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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2023 · Mathematics · Application of Derivatives

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: Under what condition does mm attain the least value?

Options

  1. An=tan⁡2θn=\tan ^2 \theta
  2. Bn=cot⁡2θn=\cot ^2 \theta
  3. Cn=sin⁡2θn=\sin ^2 \theta
  4. Dn=cos⁡2θn=\cos ^2 \theta

Correct answer

B. n=cot⁡2θn=\cot ^2 \theta

Step-by-step solution

For minimum value, cot⁡2θ+n2tan⁡2θ+2n=4n(cot⁡θ)2+(ntan⁡θ)2−2(cot⁡θ⋅ntan⁡θ)=0(cot⁡θ−ntan⁡θ)2=0cot⁡θ=ntan⁡θ∴cot⁡2θ=n\cot ^2 \theta+n^2 \tan ^2 \theta+2 n=4 n(\cot \theta)^2+(n \tan \theta)^2-2(\cot \theta \cdot n \tan \theta)=0(\cot \theta-n \tan \theta)^2=0\cot \theta=n \tan \theta\therefore \quad \cot ^2 \theta=n
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