Relations & Functions Practice Questions - DPP for JEE

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symmetric, because x
2
 + y
2
 = 1 ? y
2
 + x
2
 = 1
3. (c) Let f (x) ? 2 be true and f (y) = 2, f (z) ? 1 are false
? f (x) ? 2, f (y) ? 2, f (z) = 1
? f (x) = 3, f (y) = 3, f (z) = 1 but then function is
many one, similarly two other cases.
4. (a) f (4) = g (4) ? 8 + a = 8  ? a = 0
f (–1) = – 2 for a = 0
f (–1) > f (4)
b + 3 > 8 ? b > 5
5. (b) We have to test  the equivalencity of relation R on S.
(1) Reflexivity :
In a plane any line be parallel to itself not perpendicular. Hence ,
R is not reflexive.
(2) Symmetry :
In a plane if a line AB is perpendicular to the other
line BC, then BC is also perpendicular to AB, i.e.,
And 
Hence R is symmetric.
(3) Transitivity :
In a plane, let AB, BC and CA be three lines, such that
?  R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
6. (a) Since R is reflexive relation on A, therefore (a,a) ? R for all a
?A.
The minimum number of ordered pairs in R is n.
Hence , .
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symmetric, because x
2
 + y
2
 = 1 ? y
2
 + x
2
 = 1
3. (c) Let f (x) ? 2 be true and f (y) = 2, f (z) ? 1 are false
? f (x) ? 2, f (y) ? 2, f (z) = 1
? f (x) = 3, f (y) = 3, f (z) = 1 but then function is
many one, similarly two other cases.
4. (a) f (4) = g (4) ? 8 + a = 8  ? a = 0
f (–1) = – 2 for a = 0
f (–1) > f (4)
b + 3 > 8 ? b > 5
5. (b) We have to test  the equivalencity of relation R on S.
(1) Reflexivity :
In a plane any line be parallel to itself not perpendicular. Hence ,
R is not reflexive.
(2) Symmetry :
In a plane if a line AB is perpendicular to the other
line BC, then BC is also perpendicular to AB, i.e.,
And 
Hence R is symmetric.
(3) Transitivity :
In a plane, let AB, BC and CA be three lines, such that
?  R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
6. (a) Since R is reflexive relation on A, therefore (a,a) ? R for all a
?A.
The minimum number of ordered pairs in R is n.
Hence , .
7. (d) f (2) =    ? many to one function
and  ? into function
8. (b) We have, gof (x) = 
= 
Similarly, fog (x) = 
= 
= 
Thus, gof (x) = x,  x ? B and fog (x) = x,  x ? A, which implies
that gof = I
B
 and fog = I
A
.
9. (d) f (x) =  [x]
2
 + [x + 1] – 3 = {[x] + 2} {[x] – 1}
So,  x = 1, 1.1, 1.2, ..........   ? f (x) = 0
? f (x) is many one.
only integral values will be attained.
? f (x) is into.
10. (b)
g (x) = e
x
, x = –1
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symmetric, because x
2
 + y
2
 = 1 ? y
2
 + x
2
 = 1
3. (c) Let f (x) ? 2 be true and f (y) = 2, f (z) ? 1 are false
? f (x) ? 2, f (y) ? 2, f (z) = 1
? f (x) = 3, f (y) = 3, f (z) = 1 but then function is
many one, similarly two other cases.
4. (a) f (4) = g (4) ? 8 + a = 8  ? a = 0
f (–1) = – 2 for a = 0
f (–1) > f (4)
b + 3 > 8 ? b > 5
5. (b) We have to test  the equivalencity of relation R on S.
(1) Reflexivity :
In a plane any line be parallel to itself not perpendicular. Hence ,
R is not reflexive.
(2) Symmetry :
In a plane if a line AB is perpendicular to the other
line BC, then BC is also perpendicular to AB, i.e.,
And 
Hence R is symmetric.
(3) Transitivity :
In a plane, let AB, BC and CA be three lines, such that
?  R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
6. (a) Since R is reflexive relation on A, therefore (a,a) ? R for all a
?A.
The minimum number of ordered pairs in R is n.
Hence , .
7. (d) f (2) =    ? many to one function
and  ? into function
8. (b) We have, gof (x) = 
= 
Similarly, fog (x) = 
= 
= 
Thus, gof (x) = x,  x ? B and fog (x) = x,  x ? A, which implies
that gof = I
B
 and fog = I
A
.
9. (d) f (x) =  [x]
2
 + [x + 1] – 3 = {[x] + 2} {[x] – 1}
So,  x = 1, 1.1, 1.2, ..........   ? f (x) = 0
? f (x) is many one.
only integral values will be attained.
? f (x) is into.
10. (b)
g (x) = e
x
, x = –1
= 
?  domain = [–1, 8)
fog is decreasing in [–1, 0) and increasing in [0, 8)
 and   fog (0) = 0
As  x ? 8, fog (x) ? 8,    
?  range = [0, 8)
?
11. (b) (a) Non-reflexive because (x
3
, x
3
) ? R
1
(b) Reflexive
(c) Non-Reflexive
(d) Non-reflexive because x
4
 ? X
12. (c) Here R = {(1, 3), (2, 2); (3, 2)}, S = {(2, 1); (3, 2); (2, 3)} 
Then RoS = {(2, 3), (3, 2); (2, 2)}
13. (a) g(f(x)) = |sin x| indicates that possibly f(x) = sinx, g(x) = |x|
Assuming it correct, f(g(x)) = f(|x|) = sin |x|, which is not correct.
 indicates that possibly
  or 
Then 
(for the first combination), which is given.
Hence f(x) = sin
2
x, g (x) = 
[Students may try by checking the options one by one]
14. (b) Let f : R  R be a function defined by
f (x) = 
For any 
Let   f (x) = f (y)
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symmetric, because x
2
 + y
2
 = 1 ? y
2
 + x
2
 = 1
3. (c) Let f (x) ? 2 be true and f (y) = 2, f (z) ? 1 are false
? f (x) ? 2, f (y) ? 2, f (z) = 1
? f (x) = 3, f (y) = 3, f (z) = 1 but then function is
many one, similarly two other cases.
4. (a) f (4) = g (4) ? 8 + a = 8  ? a = 0
f (–1) = – 2 for a = 0
f (–1) > f (4)
b + 3 > 8 ? b > 5
5. (b) We have to test  the equivalencity of relation R on S.
(1) Reflexivity :
In a plane any line be parallel to itself not perpendicular. Hence ,
R is not reflexive.
(2) Symmetry :
In a plane if a line AB is perpendicular to the other
line BC, then BC is also perpendicular to AB, i.e.,
And 
Hence R is symmetric.
(3) Transitivity :
In a plane, let AB, BC and CA be three lines, such that
?  R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
6. (a) Since R is reflexive relation on A, therefore (a,a) ? R for all a
?A.
The minimum number of ordered pairs in R is n.
Hence , .
7. (d) f (2) =    ? many to one function
and  ? into function
8. (b) We have, gof (x) = 
= 
Similarly, fog (x) = 
= 
= 
Thus, gof (x) = x,  x ? B and fog (x) = x,  x ? A, which implies
that gof = I
B
 and fog = I
A
.
9. (d) f (x) =  [x]
2
 + [x + 1] – 3 = {[x] + 2} {[x] – 1}
So,  x = 1, 1.1, 1.2, ..........   ? f (x) = 0
? f (x) is many one.
only integral values will be attained.
? f (x) is into.
10. (b)
g (x) = e
x
, x = –1
= 
?  domain = [–1, 8)
fog is decreasing in [–1, 0) and increasing in [0, 8)
 and   fog (0) = 0
As  x ? 8, fog (x) ? 8,    
?  range = [0, 8)
?
11. (b) (a) Non-reflexive because (x
3
, x
3
) ? R
1
(b) Reflexive
(c) Non-Reflexive
(d) Non-reflexive because x
4
 ? X
12. (c) Here R = {(1, 3), (2, 2); (3, 2)}, S = {(2, 1); (3, 2); (2, 3)} 
Then RoS = {(2, 3), (3, 2); (2, 2)}
13. (a) g(f(x)) = |sin x| indicates that possibly f(x) = sinx, g(x) = |x|
Assuming it correct, f(g(x)) = f(|x|) = sin |x|, which is not correct.
 indicates that possibly
  or 
Then 
(for the first combination), which is given.
Hence f(x) = sin
2
x, g (x) = 
[Students may try by checking the options one by one]
14. (b) Let f : R  R be a function defined by
f (x) = 
For any 
Let   f (x) = f (y)
 x = y
 f is one – one
Let such that  f (x) = 
 
 (x – n) = x – m
 x – n = x – m
 
 x = . for a = 1, 
So,  f is not onto.
15. (a) Given  f (x) = 
 (f o f) (x) = f {f (x)} = f 
=  =  =  = x.
 (x)  = f (f o f)(x)   = f (x) = 
16. (b) By definition only f (x) = x
2
 + 4x – 5 with domain
 is one to one.
17. (b) If y = ,    10
2x
 =  
or  x =  log
10
   ? f 
–1
 (x) =  log
10
 .