Question

The set of values of p for which $x^2-px+si{n}^{-1}\left(sin4\right)>0$ for all real x is given by :

• A. $(–4,4)$
• B. $(-\infty ,-4)\cup (4,\infty )$
• C. $\varphi$
• D. None of these

Answer 1

The correct option is C $\varphi$

$x^2-px+si{n}^{-1}\left(sin4\right)>0$ for all real x.
$\Rightarrow x^2-px+si{n}^{-1}\left(sin4\right)>0$
$\Rightarrow x^2-px+(\pi -4)>0\forall xϵR$
$\Rightarrow D=p^2-4(\pi -4)<0$ $\Rightarrow p^2+16-4\pi <0$
Since $16-4\pi >0,p^2+16-4\pi$ cannot be negative for any value of $pϵR$.
$\therefore$ Set of values of $p=\varphi$