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Tardigrade
Question
Mathematics
Let f(x)= sin -1 x+ cos -1 x+π(|x|-2). The true set of values of k for which the equation f(x)=k π possesses real solution is [a, b] then the value of 2|a|+4|b| is equal to
Q. Let
f
(
x
)
=
sin
−
1
x
+
cos
−
1
x
+
π
(
∣
x
∣
−
2
)
. The true set of values of
k
for which the equation
f
(
x
)
=
kπ
possesses real solution is
[
a
,
b
]
then the value of
2∣
a
∣
+
4∣
b
∣
is equal to
310
112
Inverse Trigonometric Functions
Report Error
A
2
8%
B
3
20%
C
5
36%
D
7
36%
Solution:
f
(
x
)
=
2
π
+
π
(
∣
x
∣
−
2
)
x
∈
[
−
1
,
1
]
Equation
f
(
x
)
=
kπ
⇒
2
1
+
∣
x
∣
−
2
=
k
or
∣
x
∣
=
k
+
2
3
∈
[
0
,
1
]
⇒
0
≤
k
+
2
3
≤
1
⇒
−
2
3
≤
k
≤
−
2
1
⇒
a
=
2
−
3
and
b
=
2
−
1
∴
2∣
a
∣
+
4∣
b
∣
=
5