Using variations to order numbers
Europeansetion,season1
Eah of you has been given a number.
Rank you onthe rst lineaordingto your number.
Hey look at this ! What a beautiful variationtable.
Return your ardand rankyou aording to your new
number.
x
f
(
x)
−
10 10x
f
(
x)
−
10 10x
f
(
x)
−
10 0 10x
f
(
x)
−
10−
5 0 5 10x
f
(
x)
−
10−
7−
3 1 4 5 10Compare the numbers os
(
1)
and os(
2)
.What funtion do we need ?
x
7→
osxThe graph of this funtion :
0 1 2 3
−
1 0−
2−
3What interval do we need ?
0 1 2 3
−
1 0−
2−
3The interval
[
0, π ]
.Over the interval
[
0, π ]
, the funtions x7→
osx isdereasing.
So, as 1
<
2,os
(
1) ≥
os(
2) .
π
2 and 9sin
(
2.
3)
and sin(
2.
7)
.3
4
2and
5
4
2.
1
4
3−
1
4
and
4
17
3−
4
17 .
0
.
983−
0.
98 and 1.
023−
1.
02.2
.
17and
−
2
.
17.
( −
2.
5)
3+
2.
5 and( − π )
3π
.( −
2.
01)
2and
( −
1.
99)
2.
π
and
4 .
√
7 and
√
10 .
sin
(
0)
and sin( − π )
.√
5
.
17 and√
5
.
71.( −
2.
17)
2 and( −
1.
5)
2.sin
( −
3)
and sin( −
2)
.− √
7
and
− √
10 .
sin
( −
1)
and sin(
1)
.π
2and
3
.
152.
1
.
12 and( −
1.
1)
2.√ π +
2 and√
6.
0
.
752and
0
.
662.
Mark 1 point for every good answer.
π
2≥
9sin
(
2.
3) ≥
sin(
2.
7)
.3
4
2≤
5
4
2.
1
4
3−
1
4
≤
4
17
3−
4
17 .
0
.
983−
0.
98≤
1.
023−
1.
02.2
.
17≥ −
2
.
17.
( −
2.
5)
3+
2.
5≥ ( − π )
3π
.( −
2.
01)
2≤
( −
1.
99)
2.
π ≥
4 .
√
7
≤ √
10 .
sin
(
0) =
sin( − π )
.√
5
.
17≤ √
5
.
71.( −
2.
17)
2≥ ( −
1.
5)
2.sin
( −
3) ≥
sin( −
2)
.− √
7
≤ − √
10 .
sin
( −
1) ≤
sin(
1)
.π
2≥
3
.
152.
1
.
12= ( −
1.
1)
2.√ π +
2≤ √
6.
0
.
752≤
0
.
662.