The maximum value of the expression 1sin2θ+3sinθcosθ+5cos2θ is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
1
b
2
c
3
d
0
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
14cos2θ+1+32sin2θ=12(1+cos2θ)+1+32sin2θ=12cos2θ+32sin2θ+3Now, 3−22+322≤2cos2θ+32sin2θ+3≤3+22+322⇒12≤2cos2θ+32sin2θ+3≤112⇒211≤12cos2θ+32sin2θ+3≤2∴ maximum value is 2