Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The maximum value of 3cosθ + 5sin (θ - (π/6)) for any real value of θ is :
Q. The maximum value of
3
cos
θ
+
5
s
in
(
θ
−
6
π
)
for any real value of
θ
is :
1903
237
JEE Main
JEE Main 2019
Application of Derivatives
Report Error
A
19
56%
B
2
79
28%
C
31
6%
D
34
9%
Solution:
y
=
3
cos
θ
+
5
(
sin
θ
2
3
−
cos
θ
2
1
)
2
5
3
sin
θ
+
2
1
cos
θ
y
m
a
x
=
4
75
+
4
1
=
19