Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
(a) Rang (f) = {-4, -3, 0, 5}
(b) Clearly, pre-images of 6,-3 and 5 is ∅, {0, 2}, -2 respectively.
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,
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} = ∅
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)
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.
{(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}}
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.
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.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Chapter 3: Functions – Exercise 3.2...
Chapter 3: Functions – Exercise 3.4...
Chapter 3: Functions – Exercise 3.3...