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NDA 2022 Maths · Mathematics · Application of Derivatives · hard

The least value of f(x)=sin x+cos x on [0,(π)/(2)] is

Single correct option

The least value of f(x)=sin⁡x+cos⁡xf(x)=\sin x+\cos x on [0,π2]\left[0,\dfrac\pi2\right] is
  1. A. 00
  2. B. −2-\sqrt2
  3. C. 12\dfrac{1}{\sqrt2}
  4. D. 11
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Answer: (D) 11

f′(x)=cos⁡x−sin⁡x=0f'(x)=\cos x-\sin x=0 at x=π4x=\frac\pi4, giving the maximum 2\sqrt2. The least value is at an endpoint: f(0)=f(π2)=1f(0)=f(\frac\pi2)=1. −2-\sqrt2 is the global minimum over R\mathbb R, not on this interval.
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Original question written in the NDA 2022 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