Question
Download Solution PDFIf K + \(\frac{1}{K}\) + 2 = 0 and K < 0, then what is the value of K11 + \(\frac{1}{K^4}\)?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
K + \(\frac{1}{K}\) + 2 = 0 and K < 0
Concept used:
(a2 + 2ab + b2) = (a + b)2
Calculation:
K + \(\frac{1}{K}\) + 2 = 0
⇒ k2 + 2k + 1 = 0
⇒ (k + 1)2 = 0
⇒ (k + 1) = 0
⇒ k = -1
Now, K11 + \(\frac{1}{K^4}\)
⇒ (-1)11 + \(\frac{1}{(-1)^4}\)
⇒ -1 + 1
⇒ 0
∴ The value of K11 + \(\frac{1}{K^4}\) is 0.
Last updated on Jun 25, 2025
-> The SSC CGL Notification 2025 has been released on 9th June 2025 on the official website at ssc.gov.in.
-> The SSC CGL exam registration process is now open and will continue till 4th July 2025, so candidates must fill out the SSC CGL Application Form 2025 before the deadline.
-> This year, the Staff Selection Commission (SSC) has announced approximately 14,582 vacancies for various Group B and C posts across government departments.
-> The SSC CGL Tier 1 exam is scheduled to take place from 13th to 30th August 2025.
-> Aspirants should visit ssc.gov.in 2025 regularly for updates and ensure timely submission of the CGL exam form.
-> Candidates can refer to the CGL syllabus for a better understanding of the exam structure and pattern.
-> The CGL Eligibility is a bachelor’s degree in any discipline.
-> Candidates selected through the SSC CGL exam will receive an attractive salary. Learn more about the SSC CGL Salary Structure.
-> Attempt SSC CGL Free English Mock Test and SSC CGL Current Affairs Mock Test.
-> Candidates should also use the SSC CGL previous year papers for a good revision.
->The UGC NET Exam Analysis 2025 for June 25 is out for Shift 1.