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{\sin ^{ - 1}}\left( {\sin 5} \right) > {x^2} - 4x  holds if

Option: 1

x = 2 - \sqrt {9 - 2\pi }


Option: 2

x = 2 + \sqrt {9 - 2\pi }


Option: 3

x > 2 + \sqrt {9 - 2\pi }


Option: 4

x \in \left( {2 - \sqrt {9 - 2\pi } ,\,\,2 + \sqrt {9 - 2\pi } } \right)


Answers (1)

best_answer

As we learnt

 

Important Results of Inverse Trigonometric Functions -

 

\sin ^{-1}\left ( \sin \Theta \right )= \Theta

- wherein

if -\frac{\pi }{2}\leqslant \Theta \leqslant \frac{\pi }{2}

 

 

1 rad = 57.75° (approx.)

                                    5 rad = 288.75° (approx.)

                                    \frac{{3\pi }}{2} < 5 < \frac{{5\pi }}{2}$      {\sin ^{ - 1}}\left( {\sin 5} \right) = 5 - 2\pi $

                              Given {\sin ^{ - 1}}\left( {\sin 5} \right) > {x^2} - 4x\,\,\,\,\, \Rightarrow \,\,\,\,{x^2} - 4x < 5 - 2\pi     \Rightarrow {x^2} - 4x + \left( {2\pi - 5} \right) < 0           

                              Roots of {x^2} - 4x + 2\Pi - 5 = 0$ are 2 \pm \sqrt {9 - 2\pi }

                                    {x^2} - 4x + 2\pi - 5 < 0$

 

Posted by

Ritika Harsh

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