Relations And Functions Ii(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
Answer: Option D. -> Reflexive, transitive but not symmetric:
D
Since n | n for all n in N,
therefore R is reflexive.
Since 2 | 6 but 6 | 2, therefore R is not symmetric.
Let n R m and m R p ⇒ n|m and m|p ⇒ n|p ⇒ nRp
So. R is transitive.
Explanation:-
Answer: Option A. -> (n+1) a:
A
f(x)+f(x+a)+f(x+2a)+....+f(x+na)=k
replacing x=x+a
f(x+a)+f(x+2a)+....+f(x+(n+1)a)=k
On subtracting second equation from first one -
f(x)−f(x+(n+1)a)=0f(x)=f(x+(n+1)a∴T=(n+1)a
Explanation:-
Answer: Option B. -> 8+(5−x3)15:
B
y=f(x)=(5−(x−8)5)13
then y3=5−(x−8)5⇒(x−8)5=5−y3
⇒ x=8+(5−y3)15
Let, z=g(x)=8+(5−x3)15
To check, f(g(x))= [5−(x−8)5]13
=(5−[(5−x3)15]5)13=(5−5+x3)13=x
similarly , we can show that g(f(x))=x
Hence, g(x)=8+(5−x3)15 is the inverse of f(x)
Explanation:-
Answer: Option C. -> transitive:
C
Let R denotes the relation ‘is brother of ‘. Let X ϵ A. If x is a girl, then we cannot say that x is brother of x.
∴ (x,x) /ϵ R ∴ R is not reflexive.
Let (x, y) ϵ R ∴ x is a brother of y.
∴ y may or may not be a boy.
∴ we cannot say that (y, x) ϵ R.
∴ R is not symmetric.
Let (x, y) ϵ and (y, z) ϵ R.
∴ x is a brother of y and y is a brother of z
⇒ is brother of z ⇒ (x, z) ϵ R
∴ R is transitive.
∴ The correct answer is (c).