Example 48 - Show that f(x) = tan-1 (sin x + cos x) is always

Advertisement

Example 48 - Chapter 6 Class 12 Application of Derivatives - Part 2

Advertisement

Example 48 - Chapter 6 Class 12 Application of Derivatives - Part 3

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  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‘₯ ) = πœπ¨π¬β‘γ€–π’™ βˆ’γ€– 𝐬𝐒𝐧〗⁑𝒙 γ€—/(𝟐 + π¬π’π§β‘πŸπ’™ ) For increasing, f’(x) > 0. ∴ Numerator and denominator both must be > 0 Checking sign for denominator Since Hence, denominator is always positive for 0 < x < 𝝅/πŸ’ Checking sign for numerator cos π‘₯ – sin π‘₯ > 0 cos π‘₯ > sin π‘₯ 1 > sin⁑π‘₯/cos⁑π‘₯ 1 > tan π‘₯ π­πšπ§β‘π’™<𝟏 This is possible only if 𝟎<𝒙<𝝅/πŸ’ Thus, f’(𝒙) = ((+))/((+) ) > 0 in x ∈ (0 , πœ‹/4) Hence, f is strictly increasing function in (𝟎 , 𝝅/πŸ’)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.