Question

The minimum integral value of $x$ for which $2x^2+2x+n>9+{\sin }^{-1}\left(\sin \right(-1\left)\right)+{\cos }^{-1}\left(\cos \right(-1\left)\right)\forall x\in R,$ is

Answer 1

Given : $2x^2+2x+n>9+{\sin }^{-1}\left(\sin \right(-1\left)\right)+{\cos }^{-1}\left(\cos \right(-1\left)\right)$
We know,
${\sin }^{-1}\left(\sin x\right)=x,-\frac{\pi }{2}\le x\le \frac{\pi }{2}$,
${\cos }^{-1}\left(\cos x\right)=-x,-\pi \le x\le 0$
So inequality becomes :
$2x^2+2x+n>9-1-(-1)$
$\Rightarrow 2x^2+2x+n-9>0$ is true $\forall x\in R$
Hence, $D<0$ as coefficient of $x^2>0$
$\Rightarrow 4-8(n-9)<0$
$\Rightarrow n>\frac{19}{2}$
So, minimum integral value of $n=10$