The set of values of $\lambda$ for which $x^2-\lambda x+{\sin }^{-1}(\sin 4)>0$ for all $x\in R$, is

Solution:

We have,

$\begin{array}{l}{\sin }^{-1}(\sin 4)={\sin }^{-1}(\sin (\pi -4\left)\right)=\pi -4.\\ \therefore x^2-\lambda x+{\sin }^{-1}(\sin 4)>0\text{ for all }x\in R\\ \Rightarrow x^2-\lambda x+(\pi -4)>0\text{ for all }x\in R\\ \Rightarrow {\lambda }^2-4(\pi -4)<0\Rightarrow {\lambda }^2+16-4\pi <0\end{array}$

But, ${\lambda }^2+16-4\pi >0$ for all $\lambda \in R$.

So, there is no value of $\lambda$ for which the given in equation holds true.