Inverse Trigonometric Functions Question 118
Question: The set of values of k for which $ x^{2}-kx+{{\sin }^{-1}}(sin4)>0 $ for all real x is
Options:
A) $ \phi $
B) $ (-2,2) $
C) $ R $
D) $ (-\infty ,-2)\cup (2,\infty ) $
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Answer:
Correct Answer: A
$ \because \frac{\pi }{2}<4<\frac{3\pi }{2}, $ so $ {{\sin }^{-1}}\sin 4 $
$ ={{\sin }^{-1}}\sin (\pi -4)=\pi -4 $
The inequality becomes $ x^{2}-kx+\pi -4>0 $
The discriminant $ D=k^{2}-4(\pi -4)>0 $ for all k, That is $ x^{2}-kx+(\pi -4)>0 $ cannot hold for all x.
BETA
