wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If cos(4y3x2)cos(4y+3x+2)=2+2ln(k4255) and cos(4y3x2)+cos(4y+3x+2)=2k+8 have real solutions (x,y), then the range of k is

A
[3,4]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
[3,4]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
[4,3]
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
[4,3]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C [4,3]
cos(4y3x2)cos(4y+3x+2)=2+2ln(k4255)cos(4y3x2)+cos(4y+3x+2)=2k+8
Assuming 4y=A,3x+2=B, we get
cos(AB)cos(A+B)=2(1+ln(k4255))2sinAsinB=2(1+ln(k4255))1+ln(k4255)=sinAsinB1+ln(k4255)1ln(k4255)0k42551k4256k[4,4](1)

Also,
cos(AB)+cos(A+B)=2k+82cosAcosB=2(k+4)k+4=cosAcosBk+41k3(2)

From equation (1) and (2), we get
k[4,3]

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