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Mathematics Question on Maxima and Minima

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

A

0

B

tan−112−π4tan^{-1}\frac{1}{\sqrt{2}}-\frac{π}{4}

C

cos−113−π4cos^{-1}\frac{1}{\sqrt{3}}-\frac{π}{4}

D

−π12-\frac{π}{12}

Answer

cos−113−π4cos^{-1}\frac{1}{\sqrt{3}}-\frac{π}{4}

Explanation

Solution

f(x)=tan−1(sinx−cosx)f(x)=tan^{−1}(sinx−cosx)

Let g(x)=sinx−cosxg(x)=sinx−cosx

=2sin(x−π4)\sqrt{2}sin(x−\frac{π}{4}) and x−π4∈[−π4,3π4]x−\frac{π}{4}∈[−\frac{π}{4},\frac{3π}{4}]

∴ g(x)∈[−1,2](x)∈[−1,\sqrt2]

and tan−1xtan^{−1}x is an increasing function

∴ f(x)∈[tan−1(−1),tan−12]∈[−π4,tan−12]f(x)∈[tan^{−1}(−1),tan^{−1}\sqrt2] ∈[−\frac{π}{4},tan^{−1}\sqrt2]

∴ Sum of fmaxf_{max} and fminf_{min}=tan−12−π4tan^{−1}\sqrt{2}−\frac{π}{4}

= cos−1(13)−π4cos^{−1}(\frac{1}{\sqrt{3}})−\frac{π}{4}