12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 8 of 9 pages
Answer: Option B. -> [−2π3,π3]
:
B
Put x=sin−1x
2f(x)+f(√1−x2)=sin−1x→(1)
On Putting x=cos−1x
⇒2f(√1−x2)+f(x)=cos−1x→(2)
Eq.(1)×2⇒4f(x)+2f(√1−x2)=2sin−1x→(3)
On subtracting Eq. 2 from Eq. 3 we get -
3f(x)=2sin−1x−cos−1x
f(x)=23sin−1x−13(π2−sin−1x)
=sin−1x−π6
fmax=π2−π6=π3,fmin=−π2−π6=−4π6=−2π3
=[−2π3,π3]
:
B
Put x=sin−1x
2f(x)+f(√1−x2)=sin−1x→(1)
On Putting x=cos−1x
⇒2f(√1−x2)+f(x)=cos−1x→(2)
Eq.(1)×2⇒4f(x)+2f(√1−x2)=2sin−1x→(3)
On subtracting Eq. 2 from Eq. 3 we get -
3f(x)=2sin−1x−cos−1x
f(x)=23sin−1x−13(π2−sin−1x)
=sin−1x−π6
fmax=π2−π6=π3,fmin=−π2−π6=−4π6=−2π3
=[−2π3,π3]
Answer: Option A. -> one – one but not onto
:
A
We can draw the graph and see that the given function is one - one.
Since, Range ⊏ Codomain ⇒ f is into
∴ f is only injective.
:
A
We can draw the graph and see that the given function is one - one.
Since, Range ⊏ Codomain ⇒ f is into
∴ f is only injective.
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 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 B. -> Symmetric
:
B
Clearly, the relation is symmetric but it is neither reflexive nor transitive.
:
B
Clearly, the relation is symmetric but it is neither reflexive nor transitive.
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)
:
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)
Answer: Option B. -> f:R→[0,∞),f(x)=x2
:
B
The function f(x) = x2, x ϵR is not one – one because
f(-4)=f (4) = 16
∴ It is not invertible
:
B
The function f(x) = x2, x ϵR is not one – one because
f(-4)=f (4) = 16
∴ It is not invertible
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
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