Question
Download Solution PDFयदि , तो \((x-\frac{1}{x})^2+(x-\frac{1}{x})^4+(x-\frac{1}{x})^8\) का मान है:
Answer (Detailed Solution Below)
Detailed Solution
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दिया गया है,
समीकरण \( x^2 - x + 1 = 0 \) है,
हमें निम्नलिखित व्यंजक का मान ज्ञात करना है:
\( \left( x - \frac{1}{x} \right)^2 + \left( x - \frac{1}{x} \right)^4 + \left( x - \frac{1}{x} \right)^8 \)
समीकरण \( x^2 - x + 1 = 0 \) को निम्न प्रकार हल किया जाता है:
\( x = \frac{1 \pm \sqrt{-3}}{2} = e^{i \pi / 3} \quad \text{or} \quad x = e^{-i \pi / 3} \)
अब, \( x \) के मान को व्यंजक \( x - \frac{1}{x} \): में प्रतिस्थापित करें।
\( x - \frac{1}{x} = i\sqrt{3} \)
\( x - \frac{1}{x} \) की घातों का मूल्यांकन करें
अब, व्यंजक में प्रत्येक पद का मूल्यांकन करते हैं:
\( \left( x - \frac{1}{x} \right)^2 = (i \sqrt{3})^2 = -3 \)
\( \left( x - \frac{1}{x} \right)^4 = (-3)^2 = 9 \)
\( \left( x - \frac{1}{x} \right)^8 = 9^2 = 81 \)
अब, मानों को जोड़ें:
\( -3 + 9 + 81 = 87 \)
∴ व्यंजक का मान 87 है।
इसलिए, सही उत्तर विकल्प 3 है।
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