यदि \(a=\sqrt{3}+\sqrt{2}\) है, तो \(a^2+\frac{1}{a^2}\) का मान क्या है?
If \(a=\sqrt{3}+\sqrt{2}\), what is the value of \(a^2+\frac{1}{a^2}\)?
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C. 10
Simple Explanation
\(a^2=(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}\)। साथ ही, \((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1\), इसलिए \(\frac{1}{a}=\sqrt{3}-\sqrt{2}\) और \(\frac{1}{a^2}=5-2\sqrt{6}\)। अतः \(a^2+\frac{1}{a^2}=(5+2\sqrt{6})+(5-2\sqrt{6})=10\)। \(2\sqrt{6}\) वाले पद परस्पर कट जाते हैं। परीक्षा टिप: ऐसे प्रश्नों में पहले युग्मज सुरड का उपयोग करके व्युत्क्रम ज्ञात करें। / \(a^2=(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}\). Also, \((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1\), so \(\frac{1}{a}=\sqrt{3}-\sqrt{2}\) and \(\frac{1}{a^2}=5-2\sqrt{6}\). Therefore, \(a^2+\frac{1}{a^2}=(5+2\sqrt{6})+(5-2\sqrt{6})=10\). The \(2\sqrt{6}\) terms cancel out. Exam tip: use the conjugate surd first to find the reciprocal in such questions.
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