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Question : 103 of 150
Marks:
+1,
-0
Solution:
Given,
||+||=0Applying
C1→C1+C2+C3 and taking common from
C1⇒‌‌(1+2x)‌||‌+(3−x)||=0Applying
R2→R2−R1 and
R3→R3−R1‌⇒‌‌(1+2x)||‌+(3−x)||=0‌⇒‌‌(1+2x)(1−x)2+(3−x)x2=0‌⇒‌‌(1+2x)(1+x2−2x)+(3−x)x2=0‌⇒1+2x+x2+2x3−2x−4x2+3x2‌−x3=0⇒‌‌x3+1=0⇒x3=−1∴‌‌x=−1,−ω,−ω2∴‌‌x2005+‌=x2004⋅x+‌=(x3)668⋅x+‌=x(−1)668+‌=x+‌×‌=x+‌=x−x2 When
x=−ω‌ when ‌x−x2‌=−ω−(−ω)2‌=−ω−ω2=1x‌=−ω2x−x2‌=−ω2−(−ω2)2‌=−ω2−ω4=−ω2−ω‌=1
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