यदि \(x-\frac{1}{x}=6\), तो \(x^{2}+\frac{1}{x^{2}}\) का मान क्या है?
If \(x-\frac{1}{x}=6\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?
Explanation opens after your attempt
Correct Answer
C. (38)
Concept
We use (\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).
Why this answer is correct
The correct answer is C. (38). We use (\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).
Exam Tip
(\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2) होता है। इसलिए \(36=x^{2}+\frac{1}{x^{2}}-2\) और मान (38) है।
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