Question

If \[\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta \] then
A.\[\alpha = \dfrac{\pi }{2},\beta = \dfrac{\pi }{2}\]
B.\[\alpha = \dfrac{\pi }{2},\beta = \dfrac{{3\pi }}{2}\]
C.\[\alpha = 0,\beta = \pi \]
D.\[\alpha = 0,\beta = 2\pi \]

Answer 1

Hint: First we will take \[{\sin ^{ - 1}}\] function with its range. Then we will use the identity\[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]. With the help of this we will be able to find the values of \[\alpha \] and \[\beta \].

Complete step-by-step answer:
Given that,
\[\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta \]
We know that,
\[ - \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}x \leqslant \dfrac{\pi }{2}\]
Adding \[\dfrac{\pi }{2}\] on both sides we will get
\[\begin{gathered}
   \Rightarrow - \dfrac{\pi }{2} + \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}x + \dfrac{\pi }{2} \leqslant \dfrac{\pi }{2} + \dfrac{\pi }{2} \\
   \Rightarrow 0 \leqslant {\sin ^{ - 1}}x + \dfrac{\pi }{2} \leqslant \pi \\
\end{gathered} \]
But we know that \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
So we can write above identity as

Hence,
\[\begin{gathered}
   \Rightarrow 0 \leqslant {\sin ^{ - 1}}x + {\sin ^{ - 1}}x + {\cos ^{ - 1}}x + \leqslant \pi \\
   \Rightarrow 0 \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \pi \\
\end{gathered} \]
Comparing this equation with the given equation \[\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta \] we get the values of\[\alpha \] and \[\beta \] as
\[\alpha = 0,\beta = \pi \]
Hence option C is the correct option.

Note: Here we are given with 2 \[{\sin ^{ - 1}}\]functions in the question. So this is the key to proceed with the range of \[{\sin ^{ - 1}}\] function and not\[{\cos ^{ - 1}}\].
For the identity \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
Let
\[
  {\sin ^{ - 1}}x = \theta \\
  x = \sin \theta \\
  x = \cos \left( {\dfrac{\pi }{2} - \theta } \right) \\
  {\cos ^{ - 1}}x = \dfrac{\pi }{2} - \theta \\
  {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x \\
  {\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \dfrac{\pi }{2} \\
\]