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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

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,πœ‹/4) f(π‘₯)=tan^(βˆ’1)⁑(sin⁑π‘₯+cos⁑π‘₯ ) Finding f’(𝒙) f’(π‘₯) = (𝑑(tan^(βˆ’1)⁑〖(sin⁑π‘₯ +cos⁑π‘₯ ))γ€—)/𝑑π‘₯ = 1/(1 + (sin⁑〖π‘₯ + cos⁑π‘₯ γ€— )^2 ) Γ— 𝑑(sin⁑〖π‘₯ + cos⁑π‘₯ γ€— )/𝑑π‘₯ = 1/(1 + (〖〖𝐬𝐒𝐧〗^𝟐 𝒙〗⁑〖+γ€–πœπ¨π¬γ€—^πŸβ‘γ€–π’™ γ€—+ 2 sin⁑π‘₯ cos⁑π‘₯ γ€— ) ) Γ— (cos⁑π‘₯βˆ’sin⁑π‘₯ ) = 1/(1 + (𝟏 + 2 sin⁑π‘₯ cos⁑π‘₯ ) ) Γ— (cos⁑π‘₯βˆ’sin⁑π‘₯ ) = 1/(2 + 2 sin⁑π‘₯ cos⁑π‘₯ ) Γ— (cos⁑π‘₯βˆ’sin⁑π‘₯ ) ((tan^(βˆ’1)⁑π‘₯ )^β€²= 1/(1 + π‘₯^2 )) (sin2 x + cos2 x = 1) = cos⁑〖π‘₯ βˆ’γ€– sin〗⁑π‘₯ γ€—/(2 + sin⁑2π‘₯ ) For increasing, f’(x) > 0. ∴ Numerator and denominator both must be > 0 Checking sign for denominator Hence, denominator is always positive for 0 < x < πœ‹/4 (As sin2x = 2 sin x cos x) Checking sign for numerator cos π‘₯ – sin π‘₯ > 0. cos π‘₯ > sin π‘₯. cos⁑π‘₯/cos⁑π‘₯ > sin⁑π‘₯/cos⁑〖π‘₯.γ€— 1 > tan π‘₯ tan⁑π‘₯<1 That is possible only if 0<π‘₯<πœ‹/4 Thus, f’(π‘₯) = ((+))/((+) ) > 0 in x ∈ (0 , πœ‹/4) Hence f is strictly increasing function is (𝟎 , 𝝅/πŸ’)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.