RD Chapter 3 Functions Ex 3.1 Solutions
Question - 1 : - Define a function as a set of ordered pairs.
Answer - 1 : -
Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A├ЧB, is called a function (or a mapping) from A to B, if
(i) for each a тИИ A there exists b тИИ B such that (a, b) тИИ f
(ii) (a, b) тИИ f and (a, c) тИИ f тЗТ b = c
Question - 2 : - Define a function as a correspondence between two sets.
Answer - 2 : -
Let A and B be two non-empty sets. Then a function тАШfтАЩ from set A to B is a rule or method or correspondence which associates elements of set A to elements of set B such that:
(i) all elements of set A are associated to elements in set B.
(ii) an element of set A is associated to a unique element in set B.
Question - 3 : - What is the fundamental difference between a relation and a function? Is every relation a function?
Answer - 3 : -
Let тАШfтАЩ 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.
Question - 4 : - Let A = {тАУ2, тАУ1, 0, 1, 2} and f: A тЖТ Z be a function defined by f(x) = x2 тАУ 2x тАУ 3. Find:
(i) range of f i.e. f (A)
(ii) pre-images of 6, тАУ3 and 5
Answer - 4 : -
Given:
A = {тАУ2, тАУ1, 0, 1, 2}
f : A┬атЖТ┬аZ such that f(x) = x2┬атАУ2x тАУ 3
(i)┬аRangeof f i.e. f (A)
A is the domain of the function f. Hence, range is the set of elementsf(x) for all x┬атИИ┬аA.
Substituting x = тАУ2 in f(x), we get
f(тАУ2) = (тАУ2)2┬атАУ2(тАУ2) тАУ 3
= 4 + 4 тАУ 3
= 5
Substituting x = тАУ1 in f(x), we get
f(тАУ1) = (тАУ1)2┬атАУ2(тАУ1) тАУ 3
= 1 + 2 тАУ 3
= 0
Substituting x = 0 in f(x), we get
f(0) = (0)2┬атАУ 2(0) тАУ3
= 0 тАУ 0 тАУ 3
= тАУ 3
Substituting x = 1 in f(x), we get
f(1) = 12┬атАУ 2(1) тАУ 3
= 1 тАУ 2 тАУ 3
= тАУ 4
Substituting x = 2 in f(x), we get
f(2) = 22┬атАУ 2(2) тАУ 3
= 4 тАУ 4 тАУ 3
= тАУ3
Thus, the range of f is {-4, -3, 0, 5}.
(ii)┬аpre-imagesof 6, тАУ3 and 5
Let x be the pre-image of 6┬атЗТ┬аf(x) = 6
x2┬атАУ 2x тАУ 3 = 6
x2┬атАУ 2x тАУ 9 = 0
x = [-(-2) ┬▒┬атИЪ┬а((-2)2┬атАУ4(1) (-9))] / 2(1)
= [2 ┬▒┬атИЪ┬а(4+36)] / 2
= [2 ┬▒┬атИЪ40] / 2
= 1 ┬▒┬атИЪ10
However,┬а1 ┬▒┬атИЪ10 тИЙ A
Thus, there exists no pre-image of 6.
Now, let x be the pre-image of тАУ3┬атЗТ┬аf(x) = тАУ3
x2┬атАУ 2x тАУ 3 = тАУ3
x2┬атАУ 2x = 0
x(x тАУ 2) = 0
x = 0 or 2
Clearly, both 0 and 2 are elements of A.
Thus, 0 and 2 are the pre-images of тАУ3.
Now, let x be the pre-image of 5┬атЗТ┬аf(x) = 5
x2┬атАУ 2x тАУ 3 = 5
x2┬атАУ 2x тАУ 8= 0
x2┬атАУ 4x + 2x тАУ 8= 0
x(x тАУ 4) + 2(x тАУ 4) = 0
(x + 2)(x тАУ 4) = 0
x = тАУ2 or 4
However, 4┬атИЙ┬аA but┬атАУ2┬атИИ┬аA
Thus, тАУ2 is the pre-images of 5.
тИ┤├Ш, {0, 2}, -2 are the pre-images of 6, -3, 5
Question - 5 : - If a function f: R тЖТ R be defined by
┬а
Find: f (1), f (тАУ1), f (0), f (2).
Answer - 5 : -
Given:
Let us find f (1), f (тАУ1), 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
= 4 + 1
= 5
When x < 0, f(x) = 3x тАУ 2
Substituting x = тАУ1 in the above equation, we get
f (тАУ1) = 3(тАУ1) тАУ 2
= тАУ3 тАУ 2
= тАУ5
When x = 0, f(x) = 1
Substituting x = 0 in the above equation, we get
f (0) = 1
When x > 0, f(x) = 4x + 1
Substituting x = 2 in the above equation, we get
f (2) = 4(2) + 1
= 8 + 1
= 9
тИ┤ f (1) = 5, f (тАУ1) = тАУ5, f (0) = 1 and f (2) = 9.
Question - 6 : - A function f: R тЖТ R is defined by f(x) = x2. Determine
(i) range of f
(ii) {x: f(x) = 4}
(iii) {y: f(y) = тАУ1}
Answer - 6 : -
Given:
f : R┬атЖТ┬аR and f(x) = x2.
(i)┬аrangeof f
Domain of f = R (set of real numbers)
We know that the square of a real number is always positive or equal tozero.
тИ┤range of f = R+тИк┬а{0}
(ii)┬а{x:f(x) = 4}
Given:
f(x) = 4
we know, x2┬а= 4
x2┬атАУ 4 = 0
(x тАУ 2)(x + 2) = 0
тИ┤┬аx= ┬▒ 2
тИ┤{x: f(x) = 4} = {тАУ2, 2}
(iii)┬а{y:f(y) = тАУ1}
Given:
f(y) = тАУ1
y2┬а= тАУ1
However, the domain of f is R, and for every real number y, the value of y2┬аis non-negative.
Hence, there exists no real y for which y2┬а=тАУ1.
тИ┤{y:f(y) = тАУ1} =┬атИЕ
Question - 7 : - Let f: R+тЖТ R, where R+ is the set of all positive real numbers, be such that f(x) = loge x. Determine
(i) the image set of the domain of f
(ii) {x: f (x) = тАУ2}
(iii) whether f (xy) = f (x) + f (y) holds.
Answer - 7 : -
Given f: R+тЖТ R and f(x) = loge x.
(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.
тИ┤ The image set of f = R
(ii) {x: f(x) = тАУ2}
Given f(x) = тАУ2
loge x = тАУ2
тИ┤ x = e-2 [since, logb a = c тЗТ a = bc]
тИ┤ {x: f(x) = тАУ2} = {eтАУ2}
(iii) Whether f (xy) = f (x) + f (y) holds.
We have f (x) = loge x тЗТ f (y) = loge y
Now, let us consider f (xy)
F (xy) = loge (xy)
f (xy) = loge (x ├Ч y) [since, logb (a├Чc) = logb a + logb c]
f (xy) = loge x + loge y
f (xy) = f (x) + f (y)
тИ┤ the equation f (xy) = f (x) + f (y) holds.
Question - 8 : - Write the following relations as sets of ordered pairs and find which of them are functions:
(i) {(x, y): y = 3x, x тИИ {1, 2, 3}, y тИИ {3, 6, 9, 12}}
(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
(iii) {(x, y): x + y = 3, x, y тИИ {0, 1, 2, 3}}
Answer - 8 : -
(i) {(x, y): y = 3x, x тИИ {1, 2, 3}, y тИИ {3, 6, 9, 12}}
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
тИ┤ R = {(1, 3), (2, 6), (3, 9)}
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, y > 1 + 1 or y > 2 тЗТ y = {4, 6}
When x = 2, y > 2 + 1 or y > 3 тЗТ y = {4, 6}
тИ┤ R = {(1, 4), (1, 6), (2, 4), (2, 6)}
Hence, the given relation R is not a function.
(iii) {(x, y): x + y = 3, x, y тИИ {0, 1, 2, 3}}
When x = 0, 0 + y = 3 тЗТ y = 3
When x = 1, 1 + y = 3 тЗТ y = 2
When x = 2, 2 + y = 3 тЗТ y = 1
When x = 3, 3 + y = 3 тЗТ y = 0
тИ┤ R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Hence, the given relation R is a function.
Question - 9 : - Let f: R тЖТ R and g: C тЖТ C be two functions defined as f(x) = x2 and g(x) = x2. Are they equal functions?
Answer - 9 : -
Given:
f: R тЖТ R тИИ f(x) = x2 and g : R тЖТ R тИИ g(x) = x2
f is defined from R to R, the domain of f = R.
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 тЙа domain of g.
тИ┤ f and g are not equal functions.