Mathematics PRMO/ RMO - GIF
1 Find the values of x if
(a) [x] = 7 (b) [x] = –1 (c) [x] = 0
Solution:
[x] = 7
7 x 8 or x [7,8)
(b) [x] = -1
−1 x 0 or x [−1, 0)
(c) [x] = 0
0 x 1 or x [0,1)
2 If [x + 5] = 7 find all the possible values of x.
Solution:
[x + 5] > 7
x 2
x 3 …. ([x] is an integer and if this integer is greater than 2 it has to greater than
or equal to 3)
x [3, )
3 If 3.5 < x < 4.5 and [x] + [-x] = 0, then find the value of x.
Option:
(a) 2 (b)3 (c) 4 (d) 5
Answer:(c)
Solution:
Note that, x = [x] + {x} …(i)
( −x ) = ( − x ) + ( − x ) …(ii)
But we know that,
1 − x .... x is not an integer
− x =
0.... x is an integer
Now, x + −x = ( x − x) + ( ( −x ) − −x) …(using (i) and (ii))
= x − x − x − −x
= − x − −x
− x − (1 − x ) .... x is not an integer
=
0 .... x is an integer
1.... x is not an integer
=
0.... x is an integer
If x + −x = 0
x is on integer.
Moreover, 3.5 < x < 4.5,
So x = 4 is the only possible value for x.
4 Find the possible values of x if:
x + 25
(i) 10 (ii) x = 5 (iii) x = 5
5
Solution:
x + 25
5 10
x + 25
10
5
x
+ 5 10
5
x
5
5
x
6
5
x
6
5
x [30, )
(ii) Given, x = 5
x = 5 or x = −5
x −5, −4 ) 5, 6 )
(iii) Given, x = 5
5 | x | 6 or | x | [5, 6)
Case − I : If x 0,| x |= x
5 −x 6
Case − II : If x 0,| x |= x
5 −x 6
−5 x −6 ....(Multiplying by 1)
Combining case I and II,
x ( −6, −5 5, 6 )
5 If [x + y] = [x] + [y] then is it necessary that both x and y are integers ?
Solution:
No
e.g x = 0.1 and y = 0.2
L.H.S. x + y = 0.1 + 0.2 = 0.3 = 0
x + y = 0.1 + 0.2 = 0 + 0 = 0 .
x and y need not be integers.
1 1 1 1 2 1 3 1
6 Find value of: + + + + + + + ... + + 1 , here [.] → G.I.F.
3 3 100 3 100 3 100 3
Solution:
1 1 1 1 2 1 66 1 67 1 99 1 100
= + + + + + .... + + + + + .... + + + +
3 3 100 3 100 3 100 3 100 3 100 3 100
= 0 + (1 + 1 + .... + 1) = 34
7 Solve for x
(i) [x] = {x} + 2
(ii) 4[x] = 2{x} + 6
(iii) 2[x] = {x} + 1/2
(iv) [x] = 3{x} + 1/2
Solution:
(i) Given, [x] = {x} + 2
R.H.S should be integer
So {x} = 0
[x] = 2
x = x + x = 2
(ii) Given, 4[x] = 2{x} + 6
4[x] = 2{x} + 6
2[x] = {x} + 3
R.H.S should be integer
x = 0
2 x = 3
x = 1.5
x
(iii) Given, 2[x] = {x} + 1/2
R.H.S should be integer
x = 1/ 2
2[x] = 1, or [x] = 1/2
x
(iv) Given, [x] = 3{x} + 1/2
[x] = 3{x} + 1/2
0 x 1
0 3x 3
1
Case-I : 3x =
2
1
x =
6
1 1
x = + =1
2 2
x = x + x
1 7
x = 1+ =
6 6
3
Case-II: 3x =
2
1
x =
2
3 1
x = + =2
2 2
x = x + x
1 5
x = 2+ =
2 2
5
Case-III : 3x =
2
5
x =
6
5 1
x = + =3
2 2
x = x + x
5 5
x = 3+ =3
6 6
1 1 5
x 1 , 2 ,3
6 2 6
8 The solution set of x for 2x – 2[x] = 2
Option:
(a) x 0, 1, 2... (b) x 0,1, 2,... (c) x R (d) x
Answer:(d)
Solution:
2x – 2[x] = 2
x = [x] + {x}
Let [x] = I and {x} = f ; 0 f 1
Now, 2 ( I + f ) − 2I = 2
2f = 2 f = 1 which is not possible.
Thus, x
9 Find range of:
(i) x
(ii) 2 x
(iii) 2x
(iv) sin x
Solution:
(i) Given, x
x = 0 as 0 x 1
(ii) Given, 2 x
2 x = 0,1 as 0 2x 2
(iii) Given, 2x
2x = 0 as 0 2x 1
(iv) Given, sin x
sin x = 0 as sin x −1,0,1
10 Find number of solutions of : [x] + [2x] + [3x] = 28
Solution:
No. of solution [x] + [2x] + [3x] = 28
Putting x = I + f;f 0,1)
I + f + 2 ( I + f ) + 3 ( I + f ) = 28
I + 2I + 3I + f + 2f + 3f = 28
6I + (0 or 1 or 2 or 3) = 28
Not possible for any value.
