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

All x satisfying the inequality (cot1x)27(cot1x)+10>0, lie in the interval :

A
(cot5,cot4)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(cot2,)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(,cot5)(cot4,cot2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(,cot5)(cot2,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (cot2,)
(cot1x)27(cot1x)+10>0(cot1x)25(cot1x)2(cot1x)+10>0(cot1x)[(cot1x)5]2[(cot1x)5]>0(cot1x2)(cot1x5)>0cot1x(,2)(5,) (1)

We know that cot1x(0,π) (2)
So, from (1) and (2)
0<cot1x<2
As, cot1x is decreasing function,
x(cot2,)


flag
Suggest Corrections
thumbs-up
83
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon