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

The expression x2+2x+1x2+2x+7 lies in the interval : (xϵR).

A
[0,1]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(,0]U[1,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
[0,1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of the above.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D [0,1)
Let y=x2+2x+1x2+2x+7
(y1)x2+2(y1)x+(7y1)=0
Given that xϵR
=4(y1)24(y1)(7y1)0
(y1)(y17y+1)0
y(y1)0
yϵ[0,1]
But y ( co efficient of x20)
yϵ[0,1)
x2+2x+1x2+2x+7ϵ[0,1)

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