Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Let P(x)=(1+x+x2+......x17)2x17 has 34 complex roots of the form zk=rkei2πak,k=1,2,3.....34 that 0<a1a2a3......a34<1andrk>0.  If a1+a2+......a5=mn Where m,n are relatively prime integers then m+n equals

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

480

b

481

c

483

d

482

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

p(x)=(x181x1)2x17=(x191)(x171)(x1)2

Now gcd(17,19)=1 and also p has 34 complex roots Hence all 16.17-th roots of unity And 18,19-th roots of unity are the root of p Then,  zm=ei2ntπ19,m=1,2.....18

zn=ei2mπ17,n=1,2,.......16 thenak{m19,n17/m=1,.....18,n=1,........16}

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring