A) \[x<2-\sqrt{9-2\pi }\]
B) \[-1<x<5\]
C) \[x\in \left( -\infty ,-1 \right)\cup \left( 5,\infty \right)\]
D) \[x\in \left( 2-\sqrt{9-2\pi },2+\sqrt{9-2\pi } \right)\]
Correct Answer: D
Solution :
| \[\therefore \]\[\frac{3\pi }{2}<5<\frac{5\pi }{2}\therefore {{\sin }^{-1}}\left( \sin 5 \right)=5-2\pi \] |
| 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 \[2\pm \sqrt{9-2\pi }\] |
| \[\therefore \]\[{{x}^{2}}-4x+2\pi -5<0\] \[\Rightarrow \]\[2-\sqrt{9-2\pi }<x<2+\sqrt{9-2x}\] |
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