Question
Question: If \[\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta \] then A.\[\alpha = \dfr...
If α⩽2sin−1x+cos−1x⩽β then
A.α=2π,β=2π
B.α=2π,β=23π
C.α=0,β=π
D.α=0,β=2π
Explanation
Solution
First we will take sin−1 function with its range. Then we will use the identitysin−1x+cos−1x=2π. With the help of this we will be able to find the values of α and β.
Complete step-by-step answer:
Given that,
α⩽2sin−1x+cos−1x⩽β
We know that,
−2π⩽sin−1x⩽2π
Adding 2π on both sides we will get
{\sin ^{ - 1}}x = \theta \\
x = \sin \theta \\
x = \cos \left( {\dfrac{\pi }{2} - \theta } \right) \\
{\cos ^{ - 1}}x = \dfrac{\pi }{2} - \theta \\
{\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x \\
{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \dfrac{\pi }{2} \\