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

If α,β are the roots of the equation x2+αx+β=0 such that αβ and ||xβ|α|<1, then

A
inequality is satisfied by exactly three integral values of x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
inequality is satisfied by all values of xε(4,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
roots of the equation are opposite in sign
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
All of the above
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D All of the above
α+β=α and αβ=βα=1

β=2

(α,β)=(1,2)

||x+2|1|<1

1<|x+2|1<1

Let us inequality on left side first

1<|x+2|1

|x+2|>0x(,)

Right inequality ;

|x+2|1<1

|x+2|<2

2<x+2<2

x(4,0)

Combining both we get x(4,0)

(3,2,1) three integers satisfy the equation and all values between (4,0) satisfy the equation
α,β are of different sign

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