Relations & Functions Practice Questions - DPP for JEE

 Page 2


| x | > x,  x < 0
Hence domain of f(x) is (– , 0)
4. (d) = 
is real for 0 = |x| = 4
f(x) is real for all 0 = |x| = 2 or 3 = |x| = 4.
5. (b) In the definition of function
Putting p and q in place of x, we get
f (p) = p
and 
f (q) = q
Putting x = (p + q)
=  = p + q = f (q) + f (p)
So, f (p) + f (q) = f (p + q)
6. (a) f (2a – x) = f (a – (x – a)) = f (a) f (x – a) – f (0) f (x)
= f (a) f (x –a) – f (x) = – f (x)
[ x = 0, y = 0, 
  f (a) = 0]
Page 3


| x | > x,  x < 0
Hence domain of f(x) is (– , 0)
4. (d) = 
is real for 0 = |x| = 4
f(x) is real for all 0 = |x| = 2 or 3 = |x| = 4.
5. (b) In the definition of function
Putting p and q in place of x, we get
f (p) = p
and 
f (q) = q
Putting x = (p + q)
=  = p + q = f (q) + f (p)
So, f (p) + f (q) = f (p + q)
6. (a) f (2a – x) = f (a – (x – a)) = f (a) f (x – a) – f (0) f (x)
= f (a) f (x –a) – f (x) = – f (x)
[ x = 0, y = 0, 
  f (a) = 0]
7. (a)
8. (d) x
2
 + y
2
 = 9 ? y
2
 = 9 – x
2
 ? y = 
x = 0 ? y =  = ± 3 ? Z
x = ± 1 ? y =  =  ? Z
x = ± 2 ? y =  =  ? Z
x = ± 3 ? y =  = 0 ? Z
x = ± 4 ? y =  =  ? Z and so on.
?   R = {(0, 3), (0, –3), (3, 0), (–3, 0)}
Domain of R = {x : (x, y) ? R} = {0, 3, –3}
Range of R = {y : (x, y) ? R} = {3, –3, 0}.
9. (c)
f (x) f (y) =  = 
=  = f (xy)
? f (xy) = f (x) f ( y)
10. (d) For f (x) to be defined, we must have
 = 0 or     > 0
? x
2
 = 1 – x
2
 or   .
Also, 1 – x
2
 = 0 or x
2
 = 1.
Page 4


| x | > x,  x < 0
Hence domain of f(x) is (– , 0)
4. (d) = 
is real for 0 = |x| = 4
f(x) is real for all 0 = |x| = 2 or 3 = |x| = 4.
5. (b) In the definition of function
Putting p and q in place of x, we get
f (p) = p
and 
f (q) = q
Putting x = (p + q)
=  = p + q = f (q) + f (p)
So, f (p) + f (q) = f (p + q)
6. (a) f (2a – x) = f (a – (x – a)) = f (a) f (x – a) – f (0) f (x)
= f (a) f (x –a) – f (x) = – f (x)
[ x = 0, y = 0, 
  f (a) = 0]
7. (a)
8. (d) x
2
 + y
2
 = 9 ? y
2
 = 9 – x
2
 ? y = 
x = 0 ? y =  = ± 3 ? Z
x = ± 1 ? y =  =  ? Z
x = ± 2 ? y =  =  ? Z
x = ± 3 ? y =  = 0 ? Z
x = ± 4 ? y =  =  ? Z and so on.
?   R = {(0, 3), (0, –3), (3, 0), (–3, 0)}
Domain of R = {x : (x, y) ? R} = {0, 3, –3}
Range of R = {y : (x, y) ? R} = {3, –3, 0}.
9. (c)
f (x) f (y) =  = 
=  = f (xy)
? f (xy) = f (x) f ( y)
10. (d) For f (x) to be defined, we must have
 = 0 or     > 0
? x
2
 = 1 – x
2
 or   .
Also, 1 – x
2
 = 0 or x
2
 = 1.
Now,  ?  = 0
? 
    
or    
Also, x
2
 = 1 ? (x – 1) ( x + 1) = 0
? –1 = x = 1
Thus, x > 0,  and x
2 
= 1
? x ?
11. (c)
? f (x) is an odd function.
12. (a) f (x) is defined if – log
1/2
 
? log
1/2
 
?
?
? 0 < x < 1
13. (b) For f (x) to be defined, we must have
x
2
 – 3x + 2 = (x – 1) (x – 2) > 0 ? x < 1 or > 2
Domain of f = (– 8,1) ? ( 2, 8).
Page 5


| x | > x,  x < 0
Hence domain of f(x) is (– , 0)
4. (d) = 
is real for 0 = |x| = 4
f(x) is real for all 0 = |x| = 2 or 3 = |x| = 4.
5. (b) In the definition of function
Putting p and q in place of x, we get
f (p) = p
and 
f (q) = q
Putting x = (p + q)
=  = p + q = f (q) + f (p)
So, f (p) + f (q) = f (p + q)
6. (a) f (2a – x) = f (a – (x – a)) = f (a) f (x – a) – f (0) f (x)
= f (a) f (x –a) – f (x) = – f (x)
[ x = 0, y = 0, 
  f (a) = 0]
7. (a)
8. (d) x
2
 + y
2
 = 9 ? y
2
 = 9 – x
2
 ? y = 
x = 0 ? y =  = ± 3 ? Z
x = ± 1 ? y =  =  ? Z
x = ± 2 ? y =  =  ? Z
x = ± 3 ? y =  = 0 ? Z
x = ± 4 ? y =  =  ? Z and so on.
?   R = {(0, 3), (0, –3), (3, 0), (–3, 0)}
Domain of R = {x : (x, y) ? R} = {0, 3, –3}
Range of R = {y : (x, y) ? R} = {3, –3, 0}.
9. (c)
f (x) f (y) =  = 
=  = f (xy)
? f (xy) = f (x) f ( y)
10. (d) For f (x) to be defined, we must have
 = 0 or     > 0
? x
2
 = 1 – x
2
 or   .
Also, 1 – x
2
 = 0 or x
2
 = 1.
Now,  ?  = 0
? 
    
or    
Also, x
2
 = 1 ? (x – 1) ( x + 1) = 0
? –1 = x = 1
Thus, x > 0,  and x
2 
= 1
? x ?
11. (c)
? f (x) is an odd function.
12. (a) f (x) is defined if – log
1/2
 
? log
1/2
 
?
?
? 0 < x < 1
13. (b) For f (x) to be defined, we must have
x
2
 – 3x + 2 = (x – 1) (x – 2) > 0 ? x < 1 or > 2
Domain of f = (– 8,1) ? ( 2, 8).
14. (b) f (x)  = = ln
Clearly range is (0, 1]
15. (b) Let f (x) = log (g(x))
? f (x
1
) + f (x
2
) = log(g(x
1
)) + log(g(x
2
))
= log(g(x
1
) · g(x
2
))
? Option (b) is correct
16. (d) {x
2
} – 2 {x} = 0
? {x} ({x} – 2} = 0
? {x}= 0  or {x} = 2
Second case is not possible.
Hence {x} = 0, as {x}= [0, 1). Hence range of f (x) contains only one
element 0.
17. (b) Given 
 = 
=   (multiply and divide by 2)
=  = 
=  = 
18. (b) We have,   f (x) = exp 
i.e. ,   f (x) = 
For Domain of f (x),  should be +ve.
i.e.,