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 ) $
Show Answer
Answer:
Correct Answer: A
$ \because \frac{\pi }{2}<4<\frac{3\pi }{2}, $ so $ {{\sin }^{-1}}\sin 4 $
$ ={{\sin }^{-1}}\sin (\pi -4)=4 $
The inequality becomes $ x^{2}-kx+\pi -4>0 $
The discriminant $ D=k^{2}-4(\pi -4) $ is not greater than 0 for all k. That is, $ x^{2}-kx+(\pi -4) $ cannot hold for all x.
BETA
