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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2019 · Mathematics · Application of Derivatives

What is the minimum value of (a^2)/( ^2 x)+(b^2)/( ^2 x) where a >...

What is the minimum value of a2cos⁡2x+b2sin⁡2x\frac{a^2}{\cos ^2 x}+\frac{b^2}{\sin ^2 x} where a>0a > 0 and b>0b > 0 ?

Options

  1. A(a+b)2(a+b)^2
  2. B(a−b)2(a-b)^2
  3. Ca2+b2a^2+b^2
  4. D∣a2+b2∣\left|a^2+b^2\right|

Correct answer

A. (a+b)2(a+b)^2

Step-by-step solution

For minimum value a2cos⁡2x+b2sin⁡2x=a2sec⁡2x+b2cosec⁡2x=a2(1+tan⁡2x)+b2(1+cot⁡2x)=a2+b2+Q2tan⁡2x+b2cot⁡2x∵a2tan⁡2x+b2cot⁡2x≥2ab∴ minimum value =a2+b2+2ab=(a+b)2\begin{aligned} & \frac{a^2}{\cos ^2 x}+\frac{b^2}{\sin ^2 x} \\ & =a^2 \sec ^2 x+b^2 \operatorname{cosec}^2 x \\ & =a^2\left(1+\tan ^2 x\right)+b^2\left(1+\cot ^2 x\right) \\ & =a^2+b^2+Q^2 \tan ^2 x+b^2 \cot ^2 x \\ & \because a^2 \tan ^2 x+b^2 \cot ^2 x \geq 2 a b \\ & \therefore \text{ minimum value }=a^2+b^2+2 a b \\ & =(a+b)^2 \end{aligned}
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