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

The minimum value of the expression 9x2sin2x+4xsinx for xε(0,π) is

A
163
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
83
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 12
We know that,
APGP 3xsinx+22(3xsinx)2
Squaring both sides
9x2sin2x+4+12xsinx46xsinx9x2sin2x+4+12xsinx24xsinx9x2sin2x+412xsinx9x2sin2x+4xsinx12

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 2
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon