Question

The sum of the absolute maximum and absolute minimum values of the function \(\begin{array}{l} f\left(x\right)=\tan^{-1}\left(\sin x-\cos x\right) \end{array}\)in the interval\( [0, π]\) is

• A. 0
• B. \(tan^{-1}\frac{1}{\sqrt{2}}-\frac{π}{4}\)
• C. \(cos^{-1}\frac{1}{\sqrt{3}}-\frac{π}{4}\)
• D. \(-\frac{π}{12}\)

Answer 1

The Correct Option is C

\(f(x)=tan^{−1}(sinx−cosx)\)

Let \(g(x)=sinx−cosx\)

=\(\sqrt{2}sin(x−\frac{π}{4})\) and \(x−\frac{π}{4}∈[−\frac{π}{4},\frac{3π}{4}]\)

 ∴ g\((x)∈[−1,\sqrt2]\) 

and \(tan^{−1}x\) is an increasing function 

∴ \(f(x)∈[tan^{−1}(−1),tan^{−1}\sqrt2] ∈[−\frac{π}{4},tan^{−1}\sqrt2]\)

 ∴ Sum of \(f_{max}\) and \(f_{min}\)=\(tan^{−1}\sqrt{2}−\frac{π}{4}\)

= \(cos^{−1}(\frac{1}{\sqrt{3}})−\frac{π}{4}\)