12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 3 of 9 pages
Answer: Option C. -> (1,73]
:
C
Wehave,f(x)=x2+x+2x2+x+1=(x2+x+1)x2+x+1=1+1(x+12)2+34Wecanseeherethatasx→∞,f(x)→1whichistheminvalueoff(x).Alsof(x)ismaxwhen(x+12)2+34isminwhichissowhenx=−12andthen34.∴fmax=1+134=73∴Ri=(1,73]
:
C
Wehave,f(x)=x2+x+2x2+x+1=(x2+x+1)x2+x+1=1+1(x+12)2+34Wecanseeherethatasx→∞,f(x)→1whichistheminvalueoff(x).Alsof(x)ismaxwhen(x+12)2+34isminwhichissowhenx=−12andthen34.∴fmax=1+134=73∴Ri=(1,73]
Answer: Option D. -> [−1,54]
:
D
f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1=54−(sinx4−12)2Maximumf(x)=54Minimumf(x)=54=(−1−12)2=54−94=−1Range off(x)=[−1,54]
:
D
f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1=54−(sinx4−12)2Maximumf(x)=54Minimumf(x)=54=(−1−12)2=54−94=−1Range off(x)=[−1,54]
Answer: Option A. -> x2−y28
:
A
Let x + 2y = p
x – 2y = q
Solving we got x=p+q2
y=p−q2∴f(p,q)=p2−q28F(x,y)=x2−y28
:
A
Let x + 2y = p
x – 2y = q
Solving we got x=p+q2
y=p−q2∴f(p,q)=p2−q28F(x,y)=x2−y28
Answer: Option A. -> one-one and into
:
A
Given that f:[0,∞)→[0,∞)
s.t.f(x)=xx+1
then f′(x)=1+x−x(1+x2)=1(1+x)2 >0,∀x
∴f is an increasing function ⇒is one-one.
Also Df=[0,∞)
And for range let xx+1=y⇒x=yy+1
:
A
Given that f:[0,∞)→[0,∞)
s.t.f(x)=xx+1
then f′(x)=1+x−x(1+x2)=1(1+x)2 >0,∀x
∴f is an increasing function ⇒is one-one.
Also Df=[0,∞)
And for range let xx+1=y⇒x=yy+1
Answer: Option D. -> Not a periodic function
:
D
Try drawing cos√xgraph. It’s not periodic.
:
D
Try drawing cos√xgraph. It’s not periodic.
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 C. -> Bijective
:
C
We have seen that inverse exists only when function is one-one and onto, i.e. Bijective.
:
C
We have seen that inverse exists only when function is one-one and onto, i.e. Bijective.
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 D. -> (x−72)13
:
D
We have, f : R → R, g: R → R defined by f(x) = 2x - 3 and g (x) = x3 + 5
It can be checked that f(x) and g(x) are bijective functions
∴ fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)−3=2x3+7
(fog)(x)=y⇒2x3+7=y⇒x=(y−72)13
∴(fog)−1(x)=(x−72)13,xϵR
∴ The correct answer is (d).
:
D
We have, f : R → R, g: R → R defined by f(x) = 2x - 3 and g (x) = x3 + 5
It can be checked that f(x) and g(x) are bijective functions
∴ fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)−3=2x3+7
(fog)(x)=y⇒2x3+7=y⇒x=(y−72)13
∴(fog)−1(x)=(x−72)13,xϵR
∴ The correct answer is (d).
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
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