KeerthanaPosted on यदि x2+1x2=2x^{2}+frac{1}{x^{2}}=2x2+x21=2, तो x6?x^{6} ?x6? का मूल्य क्या है? a6 b0 c1 d3 Answer : Option CExplanation : x2+1x2=2x^{2}+frac{1}{x^{2}}=2x2+x21=2 ⇒x4+1=2x2Rightarrow x^{4}+1=2 x^{2}⇒x4+1=2x2 ⇒x4−2x2+1=0Rightarrow x^{4}-2 x^{2}+1=0⇒x4−2x2+1=0 ⇒(x2−1)2=0Rightarrowleft(x^{2}-1 ight)^{2}=0⇒(x2−1)2=0 ⇒x2−1=0⇒x2=1Rightarrow x^{2}-1=0 Rightarrow x^{2}=1⇒x2−1=0⇒x2=1 ∴x6=1 herefore x^{6}=1∴x6=1 Rate This:NaN / 5 - 1 votesAdd comment