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

If the real-valued function f(x)=px+sinx is a bijective function then the set of possible values of pR is

A
R{0}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(0,+)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
|p|>1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is A |p|>1
f(x)=px+sinx
f(x)=p+cosx
For given function to be bijective,
f(x)>0xR|p|>1

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