Question
Download Solution PDFयदि समीकरण x2−kx+k=0 का एक मूल दूसरे से 2/3 अधिक है, तो निम्नलिखित में से कौन-सा k का मान है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
द्विघात समीकरण x2 - kx + k = 0 है।
एक मूल दूसरे मूल से 2√3 अधिक है।
⇒ α - β = 2√3.
साथ ही,
मूलों का योग: α + β = k
मूलों का गुणनफल: α × β = k
गणना:
हमें निम्नलिखित सर्वसमिका ज्ञात हैं
\((\alpha + \beta )^2 = (\alpha - \beta)^2 - 4\alpha\beta \)
⇒ k2 = (2√3)2 - 4k
⇒ k2 - 12 - 4k = 0
⇒ k2 - 6k + 2k -12 = 0
⇒ k(k - 6) + 2 ( k - 6) = 0
⇒ (k - 6) (k + 2) = 0
⇒ k = 6 और k = -2
इस प्रकार, k के संभावित मान 6 और -2 हैं।
अतः सही उत्तर विकल्प 2 है।
Last updated on Jul 8, 2025
->UPSC NDA Application Correction Window is open from 7th July to 9th July 2025.
->UPSC had extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.