12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 9 of 9 pages
![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 A. -> π2
:
A
The smallest of π2,2π,4π,π is π2
Let f(x)=|sinx|+|cosx|.
∴f(x+π2)=∣∣sin(x+π2)∣∣+∣∣cos(x+π2)∣∣
=|cosx|+|−sinx|
=|cosx|+|sinx|
=f (x)
∴ The period of given function is π2
:
A
The smallest of π2,2π,4π,π is π2
Let f(x)=|sinx|+|cosx|.
∴f(x+π2)=∣∣sin(x+π2)∣∣+∣∣cos(x+π2)∣∣
=|cosx|+|−sinx|
=|cosx|+|sinx|
=f (x)
∴ The period of given function is π2
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
Answer: Option D. -> None of these
:
D
Since A⊆A.
∴ Relation ′⊆′ is reflexive.
Since A⊆B,B⊆C⇒A⊆C
∴ Relation ′⊆′ is transitive.
But If A⊆B, Doesn't imply B⊆A,
∴ Relation is not symmetric
:
D
Since A⊆A.
∴ Relation ′⊆′ is reflexive.
Since A⊆B,B⊆C⇒A⊆C
∴ Relation ′⊆′ is transitive.
But If A⊆B, Doesn't imply B⊆A,
∴ Relation is not symmetric
Answer: Option B. -> reflexive and symmetric
:
B
As (x,x) ϵ R ⇒ |x−x|≤ 1
⇒ 0 ≤ 1 (True),
Thus, reflexive.
As (x,y) ϵ R ⇒ |x−y|≤ 1
⇒|y−x||≤1⇒ (y,x) ϵ R,
Thus, symmetric.
Again, (x, y) ϵ R and (y, z) ϵ R
⇒|x−y|≤ 1 and |y−z|1/⇒|x−z|≤ 1
∴ Not transitive
:
B
As (x,x) ϵ R ⇒ |x−x|≤ 1
⇒ 0 ≤ 1 (True),
Thus, reflexive.
As (x,y) ϵ R ⇒ |x−y|≤ 1
⇒|y−x||≤1⇒ (y,x) ϵ R,
Thus, symmetric.
Again, (x, y) ϵ R and (y, z) ϵ R
⇒|x−y|≤ 1 and |y−z|1/⇒|x−z|≤ 1
∴ Not transitive
Answer: Option D. -> [−1,54]
:
D
f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1
=54−(sinx4−12)2
Maximun f(x)=54
Minimum f(x)=54−(−1−12)2=54−94=−1
Range of f(x)=[−1,54]
:
D
f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1
=54−(sinx4−12)2
Maximun f(x)=54
Minimum f(x)=54−(−1−12)2=54−94=−1
Range of f(x)=[−1,54]
Answer: Option D. -> [17,7]
:
D
y=x2−3x+4x2+3x+4
yx2+3xy+4y=x2−3x+4
x2(y−1)+3x(y+1)+4(y−1)=0
D ≥0⇒9(y+1)2−4.4(y−1)2≥0
(3(y+1)−4(y−1)) (3(y+1)+4(y−1))≥0
(−y+7)(7y−1)≥0
(y−7)(y−17)≤0
17≤y≤7
:
D
y=x2−3x+4x2+3x+4
yx2+3xy+4y=x2−3x+4
x2(y−1)+3x(y+1)+4(y−1)=0
D ≥0⇒9(y+1)2−4.4(y−1)2≥0
(3(y+1)−4(y−1)) (3(y+1)+4(y−1))≥0
(−y+7)(7y−1)≥0
(y−7)(y−17)≤0
17≤y≤7
Answer: Option B. -> Symmetric
:
B
Here we can see thatthe relation is neitherreflexive nortransitive but it is symmetric,
because x2+y2=1⇒y2+x2=1
:
B
Here we can see thatthe relation is neitherreflexive nortransitive but it is symmetric,
because x2+y2=1⇒y2+x2=1
Answer: Option D. -> an equivalence relation
:
D
(a, b) R (a, b) because a + b = b + a. So, r is reflexive.
(a, b)R (c, d) ⇒ a+d = b+c ⇒ c+b = d+a
⇒ (c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c, c + f = d + e
Adding, a + d + c + f = b + c + d +e
⇒ a + f = b + e
⇒ (a, b) R (e, f).
∴ R is transitive.
:
D
(a, b) R (a, b) because a + b = b + a. So, r is reflexive.
(a, b)R (c, d) ⇒ a+d = b+c ⇒ c+b = d+a
⇒ (c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c, c + f = d + e
Adding, a + d + c + f = b + c + d +e
⇒ a + f = b + e
⇒ (a, b) R (e, f).
∴ R is transitive.
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