CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The equation
2cos2x2sin2x=x2+x2;0<xπ2
has

A
No real solution
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
One real solution
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
more than one solution
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A No real solution

The given equation is
2cos2(x2)sin2x=x2+1x2
where 0<xπ2
L.H.S.=2cos2(x2)sin2x=(1+cosx)sin2x1+cosx<2 and sin2x1 for 0<xπ2(1+cosx)sin2x<2
and R.H.S.=x2+1x22
Therefore, for 0<xπ2, the given equation is not possible for any real value of x.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Neutralisation
Watch in App
Join BYJU'S Learning Program
CrossIcon