A) \[\phi \]
B) \[(-2,2)\]
C) \[R\]
D) \[(-\infty ,-2)\cup (2,\infty )\]
Correct Answer: A
Solution :
[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.
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