SSC संयुक्त स्नातक स्तरीय टीयर-1 प्रैक्टिस सेट 3

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Question : 64 of 100

Marks: +1, -0

यदि

x^{2}+(1 \/ x^{2})=1

, तो

x^{48}+x^{42}+x^{36}+x^{30}+x^{24}

+x^{18}+x^{12}+x^{6}+1

का मान क्या होगा?

$-9$
0
1
9

Solution:

x^{2}+{1}/{x^{2}}=1

⇒ x^{4}+1=x^{2}

⇒ x^{4}+1-x^{2}=0

⇒ (x^{2}+1)(x^{2}+1-x^{2})=0

∴(x^{6}+1)=0

प्रदत्त व्यंजक

= x^{48}+x^{42}+x^{36}+x^{30}+x^{24}

+x^{18}+x^{12}+x^{6}+1

= x^{42}(x^{6}+1)+x^{30}(x^{6}+1)

+x^{18}(x^{6}+1)+x^{6}(x^{6}+1)+1

= 0+0+0+0+1=1

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