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

Example 48 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 48 - Chapter 6 Class 12 Application of Derivatives - Part 3

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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’)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.