12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 6 of 9 pages
Answer: Option C. -> 512
:
C
Here n(A × B) = 3× 3 = 9
Since every subset of A× B defines a relationfrom A to B, the number ofrelations from A to B isequal tothe number of subsets of A ×B = 2n(A×B)
=29
= 512
:
C
Here n(A × B) = 3× 3 = 9
Since every subset of A× B defines a relationfrom A to B, the number ofrelations from A to B isequal tothe number of subsets of A ×B = 2n(A×B)
=29
= 512
Answer: Option B. -> {(1, 2), (2, 4), (3, 6)....}
:
B
R = {(2,1),(4,2),(6,3),.....}.
So, R−1 = {(1,2),(2,4),(3,6),......}.
:
B
R = {(2,1),(4,2),(6,3),.....}.
So, R−1 = {(1,2),(2,4),(3,6),......}.
Answer: Option D. -> None of these
:
D
Since A ⊆ A . ∴ Relation ' ⊆ ' is relfexive
Since A ⊆ B , B ⊆ C ⇒ A⊆ C
∴ Relation '⊆ ' is transitive.
But A ' ⊆ ' B , ⇒ B⊆ A . ∴ relation is not symmetric.
:
D
Since A ⊆ A . ∴ Relation ' ⊆ ' is relfexive
Since A ⊆ B , B ⊆ C ⇒ A⊆ C
∴ Relation '⊆ ' is transitive.
But A ' ⊆ ' B , ⇒ B⊆ A . ∴ relation is not symmetric.
Answer: Option B. -> Even
:
B
(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even
:
B
(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even
Answer: Option C. -> {(3, 3), (3, 5), (5, 3), (5, 5)}
:
C
We have, R={(1,3);(1,5);(2,3);(2,5);(3,5);(4,5)}
R−1 = {(3,1);(5,1);(3,2);(5,2);(5,3);(5,4)}
Hence RoR−1 = {(3,3);(3,5);(5,3);(5,5)}
:
C
We have, R={(1,3);(1,5);(2,3);(2,5);(3,5);(4,5)}
R−1 = {(3,1);(5,1);(3,2);(5,2);(5,3);(5,4)}
Hence RoR−1 = {(3,3);(3,5);(5,3);(5,5)}
Answer: Option A. -> {(2, 4), (3, 4)}
:
A
Given, A = {x:x2−5x+6=0}
∴ The elements of A are the roots ofx2−5x+6=0
x2−5x+6=0⇒(x−3)(x−2)=0⇒x=3and2
∴ A = {2,3} , B = {2,4}, C = {4,5}
⇒B∩C = {4}
∴ A× (B ∩ C) = {2, 3} × {4}
={(2,4),(3,4)}
:
A
Given, A = {x:x2−5x+6=0}
∴ The elements of A are the roots ofx2−5x+6=0
x2−5x+6=0⇒(x−3)(x−2)=0⇒x=3and2
∴ A = {2,3} , B = {2,4}, C = {4,5}
⇒B∩C = {4}
∴ A× (B ∩ C) = {2, 3} × {4}
={(2,4),(3,4)}
Answer: Option A. -> Even function
:
A
Let f(x), g(x) be odd
Let F(x) = f(x)g(x)
F(–x) = f(–x)g(–x) = F(x)
therefore F(x) is even
:
A
Let f(x), g(x) be odd
Let F(x) = f(x)g(x)
F(–x) = f(–x)g(–x) = F(x)
therefore F(x) is even
![Let f(x)=x−[x]1+x−[x],xϵ R, [ ]dentoes The Greatest...](https://lakshyaeducation.in/quizpics/quiz/_5bfc23fbde861c2550b8a0a66e802c90d9cf92dc20160812-3482-lww2c1.png "Click to Enlarge Image")
Answer: Option B. -> (0,1]
:
B
f(x)=ln(x2+ex2+1)=ln(x2+1−1+ex2+1)=ln(1+e−1x2+1)
0<e−1x2+1≤(e−1)⇒1<(1+e−1x2+1)≤e⇒0<ln(x2+ex2+1)≤1
Hence range is (0,1]
Hence(B) is correctanswer.
:
B
f(x)=ln(x2+ex2+1)=ln(x2+1−1+ex2+1)=ln(1+e−1x2+1)
0<e−1x2+1≤(e−1)⇒1<(1+e−1x2+1)≤e⇒0<ln(x2+ex2+1)≤1
Hence range is (0,1]
Hence(B) is correctanswer.
Answer: Option D. -> [0,1)
:
D
f(x)=x2x2+1=1−11+x2
x→0,f(x)→0
x→±∞,f(x)→1
∴Aϵ[0,1)
:
D
f(x)=x2x2+1=1−11+x2
x→0,f(x)→0
x→±∞,f(x)→1
∴Aϵ[0,1)
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email