Comprehension

Direction : Consider the following for the items that follow :

Let f(x) = cos2x + x on [-π/2, π/2]. 

What is the greatest value of f(x)? 

This question was previously asked in
NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)
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  1. \(\rm \frac{\sqrt3}{2}-\frac{\pi}{12}\)
  2. \(\rm \frac{\sqrt3}{2}+\frac{\pi}{12}\)
  3. \(\rm \frac{\sqrt3}{2}+\frac{\pi}{9}\)
  4. \(\rm \frac{\sqrt3}{2}+\frac{\pi}{6}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{\sqrt3}{2}+\frac{\pi}{12}\)
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Detailed Solution

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Explanation:

Given:

 f(x) = cos2x + x on [-π/2, π/2]. 

⇒ f'(x) = – 2 sin 2x + 1

For maxima and minima, f'(x) = 0

Sin2x = 1/2

x = π/2 or 5π/12 (not in interval)

So maximum value of f(x) occurs at x = π/12

\(f(\frac{\pi}{12}) = cos\frac{\pi }{6}+ \frac{\pi }{12} \)

\(\frac{\sqrt3}{2}+\frac{\pi}{12}\)

∴ Option (b) is correct.

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