Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Consider a real valued continuous function f such that f(x)= sin x+∫ limits-π / 2π / 2( sin x+t f(t)) d t. If and m are maximum and minimum value of the function f, then
Q. Consider a real valued continuous function
f
such that
f
(
x
)
=
sin
x
+
−
π
/2
∫
π
/2
(
sin
x
+
t
f
(
t
))
d
t
. If and
m
are maximum and minimum value of the function
f
, then
844
98
Application of Derivatives
Report Error
A
m
M
=
3
B
M
−
m
=
2
π
+
1
C
M
+
m
=
4
(
π
+
1
)
D
M
m
=
2
(
π
2
+
1
)
Solution:
f
(
x
)
=
sin
x
+
π
sin
x
+
−
π
/2
∫
π
/2
t
f
(
t
)
d
t
=
(
π
+
1
)
sin
x
+
A
A
=
−
π
/2
∫
π
/2
((
π
+
1
)
sin
t
+
A
)
t
d
t
A
=
2
(
π
+
1
)
0
∫
π
/2
t
sin
t
d
t
=
2
(
π
+
1
)
(
−
cos
t
⋅
t
+
sin
t
)
∣
0
π
/2
=
2
(
π
+
1
)
(
1
)
=
(
π
+
1
)
⋅
2
f
(
x
)
=
(
π
+
1
)
(
2
+
sin
x
)
M
=
3
(
π
+
1
)
,
m
=
(
π
+
1
)