Solution- Relations and Functions Test-3 Class 12 Notes | EduRev

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Class 12 : Solution- Relations and Functions Test-3 Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Relations and Functions) 
[ANSWERS] 
Topic:- Relations and Functions 
1. the function f is one – one, for 
f(x 1) = f(x 2) 
2x 1 = 2x 2 
x 1 = x 2 
 2. is x 1, x 2 ? R
f(x 1) = f(x 2) 
3 – 4x1 = 3 – 4x2 
x 1 = x 2 
Hence one – one 
Y = 3 – 4x 
3 4
4
3 4 3 4
3 4
4 4
x
f
- ? ?
=
? ?
? ?
- - ? ? ? ?
= - ? ? ? ?
? ? ? ?
= y  
Hence onto also. 
3. f is one – one and onto, so that f is invertible with inverse f
-1
 = {(1, 1) (2, 2) (3, 3)} 
4. fog (x) = f(g x) 
= f{|5x – 2|) 
= |5x – 2| 
5. f = {(1, a) (2, b) (3, c)} 
f
-1
 = { (a, 1) (b, 2) (c, 3)} 
(f 
-1
) 
-1
 = {(1, a) (2, b) (3, c)} 
Hence (f
-1
)
-1
 = f.  
6. (fog) (x) = f[g(x)] 
= f(x – 7)  
= x – 7 + 7 
= x  
(fog) (7) = (7) 
Page 2


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Relations and Functions) 
[ANSWERS] 
Topic:- Relations and Functions 
1. the function f is one – one, for 
f(x 1) = f(x 2) 
2x 1 = 2x 2 
x 1 = x 2 
 2. is x 1, x 2 ? R
f(x 1) = f(x 2) 
3 – 4x1 = 3 – 4x2 
x 1 = x 2 
Hence one – one 
Y = 3 – 4x 
3 4
4
3 4 3 4
3 4
4 4
x
f
- ? ?
=
? ?
? ?
- - ? ? ? ?
= - ? ? ? ?
? ? ? ?
= y  
Hence onto also. 
3. f is one – one and onto, so that f is invertible with inverse f
-1
 = {(1, 1) (2, 2) (3, 3)} 
4. fog (x) = f(g x) 
= f{|5x – 2|) 
= |5x – 2| 
5. f = {(1, a) (2, b) (3, c)} 
f
-1
 = { (a, 1) (b, 2) (c, 3)} 
(f 
-1
) 
-1
 = {(1, a) (2, b) (3, c)} 
Hence (f
-1
)
-1
 = f.  
6. (fog) (x) = f[g(x)] 
= f(x – 7)  
= x – 7 + 7 
= x  
(fog) (7) = (7) 
7. (i)  (x, x) ? R, as x and x have the same no of pages for all x?R ? R is reflexive.
(ii)  (x, y) R 
x and y have the same no. of pages 
y and x have the same no. of pages 
? (y, x) ?R
? (x, y) = (y, x) R is symmetric.   
(iii)  if (x, y) ? R, (y, y) ? R
(x, z) ?R
? R is transitive.
8. (i)  a * b = a – b + ab 
b * a = b – a + ab 
a * b ? b * a
(ii)  a * (b * c) = a * (b – c + bc) 
= a – (b – c + bc) + a. (b – c + bc)  
= a – b + c – bc + ab – ac + abc  
(a * b) * c = (a – b + ab) * c  
= [ (a – b + ab) – c ] + ( a – b + ab) 
 = a- b + ab – c + ac – bc + abc 
a * (b * c) ?  (a * b) * c.
9. (i)  gof (x) = g[f(x)] 
= g (2x + 1) 
  = (2x + 1)
2
 – 2 
(ii)  fog (x) = f (fx) 
= f (2x + 1)  
= 2(2x + 1) + 1 
= 4x + 2 + 1 = 4x + 3 
10. Let x 1 x 2 ? A
Such that f(x1) = f(x2)
1 2
1 2
1 2
2 2
3 3
x x
x x
x x
- - =
- - =
f is one – one 
2
1 3
2 2
1
y x
x
x
y
x
y
- =
- - =
- 3 2
1
y
f y
y
? ? - =
? ?
- ? ?
Hence onto 
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