CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The minimum value of a+bω+cω2, where a,band c are all not equal integers and omega ( not equal 1)is a cube root of unity, is


A

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

1


Explanation for correct option :

Given expression is a+bω+cω2where a,band c are all not equal integers and ω1

We know that cube root of unity is ω

That is x3-1=x-1x2+x+1

Then ω3=1 and the root of the equation x2+x+1=0 is ω,ω2

From the equation x2+x+1=0 we know, 1+ω+ω2=0.

Let z=a+bω+cω2

z=a+bω+cω2z2=a+bω+cω22[ω2+ω+1=0]=(a+bω+cω2)(a+bω¯+cω¯2)=a2+b2+c2-ac-bc-ab2z2=a-b2+b-c2+c-a2

Since, a,b,c are all integer not simultaneously equal

If a=b then ac and bc

Because difference of the integer is integer

b-c21

As minimum difference of two consecutive integers is +1

Similarly c-a21

Take a-b2=0

We have

2z2=a-b2+b-c2+c-a22z20+1+1z21

Hence, option C is correct.


flag
Suggest Corrections
thumbs-up
1
BNAT
mid-banner-image