12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 5 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 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 B. -> 12(1+√1+4 log2 x)
:
B
f(x)=y⇒2x(x−1)=y⇒x(x−1)log22=log2y
⇒x(x−1)=log2y⇒x2−x−log2y=0
⇒x=1±√1+4log2y2
∴x=1+√1+4log2y2
∴f−1(x)=12(1+√1+4log2x)
∴ The correct answer is (b).
:
B
f(x)=y⇒2x(x−1)=y⇒x(x−1)log22=log2y
⇒x(x−1)=log2y⇒x2−x−log2y=0
⇒x=1±√1+4log2y2
∴x=1+√1+4log2y2
∴f−1(x)=12(1+√1+4log2x)
∴ The correct answer is (b).
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 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. -> {−1,1}
:
D
f(x)=1 when x+3>0
f(x)=−1 when x+3<0
Range ={−1,1}
:
D
f(x)=1 when x+3>0
f(x)=−1 when x+3<0
Range ={−1,1}
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 C. -> f(x) = x - [x], x ϵ R
:
C
f(x) = log x, is not periodic.
f(x) = ex, is not periodic.
f(x) = x - [x] = {x}, has period 1
f(x) = x + [x], is not periodic
:
C
f(x) = log x, is not periodic.
f(x) = ex, is not periodic.
f(x) = x - [x] = {x}, has period 1
f(x) = x + [x], is not periodic
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 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
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