Function = Let A and B be two non-empty sets. A relation f from A to B i.e. a sub-set of A × B, is called a function (or a mapping or a map) from A to B, if
(i) for each a ϵ A there exists b ϵ B such that (a, b) ϵ f
(ii) (a, b) ϵ f and (a, c) ϵ f ⟹ b = c
If (a, b) ϵ f, then 'b' is called the image of 'a' under f
If a function f is expresed as the set of ordered pairs, the domain f is the set of all first components of members of f and the range of f is the set of second components of members of f.
Function = Let A and B be two non-empty sets. Then a function ' f' from set A to set 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 element in set B.
(ii) an element of se A is associated to a unique element in set B.
In other words, a function ' f' from a set A to set B associates each element of set A to a unique element of set b.
Function is a type of relation. But in a function no two ordered pairs have the same first element. For eg: R1 and R2 are two relations.
Clearly, R1 is a function, but R2 is not a function because two ordered pairs (1, 2) and (1, 4) have the same first element. This means every function is a relation but every relation is not a function.
We have,
f(x) = x2 - 2x - 3
Now,
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
= - 3
f(1) = (1)2 -2 × 1 - 3
= 1 - 2 – 3
= – 4
f(2) = (2)2 - 2 × 2 - 3
= 4 - 4 - 3
= - 3
(a) Rang (f) = {-4, -3, 0, 5}
(b) Clearly, pre-images of 6,-3 and 5 is ∅, {0, 2}, -2 respectively.
We have,
Now,
f(1) = 4 × 1 + 1 = 5,
f(-1) = 3 × (-1) - 2 = – 3 – 2 = – 5,
f(0) = 1,
and, f(2) = 4 × 2 + 1 = 9
∴ f(1) = 5, f(-1) = – 5,
f(0) = 1, f(2) = 9,
We have,
f(x) = x2 ---- (i)
(a) clearly range of f = R+ (set of all real numbers greater than or equal to zero)
(b) we have,
{x : f(x) = 4}
⟹ f(x) = 4 ---- (ii)
Using equation (i) and equation (ii), we get
x2 = 4
⟹ x = ± 2
∴ {x : f(x) = 4} = {-2, 2}
(c) {y : f(y) = -1}
⟹ f(y) = -1 ---- (iii)
Clearly, x2 ≠ -1 Or X2 ≥ 0
⟹ f(y) ≠ –1
∴ {y : f(y) = –1} = ∅
We have,
f = R+ → R
and f(x) = loge x --- (i)
(a) Now
f = R+→ R
∴ the image set of the domain of f = R
(b) Now,
{x : f (x) = – 2}
⟹ f (x) = – 2 ----(ii)
Using equation (i) and equation (ii), we get
loge x = – 2
⟹ x = e-2
∴ {x : f (x) = – 2} = {e-2}
(c) Now,
f(xy) = loge (xy) [f(x) = loge x]
= loge x + loge y [∵ logmn = logm + logn]
f(x) + f(y)
∴ f(xy) = f(x) + f(y)
Yes, f(xy) = f(x) + f(y).
(a) we have,
{(x, y) = y = 3x,x ϵ (1, 2, 3),y ϵ {3, 6, 9, 12}}
Putting x = 1,2, 3 in y = 3x, we get
y = 3, 6, 9 respectively
∴ R = {(1, 3), (2, 6), (3, 9)}
Yes, it is a function.
(b) we have,
{(x, y) : y > x + 1,x = 1, 2 and y = 2, 4, 6}
Putting x = 1, 2 in y > x + 1, we get y > 2, y > 3 respectively.
∴ R = {(1, 4), (1, 6), (2, 4), (2, 6)}
It is not a function from A to B because two ordered pairs in R have the same first element.
(c) we have,
{(x, y) = x + y = 3, x , y ϵ {0,1, 2, 3}}
Now,
y = 3 - x
Putting x = 0,1, 2, 3, we get
y = 3, 2, 1, 0 respectively
∴ R = {(0, 3), (1, 2), (2, 1,(3, 0)}
Yes, this relation is a function.
We have,
f : R → R and g : c → c
∴ Domain (f) = R and Domain (g) = c
∴ Domain (f) ≠ Domain (g) = c
∴ f(x) and g(x) are not equal functions.
(i) We have,
f(x) = x2
Range of f (x) = R+ (set of all real numbers greater than or equal to zero) = {x ϵ R|x ≥ 0}
(ii) We have,
g(x) = sinx
Range of g(x) = {x ϵ R : -1 x ≤ 1}
(iii) We have,
h(x) = x2 + 1
Range of h (x) = {x ϵ P : ≥ 1}
(a) We have,
f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
f1 is a function from X to Y.
(b) We have,
f2 = {(1, 1), (2, 7), (3, 5)}
f2 is not a function from X to Y because there is an element 4 ϵ x which is not associated to any element of Y.
(c) We have,
f3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
f3 is not a function from X to Y because an element 2 ϵ x is associated to two elements 9 and 11 in Y.