# Example 48 - Class 12

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,𝜋4 f𝑥=tan−1sin𝑥+cos𝑥 Finding f’𝑥 f’𝑥 = 𝑑(tan−1sin𝑥 +cos𝑥)𝑑𝑥 = 11+sin𝑥 + cos𝑥2 × 𝑑sin𝑥 + cos𝑥𝑑𝑥 = 11 + sin2𝑥+cos2𝑥 + 2sin𝑥 cos𝑥 × cos𝑥−sin𝑥 = 11 + 1 + 2sin𝑥 cos𝑥 × cos𝑥−sin𝑥 = 12 + 2sin𝑥 cos𝑥 × cos𝑥−sin𝑥 = cos𝑥 − sin𝑥2 + sin2𝑥 For increasing, f’(x) > 0. ∴ Numerator and denominator both must be > 0 Checking sign for denominator 0 ≤ sin θ ≤ 1 for 0 ≤ θ ≤ 𝜋2 0 ≤ sin 2𝑥 ≤ 1 for 0 ≤ 2x ≤ 𝜋2 2 + 0 ≤ 2 + sin2𝑥 ≤ 1 + 2 for 0 ≤ x ≤ 𝜋4 2 ≤ 2 + sin 2𝑥 ≤ 3 Hence, denominator is always positive for 0 < x < 𝜋4 Checking sign for numerator cos 𝑥 – sin 𝑥 > 0. cos 𝑥 > sin 𝑥. cos𝑥cos𝑥 > sin𝑥sin𝑥. 1 > tan 𝑥 tan𝑥<1 That is possible only if 0<𝑥<𝜋4 Thus, f’𝑥 = ++ > 0 in x ∈ 0 , 𝜋4 Hence f is strictly increasing function is 𝟎 , 𝝅𝟒

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Example 48 You are here

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.