Functions

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.

Let f be a function and R be a relation defined from set X to set Y.

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

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

Given A = {–2, –1, 0, 1, 2}

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

i. range of f i.e. f(A)

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

Substituting x = –2 in f(x), we get

f(–2) = (–2)² – 2(–2) – 3

⇒ f(–2) = 4 + 4 – 3

∴ f(–2) = 5

Substituting x = –1 in f(x), we get

f(–1) = (–1)² – 2(–1) – 3

⇒ f(–1) = 1 + 2 – 3

∴ f(–1) = 0

Substituting x = 0 in f(x), we get

f(0) = (0)² – 2(0) – 3

⇒ f(0) = 0 – 0 – 3

∴ f(0) = –3

Substituting x = 1 in f(x), we get

f(1) = 1² – 2(1) – 3

⇒ f(1) = 1 – 2 – 3

∴ f(1) = –4

Substituting x = 2 in f(x), we get

f(2) = 2² – 2(2) – 3

⇒ f(2) = 4 – 4 – 3

∴ f(2) = –3

Thus, the range of f is {5, 0, –3, –4}.

ii. pre-images of 6, –3 and 5

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

⇒ x² – 2x – 3 = 6

⇒ x² – 2x – 9 = 0

However,

Thus, there exists no pre-image of 6.

Now, let 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

Clearly, both 0 and 2 are elements of A.

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

Now, let 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

Thus, –2 is the pre-images of 5.

Given

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 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 f : R⁺→ R and f(x) = log_ex.

i. the image set of the domain of f

Domain of f = R⁺ (set of positive real numbers)

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

Hence, the image set of f is the set of real numbers.

Thus, the image set of f = R

ii. {x: f(x) = –2}

Given f(x) = –2

⇒ log_ex = –2

∴ x = e^-2 [∵ log_ba = c ⇒ a = b^c]

Thus, {x: f(x) = –2} = {e^–2}

iii. whether f(xy) = f(x) + f(y) holds.

We have f(x) = log_ex ⇒ f(y) = log_ey

Now, let us consider f(xy).

f(xy) = log_e(xy)

⇒ f(xy) = log_e(x × y) [∵ log_b(a×c) = log_ba + log_bc]

⇒ f(xy) = log_ex + log_ey

∴ f(xy) = f(x) + f(y)

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

i. {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}

When x = 1, we have y = 3(1) = 3

When x = 2, we have y = 3(2) = 6

When x = 3, we have y = 3(3) = 9

Thus, R = {(1, 3), (2, 6), (3, 9)}

Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components.

Hence, the given relation R is a function.

ii. {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}

When x = 1, we have y > 1 + 1 or y > 2 ⇒ y = {4, 6}

When x = 2, we have y > 2 + 1 or y > 3 ⇒ y = {4, 6}

Thus, R = {(1, 4), (1, 6), (2, 4), (2, 6)}

Every element of set x has an ordered pair in the relation. However, two ordered pairs (1, 4) and (1, 6) have the same first component but different second components.

Hence, the given relation R is not a function.

iii. {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}

When x = 0, we have 0 + y = 3 ⇒ y = 3

When x = 1, we have 1 + y = 3 ⇒ y = 2

When x = 2, we have 2 + y = 3 ⇒ y = 1

When x = 3, we have 3 + y = 3 ⇒ y = 0

Thus, R = {(0, 3), (1, 2), (2, 1), (3, 0)}

Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components.

Hence, the given relation R is a function.

Given f : R → R ∋ f(x) = x² and g : R → R ∋ g(x) = x²

As f is defined from R to R, the domain of f = R.

As g is defined from C to C, the domain of g = C.

Two functions are equal only when the domain and codomain of both the functions are equal.

In this case, the domain of f ≠ domain of g.

Thus, f and g are not equal functions.

i. f(x) = x²

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 = [0, ∞) = {y: y ≥ 0}

ii. g(x) = sin x

Domain of g = R (set of real numbers)

We know that the value of sine function always lies between –1 and 1.

Hence, the range of g is the set of all real numbers lying in the range –1 to 1.

Thus, range of g = [–1, 1] = {y: –1 ≤ y ≤ 1}

iii. h(x) = x² + 1

Domain of h = R (set of real numbers)

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

Furthermore, if we add 1 to this squared number, the result will always be greater than or equal to 1.

Hence, the range of h is the set of all real numbers greater than or equal to 1.

Thus, range of h = [1, ∞) = {y: y ≥ 1}

Given X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}

i. f₁ = {(1, 1), (2, 11), (3, 1), (4, 15)}

Every element of set X has an ordered pair in the relation f₁ and no two ordered pairs have the same first component but different second components.

Hence, the given relation f₁ is a function.

ii. f₂ = {(1, 1), (2, 7), (3, 5)}

In the relation f₂, the element 2 of set X does not have any image in set Y.

However, for a relation to be a function, every element of the domain should have an image.

Hence, the given relation f₂ is not a function.

iii. f₃ = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

Every element of set X has an ordered pair in the relation f₃. However, two ordered pairs (2, 9) and (2, 11) have the same first component but different second components.

Hence, the given relation f₃ is not a function.

Given A = {12, 13, 14, 15, 16, 17}

f : A → Z such that f(x) = highest prime factor of x.

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

We have f(12) = highest prime factor of 12

The prime factorization of 12 = 2² × 3

Thus, the highest prime factor of 12 is 3.

∴ f(12) = 3

We have f(13) = highest prime factor of 13

We know 13 is a prime number.

∴ f(13) = 13

We have f(14) = highest prime factor of 14

The prime factorization of 14 = 2 × 7

Thus, the highest prime factor of 14 is 7.

∴ f(14) = 7

We have f(15) = highest prime factor of 15

The prime factorization of 15 = 3 × 5

Thus, the highest prime factor of 15 is 5.

∴ f(15) = 5

We have f(16) = highest prime factor of 16

The prime factorization of 16 = 2⁴

Thus, the highest prime factor of 16 is 2.

∴ f(16) = 2

We have f(17) = highest prime factor of 17

We know 17 is a prime number.

∴ f(17) = 17

Thus, the range of f is {3, 13, 7, 5, 2, 17}.

Given f : R → R and f(x) = x² + 1.

We need to find f^-1{17} and f^-1{–3}.

Let f^-1{17} = x

⇒ f(x) = 17

⇒ x² + 1 = 17

⇒ x² – 16 = 0

⇒ (x – 4)(x + 4) = 0

∴ x = ±4

Clearly, both –4 and 4 are elements of the domain R.

Thus, f^-1{17} = {–4, 4}

Now, let f^-1{–3} = x

⇒ f(x) = –3

⇒ x² + 1 = –3

⇒ x² = –4

However, the domain of f is R and for every real number x, the value of x² is non-negative.

Hence, there exists no real x for which x² = –4.

Thus, f^-1{–3} = ∅

Given A = {p, q, r, s} and B = {1, 2, 3}

i. R₁ = {(p, 1), (q, 2), (r, 1), (s, 2)}

Every element of set A has an ordered pair in the relation R₁ and no two ordered pairs have the same first component but different second components.

Hence, the given relation R₁ is a function.

ii. R₂ = {(p, 1), (q, 1), (r, 1), (s, 1)}

Every element of set A has an ordered pair in the relation R_2, and no two ordered pairs have the same first component but different second components.

Hence, the given relation R₂ is a function.

iii. R₃ = {(p, 1), (q, 2), (p, 2), (s, 3)}

Every element of set A has an ordered pair in the relation R₃. However, two ordered pairs (p, 1) and (p, 2) have the same first component but different second components.

Hence, the given relation R₃ is not a function.

iv. R₄ = {(p, 2), (q, 3), (r, 2), (s, 2)}

Every element of set A has an ordered pair in the relation R_4, and no two ordered pairs have the same first component but different second components.

Hence, the given relation R₄ is a function.

Given A = {9, 10, 11, 12, 13}

f : A → Z such that f(n) = the highest prime factor of n.

A is the domain of the function f. Hence, the range is the set of elements f(n) for all n ∈ A.

We have f(9) = highest prime factor of 9

The prime factorization of 9 = 3²

Thus, the highest prime factor of 9 is 3.

∴ f(9) = 3

We have f(10) = highest prime factor of 10

The prime factorization of 10 = 2 × 5

Thus, the highest prime factor of 10 is 5.

∴ f(10) = 5

We have f(11) = highest prime factor of 11

We know 11 is a prime number.

∴ f(11) = 11

We have f(12) = highest prime factor of 12

The prime factorization of 12 = 2² × 3

Thus, the highest prime factor of 12 is 3.

∴ f(12) = 3

We have f(13) = highest prime factor of 13

We know 13 is a prime number.

∴ f(13) = 13

Thus, the range of f is {3, 5, 11, 13}.

Given and

Let us first show that f is a function.

When 0 ≤ x ≤ 3, f(x) = x².

The function x² associates all the numbers 0 ≤ x ≤ 3 to unique numbers in R.

Hence, the images of {x ∈ Z: 0 ≤ x ≤ 3} exist and are unique.

When 3 ≤ x ≤ 10, f(x) = 3x.

The function x² associates all the numbers 3 ≤ x ≤ 10 to unique numbers in R.

Hence, the images of {x ∈ Z: 3 ≤ x ≤ 10} exist and are unique.

When x = 3, using the first definition, we have

f(3) = 3² = 9

When x = 3, using the second definition, we have

f(3) = 3(3) = 9

Hence, the image of x = 3 is also distinct.

Thus, as every element of the domain has an image and no element has more than one image, f is a function.

Now, let us show that g is not a function.

When 0 ≤ x ≤ 2, g(x) = x².

The function x² associates all the numbers 0 ≤ x ≤ 2 to unique numbers in R.

Hence, the images of {x ∈ Z: 0 ≤ x ≤ 2} exist and are unique.

When 2 ≤ x ≤ 10, g(x) = 3x.

The function x² associates all the numbers 2 ≤ x ≤ 10 to unique numbers in R.

Hence, the images of {x ∈ Z: 2 ≤ x ≤ 10} exist and are unique.

When x = 2, using the first definition, we have

g(2) = 2² = 4

When x = 2, using the second definition, we have

g(2) = 3(2) = 6

Here, the element 2 of the domain is associated with two elements distinct elements 4 and 6.

Thus, g is not a function.

Given f(x) = x².

We need to find the value of

Thus,

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(x) = x² – 3x + 4.

We need to find x satisfying f(x) = f(2x + 1).

We have f(2x + 1) = (2x + 1)² – 3(2x + 1) + 4

⇒ f(2x + 1) = (2x)² + 2(2x)(1) + 1² – 6x – 3 + 4

⇒ f(2x + 1) = 4x² + 4x + 1 – 6x + 1

∴ f(2x + 1) = 4x² – 2x + 2

Now, f(x) = f(2x + 1)

⇒ x² – 3x + 4 = 4x² – 2x + 2

⇒ 3x² + x – 2 = 0

⇒ 3x² + 3x – 2x – 2 = 0

⇒ 3x(x + 1) – 2(x + 1) = 0

⇒ (x + 1)(3x – 2) = 0

⇒ x + 1 = 0 or 3x – 2 = 0

⇒ x = –1 or 3x = 2

∴ x = –1 or

Thus, the required values of x are –1 and.

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

Given

We need to prove that x = f(y).

We have

⇒ y(bx – a) = ax – b

⇒ bxy – ay = ax – b

⇒ bxy – ax = ay – b

⇒ x(by – a) = ay – b

∴ x = f(y)

Thus, x = f(y).

Given

We need to prove that f[f{f(x)}] = x.

First, we will evaluate f{f(x)}.

Now, we will evaluate f[f{f(x)}]

∴ f[f{f(x)}] = x

Thus, f[f{f(x)}] = x

Given

We need to prove that f[f(x)] = x.

∴ f[f(x)] = x

Thus, f[f(x)] = x

Given

i.

When 0 ≤ x ≤ 1, f(x) = x

ii. f(–2)

When x < 0, f(x) = x²

⇒ f(–2) = (–2)²

∴ f(–2) = 4

iii. f(1)

When x ≥ 1,

∴ f(1) = 1

iv.

We have

When x ≥ 1,

v.

We know is not a real number and the function f(x) is defined only when x ∈ R.

Thus, does not exist.

Given

We need to prove that

We have,

Thus,

Given

We need to prove that f(tanθ) = sin2θ.

We have

We know

However, cos²θ + sin²θ = 1

⇒ f(tanθ) = 2sinθcosθ

∴ f(tanθ) = sin2θ

Thus, f(tanθ) = sin2θ

Given

i. We need to prove that

We have

Thus,

ii. We need to prove that

We have

Thus,

Given, where a > 0 and n ∈ N.

We need to prove that f[f(x)] = x.

[∵ (a^m)ⁿ = a^mn]

[∵ (a^m)ⁿ = a^mn]

⇒ f[f(x)] = x¹

∴ f[f(x)] = x

Thus, f[f(x)] = x for all x.

Given x ≠ 0 and a ≠ b such that

..... (1)

Substituting in place of x, we get

.....-- (2)

On adding equations (1) and (2), we get

..... (3)

On subtracting equations (1) and (2), we get

..... (4)

On adding equations (3) and (4), we get

Thus,

i.

Clearly, f(x) is defined for all real values of x, except for the case when x = 0.

When x = 0, f(x) will be undefined as the division result will be indeterminate.

Thus, domain of f = R – {0}

ii.

Clearly, f(x) is defined for all real values of x, except for the case when x – 7 = 0 or x = 7.

When x = 7, f(x) will be undefined as the division result will be indeterminate.

Thus, domain of f = R – {7}

iii.

Clearly, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = –1.

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

Thus, domain of f = R – {–1}

iv.

Clearly, f(x) is defined for all real values of x, except for the case when x² – 9 = 0.

x² – 9 = 0

⇒ x² – 3² = 0

⇒ (x + 3)(x – 3) = 0

⇒ x + 3 = 0 or x – 3 = 0

⇒ x = ±3

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

Thus, domain of f = R – {–3, 3}

v.

Clearly, f(x) is defined for all real values of x, except for the case when x² – 8x + 12 = 0.

x² – 8x + 12 = 0

⇒ x² – 2x – 6x + 12 = 0

⇒ x(x – 2) – 6(x – 2) = 0

⇒ (x – 2)(x – 6) = 0

⇒ x – 2 = 0 or x – 6 = 0

⇒ x = 2 or 6

When x = 2 or 6, f(x) will be undefined as the division result will be indeterminate.

Thus, domain of f = R – {2, 6}

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x – 2 ≥ 0

⇒ x ≥ 2

∴ x ∈ [2, ∞)

Thus, domain of f = [2, ∞)

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x² – 1 ≥ 0

⇒ x² – 1² ≥ 0

⇒ (x + 1)(x – 1) ≥ 0

⇒ x ≤ –1 or x ≥ 1

∴ x ∈ (–∞, –1] ∪ [1, ∞)

In addition, f(x) is also undefined when x² – 1 = 0 because denominator will be zero and the result will be indeterminate.

x² – 1 = 0 ⇒ x = ±1

Hence, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}

∴ x ∈ (–∞, –1) ∪ (1, ∞)

Thus, domain of f = (–∞, –1) ∪ (1, ∞)

We know the square of a real number is never negative.

Clearly, f(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]

Thus, domain of f = [–3, 3]

We know the square root of a real number is never negative.

Clearly, f(x) takes real values only when x – 2 and 3 – x are both positive or negative.

(a) Both x – 2 and 3 – x are positive

x – 2 ≥ 0

⇒ x ≥ 2

3 – x ≥ 0

⇒ x ≤ 3

Hence, x ≥ 2 and x ≤ 3

∴ x ∈ [2, 3]

(b) Both x – 2 and 3 – x are negative

x – 2 ≤ 0 ⇒ x ≤ 2

3 – x ≤ 0 ⇒ x ≥ 3

Hence, x ≤ 2 and x ≥ 3

However, the intersection of these sets in null set. Thus, this case is not possible.

In addition, f(x) is also undefined when 3 – x = 0 because the denominator will be zero and the result will be indeterminate.

3 – x = 0

⇒ x = 3

Hence, x ∈ [2, 3] – {3}

∴ x ∈ [2, 3)

Thus, domain of f = [2, 3)

Clearly, f(x) is defined for all real values of x, except for the case when bx – a = 0 or.

When, f(x) will be undefined as the division result will be indeterminate.

Thus, domain of f = R –

Let f(x) = y

⇒ ax + b = y(bx – a)

⇒ ax + b = bxy – ay

⇒ ax – bxy = –ay – b

⇒ x(a – by) = –(ay + b)

Clearly, when a – by = 0 or, x will be undefined as the division result will be indeterminate.

Hence, f(x) cannot take the value.

Thus, range of f = R –

Clearly, f(x) is defined for all real values of x, except for the case when cx – d = 0 or.

When, f(x) will be undefined as the division result will be indeterminate.

Thus, domain of f = R –

Let f(x) = y

⇒ ax – b = y(cx – d)

⇒ ax – b = cxy – dy

⇒ ax – cxy = b – dy

⇒ x(a – cy) = b – dy

Clearly, when a – cy = 0 or, x will be undefined as the division result will be indeterminate.

Hence, f(x) cannot take the value.

Thus, range of f = R –

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x – 1 ≥ 0

⇒ x ≥ 1

∴ x ∈ [1, ∞)

Thus, domain of f = [1, ∞)

When x ≥ 1, we have x – 1 ≥ 0

Hence,

∴ f(x) ∈ [0, ∞)

Thus, range of f = [0, ∞)

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x – 3 ≥ 0

⇒ x ≥ 3

∴ x ∈ [3, ∞)

Thus, domain of f = [3, ∞)

When x ≥ 3, we have x – 3 ≥ 0

Hence,

∴ f(x) ∈ [0, ∞)

Thus, range of f = [0, ∞)

Clearly, f(x) is defined for all real values of x, except for the case when 2 – x = 0 or x = 2.

When x = 2, f(x) will be undefined as the division result will be indeterminate.

Thus, domain of f = R – {2}

We have

∴ f(x) = –1

Clearly, when x ≠ 2, f(x) = –1

Thus, range of f = {–1}

f(x) = |x – 1|

We know

Now, we have

Hence, f(x) is defined for all real numbers x.

Thus, domain of f = R

When x < 1, we have x – 1 < 0 or 1 – x > 0.

Hence, |x – 1| > 0 ⇒ f(x) > 0

When x ≥ 1, we have x – 1 ≥ 0.

Hence, |x – 1| ≥ 0 ⇒ f(x) ≥ 0

∴ f(x) ≥ 0 or f(x) ∈ [0, ∞)

Thus, range of f = [0, ∞)

f(x) = –|x|

We know

Now, we have

Hence, f(x) is defined for all real numbers x.

Thus, domain of f = R

When x < 0, we have –|x| < 0

Hence, f(x) < 0

When x ≥ 0, we have –x ≤ 0.

Hence, –|x| ≤ 0 ⇒ f(x) ≤ 0

∴ f(x) ≤ 0 or f(x) ∈ (–∞, 0]

Thus, range of f = [0, ∞)

We know the square of a real number is never negative.

Clearly, f(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]

Thus, domain of f = [–3, 3]

When x ∈ [–3, 3], we have 0 ≤ 9 – x² ≤ 9

Hence,

∴ f(x) ∈ [0, 3]

Thus, range of f = [0, 3]

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when 16 – x² ≥ 0

⇒ 16 ≥ x²

⇒ x² ≤ 16

⇒ x² – 16 ≤ 0

⇒ x² – 4² ≤ 0

⇒ (x + 4)(x – 4) ≤ 0

⇒ x ≥ –4 and x ≤ 4

∴ x ∈ [–4, 4]

In addition, f(x) is also undefined when 16 – x² = 0 because denominator will be zero and the result will be indeterminate.

16 – x² = 0 ⇒ x = ±4

Hence, x ∈ [–4, 4] – {–4, 4}

∴ x ∈ (–4, 4)

Thus, domain of f = (–4, 4)

Let f(x) = y

⇒ 1 = (16 – x²)y²

⇒ 1 = 16y² – x²y²

⇒ x²y² + 1 – 16y² = 0

⇒ (y²)x² + (0)x + (1 – 16y²) = 0

As x ∈ R, the discriminant of this quadratic equation in x must be non-negative.

⇒ 0² – 4(y²)(1 – 16y²) ≥ 0

⇒ –4y²(1 – 16y²) ≥ 0

⇒ 4y²(1 – 16y²) ≤ 0

⇒ 1 – 16y² ≤ 0 [∵ y² ≥ 0]

⇒ 16y² – 1 ≥ 0

⇒ (4y)² – 1² ≥ 0

⇒ (4y + 1)(4y – 1) ≥ 0

⇒ 4y ≤ –1 and 4y ≥ 1

However, y is always positive because it is the reciprocal of a non-zero square root.

Thus, range of f =

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x² – 16 ≥ 0

⇒ x² – 4² ≥ 0

⇒ (x + 4)(x – 4) ≥ 0

⇒ x ≤ –4 or x ≥ 4

∴ x ∈ (–∞, –4] ∪ [4, ∞)

Thus, domain of f = (–∞, –4] ∪ [4, ∞)

When x ∈ (–∞, –4] ∪ [4, ∞), we have x² – 16 ≥ 0

Hence,

∴ f(x) ∈ [0, ∞)

Thus, range of f = [0, ∞)

i. f(x) = x³ + 1 and g(x) = x + 1

We have f(x) : R → R and g(x) : R → R

(a) f + g

We know (f + g)(x) = f(x) + g(x)

⇒ (f + g)(x) = x³ + 1 + x + 1

∴ (f + g)(x) = x³ + x + 2

Clearly, (f + g)(x) : R → R

Thus, f + g : R → R is given by (f + g)(x) = x³ + x + 2

(b) f – g

We know (f – g)(x) = f(x) – g(x)

⇒ (f – g)(x) = x³ + 1 – (x + 1)

⇒ (f – g)(x) = x³ + 1 – x – 1

∴ (f – g)(x) = x³ – x

Clearly, (f – g)(x) : R → R

Thus, f – g : R → R is given by (f – g)(x) = x³ – x

(c) cf (c ∈ R, c ≠ 0)

We know (cf)(x) = c × f(x)

⇒ (cf)(x) = c(x³ + 1)

∴ (cf)(x) = cx³ + c

Clearly, (cf)(x) : R → R

Thus, cf : R → R is given by (cf)(x) = cx³ + c

(d) fg

We know (fg)(x) = f(x)g(x)

⇒ (fg)(x) = (x³ + 1)(x + 1)

⇒ (fg)(x) = (x + 1)(x² – x + 1)(x + 1)

∴ (fg)(x) = (x + 1)²(x² – x + 1)

Clearly, (fg)(x) : R → R

Thus, fg : R → R is given by (fg)(x) = (x + 1)²(x² – x + 1)

(e)

We know

Observe that is undefined when f(x) = 0 or when x = – 1.

Thus, : R – {–1} → R is given by

(f)

We know

Observe that is undefined when g(x) = 0 or when x = –1.

Using x³ + 1 = (x + 1)(x² – x + 1), we have

Thus, : R – {–1} → R is given by

and

We have f(x) : [1, ∞) → R⁺ and g(x) : [–1, ∞) → R⁺ as real square root is defined only for non-negative numbers.

(a) f + g

We know (f + g)(x) = f(x) + g(x)

Domain of f + g = Domain of f ∩ Domain of g

⇒ Domain of f + g = [1, ∞) ∩ [–1, ∞)

∴ Domain of f + g = [1, ∞)

Thus, f + g : [1, ∞) → R is given by

(b) f – g

We know (f – g)(x) = f(x) – g(x)

Domain of f – g = Domain of f ∩ Domain of g

⇒ Domain of f – g = [1, ∞) ∩ [–1, ∞)

∴ Domain of f – g = [1, ∞)

Thus, f – g : [1, ∞) → R is given by

(c) cf (c ∈ R, c ≠ 0)

We know (cf)(x) = c × f(x)

Domain of cf = Domain of f

∴ Domain of cf = [1, ∞)

Thus, cf : [1, ∞) → R is given by

(d) fg

We know (fg)(x) = f(x)g(x)

Domain of fg = Domain of f ∩ Domain of g

⇒ Domain of fg = [1, ∞) ∩ [–1, ∞)

∴ Domain of fg = [1, ∞)

Thus, fg : [1, ∞) → R is given by

(e)

We know

Domain of = Domain of f

∴ Domain of = [1, ∞)

Observe that is also undefined when x – 1 = 0 or x = 1.

Thus, : (1, ∞) → R is given by

(f)

We know

Domain of = Domain of f ∩ Domain of g

⇒ Domain of = [1, ∞) ∩ [–1, ∞)

∴ Domain of = [1, ∞)

Thus, : [1, ∞) → R is given by

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.

We know

Clearly, 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, will be undefined as the division result will be indeterminate.

Thus, domain of = R – {–1, 0}

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

Now, we have

However, |x| ≥ 0 ⇒ f(|x|) = |x| – 1 when 0 ≤ |x| ≤ 2

We also have

We know

Here, we are interested only in the range [0, 2].

Substituting this value of |x – 1| in |f(x)|, we get

We need to find g(x).

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

Thus,

Given and

We know the square of a real number is never negative.

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

⇒ x ≥ –1

∴ x ∈ [–1, ∞)

Thus, 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]

Thus, domain of g = [–3, 3]

i. f + g

We know (f + g)(x) = f(x) + g(x)

Domain of f + g = Domain of f ∩ Domain of g

⇒ Domain of f + g = [–1, ∞) ∩ [–3, 3]

∴ Domain of f + g = [–1, 3]

Thus, f + g : [–1, 3] → R is given by

ii. f – g

We know (f – g)(x) = f(x) – g(x)

Domain of f – g = Domain of f ∩ Domain of g

⇒ Domain of f – g = [–1, ∞) ∩ [–3, 3]

∴ Domain of f – g = [–1, 3]

Thus, f – g : [–1, 3] → R is given by

iii. fg

We know (fg)(x) = f(x)g(x)

As earlier, domain of fg = [–1, 3]

Thus, f – g : [–1, 3] → R is given by

iv.

We know

As earlier, domain of = [–1, 3]

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

When x = ±3, will be undefined as the division result will be indeterminate.

⇒ Domain of = [–1, 3] – {–3, 3}

∴ Domain of = [–1, 3)

Thus, : [–1, 3) → R is given by

v.

We know

As earlier, domain of = [–1, 3]

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

When x = –1, will be undefined as the division result will be indeterminate.

⇒ Domain of = [–1, 3] – {–1}

∴ Domain of = (–1, 3]

Thus, : (–1, 3] → R is given by

vi.

We know (f – g)(x) = f(x) – g(x) and (cf)(x) = cf(x)

As earlier, Domain of = [–1, 3]

Thus, : [–1, 3] → R is given by

vii. f² + 7f

We know (f² + 7f)(x) = f²(x) + (7f)(x)

⇒ (f² + 7f)(x) = f(x)f(x) + 7f(x)

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

∴ Domain of f² + 7f = [–1, ∞)

Thus, f² + 7f : [–1, ∞) → R is given by

viii.

We know and (cg)(x) = cg(x)

Domain of = Domain of g = [–3, 3]

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

When x = ±3, will be undefined as the division result will be indeterminate.

⇒ Domain of = [–3, 3] – {–3, 3}

∴ Domain of = (–3, 3)

Thus, : (–3, 3) → R is given by

Given f(x) = log_e(1 – x) and g(x) = [x]

Clearly, f(x) takes real values only when 1 – x > 0

⇒ 1 > x

⇒ x < 1

∴ x ∈ (–∞, 1)

Thus, domain of f = (–∞, 1)

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

Thus, domain of g = R

i. f + g

We know (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = log_e(1 – x) + [x]

Domain of f + g = Domain of f ∩ Domain of g

⇒ Domain of f + g = (–∞, 1) ∩ R

∴ Domain of f + g = (–∞, 1)

Thus, f + g : (–∞, 1) → R is given by (f + g)(x) = log_e(1 – x) + [x]

ii. fg

We know (fg)(x) = f(x)g(x)

⇒ (fg)(x) = log_e(1 – x) × [x]

∴ (fg)(x) = [x]log_e(1 – x)

Domain of fg = Domain of f ∩ Domain of g

⇒ Domain of fg = (–∞, 1) ∩ R

∴ Domain of fg = (–∞, 1)

Thus, f – g : (–∞, 1) → R is given by (fg)(x) = [x]log_e(1 – x)

iii.

We know

As earlier, domain of = (–∞, 1)

However, is defined for all real values of x ∈ (–∞, 1), except for the case when [x] = 0.

We have [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

When 0 ≤ x < 1, will be undefined as the division result will be indeterminate.

⇒ Domain of = (–∞, 1) – [0, 1)

∴ Domain of = (–∞, 0)

Thus, : (–∞, 0) → R is given by

iv.

We know

As earlier, domain of = (–∞, 1)

However, is defined for all real values of x ∈ (–∞, 1), except for the case when log_e(1 – x) = 0.

log_e(1 – x) = 0 ⇒ 1 – x = 1 or x = 0

When x = 0, will be undefined as the division result will be indeterminate.

⇒ Domain of = (–∞, 1) – {0}

∴ Domain of = (–∞, 0) ∪ (0, ∞)

Thus, : (–∞, 0) ∪ (0, ∞) → R is given by

We have (f + g)(x) = log_e(1 – x) + [x], x ∈ (–∞, 1)

We need to find (f + g)(–1).

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

(f + g)(–1) = log_e(1 – (–1)) + [–1]

⇒ (f + g)(–1) = log_e(1 + 1) + (–1)

∴ (f + g)(–1) = log_e2 – 1

Thus, (f + g)(–1) = log_e2 – 1

We have (fg)(x) = [x]log_e(1 – x), x ∈ (–∞, 1)

We need to find (fg)(0).

Substituting x = 0 in the above equation, we get

(fg)(0) = [0]log_e(1 – 0)

⇒ (fg)(0) = 0 × log_e1

∴ (fg)(0) = 0

Thus, (fg)(0) = 0

We have, x ∈ (–∞, 0)

We need to find

However, is not in the domain of.

Thus, does not exist.

We have, x ∈ (–∞, 0) ∪ (0, ∞)

We need to find

Substituting in the above equation, we get

Thus,

Given, and h(x) = 2x³ – 3

We know the square of a real number is never negative.

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

⇒ x ≥ –1

∴ x ∈ [–1, ∞)

Thus, domain of f = [–1, ∞)

g(x) is defined for all real values of x, except for 0.

Thus, domain of g = R – {0}

h(x) is defined for all real values of x.

Thus, domain of h = R

We know (2f + g – h)(x) = (2f)(x) + g(x) – h(x)

⇒ (2f + g – h)(x) = 2f(x) + g(x) – h(x)

Domain of 2f + g – h = Domain of f ∩ Domain of g ∩ Domain of h

⇒ Domain of 2f + g – h = [–1, ∞) ∩ R – {0} ∩ R

∴ Domain of 2f + g – h = [–1, ∞) – {0}

i. (2f + g – h)(1)

We have

ii. (2f + g – h)(0)

0 is not in the domain of (2f + g – h)(x).

Hence, (2f + g – h)(0) does not exist.

Thus, and (2f + g – h)(0) does not exist as 0 is not in the domain of (2f + g – h)(x).

Given

When x < 0, we have f(x) = 1 – x

f(–4) = 1 – (–4) = 1 + 4 = 5

f(–3) = 1 – (–3) = 1 + 3 = 4

f(–2) = 1 – (–2) = 1 + 2 = 3

f(–1) = 1 – (–1) = 1 + 1 = 2

When x = 0, we have f(x) = f(0) = 1

When x > 0, we have f(x) = 1 + x

f(1) = 1 + 1 = 2

f(2) = 1 + 2 = 3

f(3) = 1 + 3 = 4

f(4) = 1 + 4 = 5

Plotting these points on a graph sheet, we get

Given f(x) = x + 1 and g(x) = 2x – 3

Clearly, both f(x) and g(x) exist for all real values of x.

Hence, Domain of f = Domain of g = R

Range of f = Range of g = R

i. f + g

We know (f + g)(x) = f(x) + g(x)

⇒ (f + g)(x) = x + 1 + 2x – 3

∴ (f + g)(x) = 3x – 2

Domain of f + g = Domain of f ∩ Domain of g

⇒ Domain of f + g = R ∩ R

∴ Domain of f + g = R

Thus, f + g : R → R is given by (f + g)(x) = 3x – 2

ii. f – g

We know (f – g)(x) = f(x) – g(x)

⇒ (f – g)(x) = x + 1 – (2x – 3)

⇒ (f – g)(x) = x + 1 – 2x + 3

∴ (f – g)(x) = –x + 4

Domain of f – g = Domain of f ∩ Domain of g

⇒ Domain of f – g = R ∩ R

∴ Domain of f – g = R

Thus, f – g : R → R is given by (f – g)(x) = –x + 4

iii.

We know

Clearly, is defined for all real values of x, except for the case when 2x – 3 = 0 or.

When, will be undefined as the division result will be indeterminate.

Thus, domain of = R –

Thus, : R – → R is given by

Given and g(x) = x

Domain of f = [0, ∞)

Domain of g = R

i. f + g

We know (f + g)(x) = f(x) + g(x)

Domain of f + g = Domain of f ∩ Domain of g

⇒ Domain of f + g = [0, ∞) ∩ R

∴ Domain of f + g = [0, ∞)

Thus, f + g : [0, ∞) → R is given by

ii. f – g

We know (f – g)(x) = f(x) – g(x)

Domain of f – g = Domain of f ∩ Domain of g

⇒ Domain of f – g = [0, ∞) ∩ R

∴ Domain of f – g = [0, ∞)

Thus, f – g : [0, ∞) → R is given by

iii. fg

We know (fg)(x) = f(x)g(x)

Clearly, (fg)(x) is also defined only for non-negative real numbers x as square of a real number is never negative.

Thus, fg : [0, ∞) → R is given by

iv.

We know

Clearly, is defined for all positive real values of x, except for the case when x = 0.

When x = 0, will be undefined as the division result will be indeterminate.

⇒ Domain of = [0, ∞) – {0}

∴ Domain of = (0, ∞)

Thus, : (0, ∞) → R is given by

Given f(x) = x² and g(x) = 2x + 1

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) = x² + 2x + 1

∴ (f + g)(x) = (x + 1)²

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

∴ Domain of (f + g) is R

Thus, f + g : R → R is given by (f + g)(x) = (x + 1)²

ii. f – g

We know (f – g)(x) = f(x) – g(x)

⇒ (f – g)(x) = x² – (2x + 1)

∴ (f – g)(x) = x² – 2x – 1

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

∴ Domain of (f – g) is R

Thus, f – g : R → R is given by (f – g)(x) = x² – 2x – 1

iii. fg

We know (fg)(x) = f(x)g(x)

⇒ (fg)(x) = x²(2x + 1)

∴ (fg)(x) = 2x³ + x²

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

∴ Domain of fg is R

Thus, fg : R → R is given by (fg)(x) = 2x³ + x²

iv.

We know

Clearly, is defined for all real values of x, except for the case when 2x + 1 = 0.

2x + 1 = 0

⇒ 2x = –1

When, will be undefined as the division result will be indeterminate.

Thus, the domain of = R –

f(x) = |x|

f(-x) = |-x|

therefore, f(x) will always be 0 or positive.

Thus, range of f(x) ϵ [0,).

{since, (a + b)² = a² +b² +2ab}

f(y) =

x + 1 = xy

x² -yx+1= 0

for x to be real

y ∈(-∞,2]∪[2,∞)

|y|>2Ans.

F(x) =

=-sin 1

= 0

Using properties of greatest integer function:

[1] = 1; [0.5] = 0; [-0.5] = -1

Therefore, R(f) = {-}

π² ≈ 9.8596

So, we have [π² ]=9 and [-π² ]=-10

f(x)=cos 18x+ cos (-10)x

=cos 18x + cos 10x

=2 cos 14x cos4x

f(π)=2 cos 14π cos 4π

=2×1×1

Therefore, f(π)= 2

[x]= -2

f(x)= cos [x]= cos (-2)

= cos 2

because cos(-x) = cos(x)

for-1 ≤x<0

[x]=-1

f(x)= cos[x]=cos (-1)

= cos 1

for 0 ≤x< 1

[x]=0

f(x)=cos 0 =1

for 1 ≤x<π/2

[x]=1

f(x)=cos1

Therefore, R(f) = {1, cos 1, cos 2}

0 ≤x-[x]<1

e⁰ ≤ e^x-[x] <e¹

1 ≤ e^x-[x] < e

Therefore, R(f) = [1, e)

If

a²x=ax²+x²+x

x² (a+1)+x(1- a²)=0

x² (a+1)+x(1-a)(1+a)=0

(a+1)( x²+x(1-a))=0

a+1=0

Therefore, a=-1

For function to be defined,

x≠2

Therefore, D(f)= R-{2}.

Let

y=-1

Therefore, R(f) = {-1}.

f(a+1)-f(a-1) =[4(a+1) -(a+1)²]-[4(a-1)-(a-1)² ]

=4[(a+1) -(a-1)]- [ (a+1)² -(a+1)²]

=4(2)-[(a+1+a-1) (a+1-a+1)]

Using: a² -b²= (a + b)(a-b)

f(a+1)-f(a-1)=4(2)-2a(2)

=4(2-a)

Therefore, answer = 0.

For f(x) to be defined,

x-|x|>0

But x-|x|≤0

So, f(x) does not exist..

Therefore,

For f(x) to be defined,

x-[x]≥0

We know that, where {x} is fractional part function and [x] is greatest integer function.

{x}≥0

Also,

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

For function to be defined,

[x]-x≥0

-{x}≥

Therefore, domain of f(x) is integers.

D(f)∈I

Range = {0}.

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

D(f) = {2, 5, 8, 10}

D(g) = {2, 7, 8, 10, 11}

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

f(x)=3x² -1;g(x)=3+x

For

3x² -1=3+x

3x² -x-4=0

(3x-4)(x+1)=0

3x-4=0 orx+1=0

D(f) = {0, 2, 3, 4, 5}

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

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

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.

Replace f(x) by y,

Replace x by and y by x.

So,

Option C is correct.

A function is said to exist when we get a unique value for any value of x..

Therefore, option B is correct.. is not a function as for each value of x, we will get 2 values of y..which is not as per the definition of a function..

Now,

Using:

=cos(2 log x)〗 cos(2 log y)-cos(2 log x) cos(2 log y)

=0

Now,

Using:

=cos(log x) cos(log y)-cos(log x) cos(log y)

=0

f(x)=|x-1|

f(x² )=| x²-1|

[f(x)²=(x-1)²

= x²+1-2x

So, f(x²)≠[f(x)]²

f(x + y)=|x+y-1|

f(x)f(y)=(x-1)(y-1)

So, f(x + y) ≠ f(x)f(y)

f(|x|)=||x|-1|

Therefore, option D is correct.

[x]= -2

f(x)= cos[x]= cos(-2)

= cos2

because cos(-x)= cos(x)

for-1 ≤x<0

[x]=-1

f(x)= cos[x]

=cos(-1)

= cos1

for 0 ≤x< 1

[x]=0

f(x) = cos 0

=1

[x]=1

f(x)=cos 1

Therefore, R(f) = {1, cos 1,cos 2}

Option B is correct.

A function is said to exist when we get a unique value of y for any value of x..If we get 2 values of y for any value of x, then it is not a function..

Therefore, option B is correct .

NOTE: To check if a given curve is a function or not, draw the curve and then draw a line parallel to y-axis..If it intersects the curve at only one point, then it is a function, else not..

Using: (1+x)³=1+3x+3x²+x³

And (1-x)³=1-3x+3x²-x³

f(g(x))=3f(x)

Option C is correct.

Since A has 3 elements and B has 2..then number of functions from A to B

Option B is correct.

=2f(x)

Option C is correct..

Now,

Using:

=cos(log x) cos(log 4)-cos(log x)cos(4)

=0

Option C is correct..

Option A is correct.

eqn.1

Replace x by 1/x in eqn.1;

eqn.2

Multiply eqn.1 by 2 and eqn.2 by 3 and add them..

On adding, we get

Option A is correct.

f(2x)=2(2x)+|2x|=4x+2|x|

f(-x)=2(-x)+|-x|

f(2x)+f(-x)-f(x)=4x+2|x|-2x+|-x|-(2x+|x|)

=4x+2|x|-2x+|x|-2x-|x|=2|x|

Option B is correct..

Let

y(x²+2x)= x²-x

yx(x+2)=x(x-1)

y(x+2)=x-1

Value of x can’t be zero or it cannot be not defined..

y≠1, -1/2

So, range= R-{-1/2, 1}

=3

Option C is correct..

Now,

Using:

=cos(log_e x) cos(log_e x y)-{cos(log_e x x ) cos(log_e x y)〗}

=0

h(x)=1

f(x)g(x)=1

x≠0

Option D is correct.

Now, f(2002)=1

Option A is correct..

f(x)=sin⁴ x+1-sin² x

Minimum value of f(x)=3/4

0≤sin² x≤1

So, maximum value of

=1

R(f)=[3/4, 1]

Answer is C.

When -4≤x<0

=-1

When 0<x≤4

=1

R(f)={-1, 1}

Option A is correct..

g(f(x))=8

(f(x))²+7=8

(2x+3)²=1

4x²+12x+9=1

4 x²+12x+8=0

x²+3x+2=0

(x+1)(x+2)=0

x+1=0 or x+2=0

x=-1 or x=-2

Option C is correct..

f(|x|)=|x|-1

f(|x|)=x

We have, |x|=x ;x≥0

And |x|=-x ;x≤0

So, -x-1=x

2x=-1

Option C…

2k=1;

=0.5

Option A is correct.

There is a mistake in the question…

Now,

Since, α, β are roots of

So, f(α)=f(β)

=(2)³-9(2)

=8-18

=-10

Option C…

Since, and α, β are its roots,

f(x)=3³-12(3)

=27-36

=-9

So, f (α) = f (β) = - 9

Option A is correct..

eqn. 1

Replacing x by 1/x;

eqn. 2

Multiply eqn. 1 by 3 and eqn. 2 by 5, and then subtract them

We get,

[]=9 and [-]=-10

Now, f(x)=sin[π² ] x + sin[-π²]x

=sin 9x-sin 10x

Now, checking values of f(x) at given points..

=1-0

=1

Option A is correct..

f(π)=sin 9π-sin 10π

=0-0

=0

Option B & C are incorrect..

for f(x) to be defined,

2-2x-x²≥0

x²+2x-2≤0

(x-(1-√3))(x-(-1+√3))≤0

x∈[-1-√3,-1+√3]

Option B is correct..

for given function,

x≠2, 5

Therefore, x∈(-∞,-3]∪(2, 5)

Option B is correct..

Here,

But

So,

Option A is correct..

Here, -1≥0 and 3-x≥0

So, x≥1 and x≤3

Therefore, x∈[1, 3]
option D is correct..

For function to be defined,

x∈(-∞,-2)∪[2, ∞) …(1)

And

So, …(2)

Taking common of both the solutions, we get

Option C is correct..

For f(x) =log|x|;

It is defined at all positive values of x except 0..

But since we have |x|;

So, |x|>0;

x ∈ R-{0}

Here, 4x-x²≥0

x²-4x≤0

x(x-4)≤

So,

Option D is correct..

Here,

x-4≥0; x≥4 …..(1)

Also,

x≥4

Option A is correct..

5|x|-x²-6≥0

x²-5|x|+6≤0

(|x|-2)(|x|-3)≤0

So, |x|∈[2, 3]

Therefore,x∈[-3, -2]∪[2, 3]

Option C is correct.

We know that

|x| = -x in (-∞, 0) and |x| = x in [0, ∞)

So, in (-∞, 0)

And in (0, ∞)

As clearly shown above f(x) has only two values 1 and -1

So, range of f(x) = {-1, 1}

When x>-2,

We have

=1

When x<-2,

We have

=-1

R(f)={-1, 1}

Option A is correct..

A modulus function always gives a positive value..

Option B..

So, comparing, f(xy) and f(x)f(y);

We get f(xy)≤f(x)f(y)

Option C..

[x]²-5[x]+6=0

([x]-2)([x]-3)=0

if [x]=2

2≤x<3

and if [x]=3

3≤x<4

Therefore,

Option D..

we know, -1≤cosx≤1

-2≤-2cosx≤2

-1≤(1-2cosx)≤3

So, R(f)=[-1, 1/3]

Option ..B