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Question: If \[\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta \] then A.\[\alpha = \dfr...

If α2sin1x+cos1xβ\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta then
A.α=π2,β=π2\alpha = \dfrac{\pi }{2},\beta = \dfrac{\pi }{2}
B.α=π2,β=3π2\alpha = \dfrac{\pi }{2},\beta = \dfrac{{3\pi }}{2}
C.α=0,β=π\alpha = 0,\beta = \pi
D.α=0,β=2π\alpha = 0,\beta = 2\pi

Explanation

Solution

First we will take sin1{\sin ^{ - 1}} function with its range. Then we will use the identitysin1x+cos1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}. With the help of this we will be able to find the values of α\alpha and β\beta .

Complete step-by-step answer:
Given that,
α2sin1x+cos1xβ\alpha \leqslant 2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x \leqslant \beta
We know that,
π2sin1xπ2- \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}x \leqslant \dfrac{\pi }{2}
Adding π2\dfrac{\pi }{2} on both sides we will get

\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} \\