Relations and Functions
Get insights from 232 questions on Relations and Functions, answered by students, alumni, and experts. You may also ask and answer any question you like about Relations and Functions
- 232
Questions
- 0
Discussions
- 0
Active Users
- 1
Followers
Have a question about Relations and Functions? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
For a suitably chosen real constant a, let a function, f: R − {−a} → R be defined by f(x) = (a-x)/(a+x). Further suppose that for any real number x ≠ −a and f(x) ≠ −a, (f ? f)(x) = x. Then f(-1/3) is equal to:
For a suitably chosen real constant a, let a function, f: R − {−a} → R be defined by f(x) = (a-x)/(a+x). Further suppose that for any real number x ≠ −a and f(x) ≠ −a, (f ? f)(x) = x. Then f(-1/3) is equal to:
f (f (x) = (a-f (x)/ (a+f (x) = x
Let f (x) = y. (a-y)/ (a+y) = x ⇒ a-y = ax + xy ⇒ a (1-x) = y (1+x) ⇒ y = a (1-x)/ (1+x)
⇒ f (x) = a (1-x)/ (1+x)
From the given options, we infer that comparing the derived f (x) leads to a=1.
⇒ a = 1
So f (x) = (1-x)/ (1+x)
f (-1/2) = (1 - (-1/2)/ (1 + (-1/2) = (3/2)/ (
Upvote 
Let R? and R? be two relations defined as follows:
R? = {(a, b) ∈ R²: a² + b² ∈ Q} and
R? = {(a, b) ∈ R²: a² + b² ∉ Q}, where Q is the set of all rational numbers. Then:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Let R? and R? be two relations defined as follows:
R? = {(a, b) ∈ R²: a² + b² ∈ Q} and
R? = {(a, b) ∈ R²: a² + b² ∉ Q}, where Q is the set of all rational numbers. Then:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
For R? let a = 1+√2, b=1-√2, c = 8¹/?
aR? b => a² + b² = (1+√2)² + (1-√2)² = 6 ∈ Q
aR? c => b² + c² = (1-√2)² + (8¹/? )² = 3 ∈ Q
aR? c => a² + c² = (1+√2)² + (8¹/? )² = 3 + 4√2 ∉ Q
∴ R? is not transitive.
For R? let a = 1+√2, b=√2, c=1-√2
aR? B => a² + b² = (1+√2)² + (√2)² = 5+2√2 ∉ Q
bR? b => b² + c² = (
Upvote 
If R={(x,y): x,y∈Z, x²+3y²≤8} is a relation on the set of integers Z, then the domain R?¹ is:
If R={(x,y): x,y∈Z, x²+3y²≤8} is a relation on the set of integers Z, then the domain R?¹ is:
R { (x, y):x, y∈z, x²+3y²≤8}
For domain of R? ¹
Collection of all integral of 's
For x=0, 3y²≤8
⇒ y∈ {-1,0,1}
Upvote Let g: N → N be defined as g(3n+1)=3n+2, g(3n+2)=3n+3, g(3n+3)=3n+1, for all n ≥ 0. Then which of the following statements is true?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Let g: N → N be defined as g(3n+1)=3n+2, g(3n+2)=3n+3, g(3n+3)=3n+1, for all n ≥ 0. Then which of the following statements is true?
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.
Upvote Let A=(a,b,c) and B=(1,2,3,4). Then the number of elements in the set C={f: A→B | 2∈f(A) and f is not one-one} is
Let A=(a,b,c) and B=(1,2,3,4). Then the number of elements in the set C={f: A→B | 2∈f(A) and f is not one-one} is
19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? × (3!/(2!1!)) × 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19
Upvote