Question

The set of value(s) of $k$ for which $x^2-kx+{\sin }^{-1}\left(\sin 4\right)>0$ for all real $x$ is

• A. $\varphi$
• B. $(-2,2)$
• C. $R$
• D. $(-1,3)$

Answer 1

The correct option is A $\varphi$

Given, $x^2-kx+{\sin }^{-1}\left(\sin 4\right)>0$
Now,
${\sin }^{-1}\left(\sin 4\right)={\sin }^{-1}\left(\sin (\pi +(4-\pi ))\right)=\pi -4$

So, the given inequality becomes
$x^2-kx+\pi -4>0\Rightarrow D<0\Rightarrow k^2-4(\pi -4)<0\Rightarrow k^2+4(4-\pi )<0$
As $k^2+4(4-\pi )>0$, so above inequality is not valid.
Hence, no value of $k$ is possible.