Question
Download Solution PDFयदि \(k<(\sqrt{2}+1)^3<k+2,\) है, जहाँ k एक धनपूर्णांक है, तो k का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
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दिया गया है,
असमिका इस प्रकार है: \( k < (√{2} + 1)^3 < k + 2 \), जहाँ k एक प्राकृत संख्या है।
हमें \((√{2} + 1)^3\) की गणना करने की आवश्यकता है।
\( (√{2} + 1)^3 = 2√{2} + 6 + 3√{2} + 1 = 5√{2} + 7 \)
√ 2≈ 1.414 का सन्निकटन करते हुए, हम पाते हैं:
\( 5√{2} + 7 ≈ 5 \times 1.414 + 7 = 7.07 + 7 = 14.07 \)
असमिका निम्नवत हो जाती है:
\( k < 14.07 < k + 2 \)
इसका अर्थ है कि k > 12.07, इसलिए k का सबसे छोटा पूर्णांक मान है:
\( k = 13 \)
∴ k का मान 13 है।
इसलिए, सही उत्तर विकल्प 2 है।
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