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NDA 2020 Maths · Mathematics · Application of Derivatives · hard

The minimum value of 4sec ^2x+9csc ^2x is

Single correct option

The minimum value of 4sec⁡2x+9csc⁡2x4\sec^2x+9\csc^2x is
  1. A. 2525
  2. B. 1313
  3. C. 1212
  4. D. 3636
Show answer and solution

Answer: (A) 2525

Write the expression as 4(1+tan⁡2x)+9(1+cot⁡2x)=13+4tan⁡2x+9cot⁡2x4(1+\tan^2x)+9(1+\cot^2x)=13+4\tan^2x+9\cot^2x. By AM-GM, 4tan⁡2x+9cot⁡2x≥236=124\tan^2x+9\cot^2x\ge2\sqrt{36}=12, with equality when tan⁡2x=32\tan^2x=\tfrac32. So the minimum is 13+12=2513+12=25. The value 1313 ignores the positive extra part, and 3636 would come from (2+3)2(2+3)^2 computed wrongly. In general a2sec⁡2x+b2csc⁡2x≥(a+b)2a^2\sec^2x+b^2\csc^2x\ge(a+b)^2.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake