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

Created by: Arshdeep Kaur

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

``` Page 1

CBSE TEST PAPER-03
CLASS - XII MATHEMATICS (Relations and Functions)
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)
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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