web analytics

Example 48 - Show that f(x) = tan-1 (sin x + cos x) is always - To show increasing/decreasing in intervals

Slide36.JPG
Slide37.JPG

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

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﷮4﷯﷯ f﷐𝑥﷯=﷐﷐tan﷮−1﷯﷮﷐﷐sin﷮𝑥﷯+﷐cos﷮𝑥﷯﷯﷯ Finding f’﷐𝑥﷯ f’﷐𝑥﷯ = ﷐𝑑(﷐﷐tan﷮−1﷯﷮﷐﷐sin﷮𝑥﷯ +﷐cos﷮𝑥﷯﷯)﷯﷮𝑑𝑥﷯ = ﷐1﷮1+﷐﷐﷐sin﷮𝑥 + ﷐cos﷮𝑥﷯﷯﷯﷮2﷯﷯ × ﷐𝑑﷐﷐sin﷮𝑥 + ﷐cos﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ = ﷐1﷮1 + ﷐﷐﷐sin﷮2﷯𝑥﷮+﷐﷐cos﷮2﷯﷮𝑥 + 2﷐sin﷮𝑥﷯ ﷐cos﷮𝑥﷯﷯﷯﷯﷯ × ﷐﷐cos﷮𝑥﷯−﷐sin﷮𝑥﷯﷯ = ﷐1﷮1 + ﷐1 + 2﷐sin﷮𝑥﷯ ﷐cos﷮𝑥﷯﷯﷯ × ﷐﷐cos﷮𝑥﷯−﷐sin﷮𝑥﷯﷯ = ﷐1﷮2 + 2﷐sin﷮𝑥﷯ ﷐cos﷮𝑥﷯﷯ × ﷐﷐cos﷮𝑥﷯−﷐sin﷮𝑥﷯﷯ = ﷐﷐cos﷮𝑥 −﷐ sin﷮𝑥﷯﷯﷮2 + ﷐sin﷮2𝑥﷯﷯ For increasing, f’(x) > 0. ∴ Numerator and denominator both must be > 0 Checking sign for denominator 0 ≤ sin θ ≤ 1 for 0 ≤ θ ≤ ﷐𝜋﷮2﷯ 0 ≤ sin 2𝑥 ≤ 1 for 0 ≤ 2x ≤ ﷐𝜋﷮2﷯ 2 + 0 ≤ 2 + sin2𝑥 ≤ 1 + 2 for 0 ≤ x ≤ ﷐𝜋﷮4﷯ 2 ≤ 2 + sin 2𝑥 ≤ 3 Hence, denominator is always positive for 0 < x < ﷐𝜋﷮4﷯ Checking sign for numerator cos 𝑥 – sin 𝑥 > 0. cos 𝑥 > sin 𝑥. ﷐﷐cos﷮𝑥﷯﷮﷐cos﷮𝑥﷯﷯ > ﷐﷐sin﷮𝑥﷯﷮﷐sin﷮𝑥.﷯﷯ 1 > tan 𝑥 ﷐tan﷮𝑥﷯<1 That is possible only if 0<𝑥<﷐𝜋﷮4﷯ Thus, f’﷐𝑥﷯ = ﷐﷐+﷯﷮﷐+﷯﷯ > 0 in x ∈ ﷐0 , ﷐𝜋﷮4﷯﷯ Hence f is strictly increasing function is ﷐𝟎 , ﷐𝝅﷮𝟒﷯﷯

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail