Substitutex=cos\alp,\alp∈[0:;π] T
h
e
n
∣
x
+
1
−
x
2
∣
=
2
(
2
x
2
−
1
)
\Leftright
∣
c
o
s
\alp
+
s
i
n
\alp
∣
=
2
(
2
c
o
s
2
\alp
−
1
)
Then∣x+
1−x
2

​
∣=
2
​
(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2
​
(2cos
2
\alp−1)
∣
N
2
c
o
s
(
\alp
−
π
4
)
∣
=
N
2
c
o
s
(
2
\alp
)
\Right
\alp
∈
[
0
;
π
4
]
∪
[
3
π
4
;
π
]
∣N
2
​
cos(\alp−
4
π
​
)∣=N
2
​
cos(2\alp)\Right\alp∈[0;
4
π
​
]∪[
4
3π
​
;π]
1)
\alp
∈
[
0
;
π
4
]
\alp∈[0;
4
π
​
]
c
o
s
(
\alp
−
π
4
)
=
c
o
s
(
2
\alp
)
…
cos(\alp−
4
π
​
)=cos(2\alp)…
2.
\alp
∈
[
3
π
4
;
π
]
\alp∈[
4
3π
​
;π]
−
c
o
s
(
\alp
−
π
4
)
=
c
o
s
(
2
\alp
)
…
−cos(\alp−
4
π
​
)=cos(2\alp)…

this is joke just answer 1+1