Question
Download Solution PDF\(f(x) = \frac{1}{\tan x+\cot x},\) ప్రమేయాల యొక్క గరిష్ట విలువ ఎంత, ఇక్కడ \(0 < x < \frac{\pi}{2}?\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFఉపయోగించిన సూత్రం:
- sin θ/cos θ = tan θ
- cos θ/sin θ = cot θ
- sin2θ + cos2θ = 1
- 2sin θ cos θ = sin 2θ
గణన:
\(f(x) = \frac{1}{\tan x+\cot x},\) \(0 < x < \frac{\pi}{2}?\)
⇒ \(f(x) = \frac{1}{\tan x+\cot x},\)
⇒ \(f(x) = \frac{1}{\frac{sin x}{cosx}+\frac{cosx}{sinx}}\)
⇒ \(f(x) =\frac{sinx\ cosx}{sin^2x+cos^2x}\)
⇒ f(x) = sin x cos x (\(\frac{2}{2}\)) [∵ sin2θ + cos2θ = 1]
⇒ f(x) = \(\frac{1}{2}\)sin 2x [∵ 2sin θ cosθ = sin 2θ]
మనకు తెలుసు, -1 ≤ sin θ ≤ 1
⇒ -1 ≤ sin 2x ≤ 1
∴ f(x)max = \(\frac{1}{2}\)(1) = 1/2
Last updated on May 30, 2025
->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.
-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.
-> 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.