Question

The least integral value of $k$ for which $(k-2)x^2+8x+k+4>{\sin }^{-1}\left(\sin 12\right)+{\cos }^{-1}\left(\cos 12\right)$ for all $x\in R$, is

Answer 1

$(k-2)x^2+8x+k+4>{\sin }^{-1}\left(\sin 12\right)+{\cos }^{-1}\left(\cos 12\right)$

Now,
${\sin }^{-1}\left(\sin 12\right)={\sin }^{-1}\left(\sin (4\pi +(12-4\pi ))\right)={\sin }^{-1}\left(\sin (12-4\pi )\right)=12-4\pi$
Similarly,
${\cos }^{-1}\left(\cos 12\right)={\cos }^{-1}\left(\cos (4\pi -(4\pi -12))\right)={\cos }^{-1}\left(\cos (4\pi -12)\right)=4\pi -12$
So,
$(k-2)x^2+8x+k+4>0$
This is possible when
$D<0$ and $k-2>0$
$\Rightarrow 64-4(k-2)(k+4)<0\Rightarrow k^2+2k-24>0\Rightarrow (k+6)(k-4)>0\Rightarrow k\in (-\infty ,-6)\cup (4,\infty )$
And
$k>2k\in (4,\infty )$
$\therefore$ The least integral value of $k$ is $5$