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

The domain of f(x)=x42x3+3x22x+22x22x+1 is

A
[0,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
R
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
R{12}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
R{12,1}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B R
Given denominator
2x22x+1
=2[x2x]+1
=2[x2x+1414]+1
=2[(x12)2]+120 xR

Now for f(x) to be defined
x42x3+3x22x+20
(x4+3x2+2)(2x3+2x)0
(x2+1)(x2+2)2x(x2+1)0
(x2+1)(x22x+2)0
(x2+1)((x1)2+1)0,xR

Hence, Domain is R

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