Q203 / 270 · NDA 2023 (Phase 1) · Mathematics

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

  • A
    $n$
  • B
    $2 n$
  • C
    $3 n$
  • Correct answer
    D
    $4 n$

Solution

$m(\theta)=\cot ^2 \theta+n^2 \tan ^2 \theta+2 n$ $\mathrm{AM} \geq \mathrm{GM}$ $\frac{\cot ^2 \theta+n^2 \tan ^2 \theta}{2} \geq \sqrt{\cot ^2 \theta \cdot n^2 \tan ^2 \theta}$ $\cot ^2 \theta+n^2 \tan ^2 \theta \geq 2 n$ For minimum value, $\cot ^2 \theta+n^2 \tan ^2 \theta=2 n$ $\therefore \quad m(\theta)_{\min }=2 n+2 n=4 n$

Hence, the correct answer is option: 'D'.

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