Answer: Option B : B Put x=sin−1x 2f(x)+f(√1−x2)=sin−1x→(1) On Putting x=cos−1x ⇒2f(√1−x2)+f(x)=cos−1x→(2) Eq.(1)×2⇒4f(x)+2f(√1−x2)=2sin−1x→(3) On subtracting Eq. 2 from Eq. 3 we get - 3f(x)=2sin−1x−cos−1x f(x)=23sin−1x−13(π2−sin−1x) =sin−1x−π6 fmax=π2−π6=π3,fmin=−π2−π6=−4π6=−2π3 =[−2π3,π3]
Submit Your Solution hear: