Question

8. If the remainder R(x) = ax + b is obtained by dividing the polynomial x100 by the polynomial x2 - 3x + 2 then (A) a = 2100 - 1 (B) b = 2(299 – 1) (C) b = -2(299 - 1) (D) a = 2100

JEE/Engineering Exams

Maths

Solution

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( x^{100}=left(x^{2}-3 x+2right) ) This question is from foom, resonance module Remainden theorm ( rightarrow ) ( x^{100}=g(x) cdotleft(x^{2}-3 x+2right)+R(x) ) whey ( Q(n) ) is nome quatient ( Rightarrow x^{100}=varphi(x) cdot(x-2)(x-1)+a x+b ) Put ( x=1 ) ( Rightarrow quad begin{aligned} Rightarrow quad 1 &=0+a+b & Rightarrow a+b=1 end{aligned} ) Put ( n=2 rightarrow ) ( Rightarrow quad 2^{100}=0+2 a+b ) ( =2 a+b=2^{100}=2left(2^{20}-1right) ) ( Rightarrow a+a+b=2^{100} ) ( Rightarrow a=2-1 quad therefore b=1-aleft(2^{100}right)=2 )