given,

f(x) = (x – a)²(x – b)² 

We need to find,

f(a + b). 

We have, 

f(a + b) = (a + b – a)²(a + b – b)² 

⇒ f(a + b) = (b)²(a)² 

∴ f(a + b) = a²b² 

Thus,

f(a + b) = a²b²