# Example 48 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 48 Show that the function f given by f (x) = tan 1(sin x + cos x), x > 0 is always an strictly increasing function in 0, 4 f = tan 1 sin + cos Finding f f = ( tan 1 sin + cos ) = 1 1+ sin + cos 2 sin + cos = 1 1 + sin 2 + cos 2 + 2 sin cos cos sin = 1 1 + 1 + 2 sin cos cos sin = 1 2 + 2 sin cos cos sin = cos sin 2 + sin 2 For increasing, f (x) > 0. Numerator and denominator both must be > 0 Checking sign for denominator 0 sin 1 for 0 2 0 sin 2 1 for 0 2x 2 2 + 0 2 + sin2 1 + 2 for 0 x 4 2 2 + sin 2 3 Hence, denominator is always positive for 0 < x < 4 Checking sign for numerator cos sin > 0. cos > sin . cos cos > sin sin . 1 > tan tan <1 That is possible only if 0< < 4 Thus, f = + + > 0 in x 0 , 4 Hence f is strictly increasing function is ,

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Example 48 You are here

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.