यदि \(a=2+\sqrt{3}\), तो \(a^{2}+\frac{1}{a^{2}}\) का मान क्या होगा?
If \(a=2+\sqrt{3}\), what is the value of \(a^{2}+\frac{1}{a^{2}}\)?
Explanation opens after your attempt
Correct Answer
A. (14)
Concept
Here \(\frac{1}{a}=2-\sqrt{3}\), so \(a+\frac{1}{a}=4\) and \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\). In exams, use the identity (\left\(a+\frac{1}{a}\right\)^{2}).
Why this answer is correct
The correct answer is A. (14). Here \(\frac{1}{a}=2-\sqrt{3}\), so \(a+\frac{1}{a}=4\) and \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\). In exams, use the identity (\left\(a+\frac{1}{a}\right\)^{2}).
Exam Tip
\(\frac{1}{a}=2-\sqrt{3}\), इसलिए \(a+\frac{1}{a}=4\) और \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\)। परीक्षा में (\left\(a+\frac{1}{a}\right\)^{2}) पहचान लगाएं।
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