CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The maximum value of the expression 1sin2θ+3sinθcosθ+5cos2θ is

A
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2
Ans. Ans. (a).
f(θ)=1sin2θ+3sinθcosθ+5cos2θ
=11cos2θ2+32sin2θ+5(1+cos2θ)2
=26+3sin2θ+4cos2θ
Hence, [f(θ)]maximum=265=2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Principal Solution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon