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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2022 · Mathematics · Trigonometric Ratios & Identities

What is the maximum value of the function f(x)=(1)/( x+ x) , where...

What is the maximum value of the function f(x)=1tan⁡x+cot⁡xf(x)=\frac{1}{\tan x+\cot x}, where 0<x<π20 < x < \frac{\pi}{2} ?

Options

  1. A14\frac{1}{4}
  2. B12\frac{1}{2}
  3. C1
  4. D2

Correct answer

B. 12\frac{1}{2}

Step-by-step solution

f(x)=1tan⁡x+cot⁡x=sin⁡x−cos⁡xsin⁡2x+cos⁡2xf(x)=\frac{1}{\tan x+\cot x}=\frac{\sin x-\cos x}{\sin ^2 x+\cos ^2 x}=12sin⁡2xf′(x)=cos⁡2x=0∴x=π4f′(x)=−2sin⁡2x=−2sin⁡π4=−2<0∴f(x) is maximum at x=π4∴ Maximum value is f(π4)=12sin⁡(π2)=12\begin{aligned} & =\frac{1}{2} \sin 2 \mathrm{x} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\cos 2 \mathrm{x}=0 \\ & \therefore \mathrm{x}=\frac{\pi}{4} \\ & \mathrm{f}^{\prime}(\mathrm{x})=-2 \sin 2 \mathrm{x}=-2 \sin \frac{\pi}{4}=-2 < 0 \\ & \therefore \mathrm{f}(\mathrm{x}) \text{ is maximum at } \mathrm{x}=\frac{\pi}{4} \\ & \therefore \text{ Maximum value is } \\ & \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{2} \sin \left(\frac{\pi}{2}\right)=\frac{1}{2}\end{aligned}$$
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