12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 4 of 9 pages
Answer: Option D. -> R
:
D
g(f(x))≤f(g(x))⇒g(|x|)≤f[x]⇒[|x|]≤|[x]|
This is true for xϵR..
:
D
g(f(x))≤f(g(x))⇒g(|x|)≤f[x]⇒[|x|]≤|[x]|
This is true for xϵR..
Answer: Option A. -> many – one and into
:
A
Since, f (x + iy) = f (x - iy) ∴ f is many – one
Also, Range =R+ and codomain = R.
∴ Range ⊏ codomain ⇒ f is into.
Hence, f is many – one into.
:
A
Since, f (x + iy) = f (x - iy) ∴ f is many – one
Also, Range =R+ and codomain = R.
∴ Range ⊏ codomain ⇒ f is into.
Hence, f is many – one into.
Answer: Option B. -> 3f(x)
:
B
(fog)(x)=f(g(x))=f(3x+x31+3x2)=log1+3x2+3x+x31+3x2−3x−x3=log(1+x)3(1−x)3=3f(x)
:
B
(fog)(x)=f(g(x))=f(3x+x31+3x2)=log1+3x2+3x+x31+3x2−3x−x3=log(1+x)3(1−x)3=3f(x)
Answer: Option A. -> xR1 y⇔|x|=|y|
:
A
It is simple to check that only, R1is an equivalence relation.
:
A
It is simple to check that only, R1is an equivalence relation.
Answer: Option D. -> Reflexive, transitive but not symmetric
:
D
Since n | n for all n inN,
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.
:
D
Since n | n for all n inN,
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.
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).
:
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).
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
:
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
Answer: Option A. -> many – one and into
:
A
Since, f (x + iy) = f (x - iy) ∴ f is many – one
Also, Range =R+ and codomain = R.
∴ Range ⊏ codomain ⇒ f is into.
Hence, f is many – one into.
:
A
Since, f (x + iy) = f (x - iy) ∴ f is many – one
Also, Range =R+ and codomain = R.
∴ Range ⊏ codomain ⇒ f is into.
Hence, f is many – one into.
![Let F(x)=x−[x]1+x−[x],x ϵ R, Where [ X] Denotes The G...](https://lakshyaeducation.in/quizpics/quiz/content_24.png "Click to Enlarge Image")
Answer: Option D. -> {0}
:
D
f(x)=tan(π[x2−x])1+sin(cosx)={0} because of [x2−x] is integer
:
D
f(x)=tan(π[x2−x])1+sin(cosx)={0} because of [x2−x] is integer
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