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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Sequences and Series

NDA 2023 · Mathematics · Sequences and Series

Paragraph: Given that m(θ)= ^2 θ+n^2 ^2 θ+2 n , where n is a fixed...

Paragraph: Given that m(θ)=cot⁡2θ+n2tan⁡2θ+2nm(\theta)=\cot ^2 \theta+n^2 \tan ^2 \theta+2 n, where nn is a fixed positive real number.
Question: What is the least value of m(θ)\mathrm{m}(\theta) ?

Options

  1. Ann
  2. B2n2 n
  3. C3n3 n
  4. D4n4 n

Correct answer

D. 4n4 n

Step-by-step solution

m(θ)=cot⁡2θ+n2tan⁡2θ+2nAM≥GMcot⁡2θ+n2tan⁡2θ2≥cot⁡2θ⋅n2tan⁡2θcot⁡2θ+n2tan⁡2θ≥2nm(\theta)=\cot ^2 \theta+n^2 \tan ^2 \theta+2 n\mathrm{AM} \geq \mathrm{GM}\frac{\cot ^2 \theta+n^2 \tan ^2 \theta}{2} \geq \sqrt{\cot ^2 \theta \cdot n^2 \tan ^2 \theta}\cot ^2 \theta+n^2 \tan ^2 \theta \geq 2 n For minimum value, cot⁡2θ+n2tan⁡2θ=2n∴m(θ)min⁡=2n+2n=4n\cot ^2 \theta+n^2 \tan ^2 \theta=2 n\therefore \quad m(\theta)_{\min }=2 n+2 n=4 n
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