TS Police SI Mains 2023 Paper 3 · Question 5
Question 5 of the TSLPRB SI Final Written Examination 2023, Paper III (Arithmetic & Reasoning) (8 April 2023). Try it first, then open the answer and explanation.
(2^√(7 + 4√3) + 2^√(12 + 6√3)) / (2^√(12 − 6√3) − 2^√(7 − 4√3)) =
- 3
- 3(2^(2√3))
- 2³
- √3
(2^√(7 + 4√3) + 2^√(12 + 6√3)) / (2^√(12 − 6√3) − 2^√(7 − 4√3)) =
- 3
- 3(2^(2√3))
- 2³
- √3
Show answer
Answer: (2) 3(2^(2√3))
సమాధానం: (2) 3(2^(2√3))
√(7 + 4√3) = 2 + √3, √(12 + 6√3) = 3 + √3, √(12 − 6√3) = 3 − √3 and √(7 − 4√3) = 2 − √3. The numerator is 2^√3(2² + 2³) = 12 × 2^√3 and the denominator is 2^(−√3)(2³ − 2²) = 4 × 2^(−√3), so the value is 3 × 2^(2√3).
√(7 + 4√3) = 2 + √3, √(12 + 6√3) = 3 + √3, √(12 − 6√3) = 3 − √3 మరియు √(7 − 4√3) = 2 − √3. లవం 2^√3(2² + 2³) = 12 × 2^√3 మరియు హారం 2^(−√3)(2³ − 2²) = 4 × 2^(−√3), కాబట్టి విలువ 3 × 2^(2√3).