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 $xϵR$, is

• A. $-7$
• B. $-5$
• C. $-3$
• D. $5$

Answer 1

The correct option is D $5$

$(k-2)x^2+8x+k+4>{\sin }^{-1}\left(\sin 12\right)+{\cos }^{-1}\left(\cos 12\right)$
Now,
${\sin }^{-1}\left(\sin 12\right)={\sin }^{-1}\left(\sin (4\pi +(12-4\pi \right)\left)\right)={\sin }^{-1}\left(\sin (12-4\pi \right))=12-4\pi$
${\cos }^{-1}\left(\cos 12\right)={\cos }^{-1}(\cos (4\pi -(4\pi -12)\left)\right)={\cos }^{-1}\left(\cos (4\pi -12\right))=4\pi -12$
$\Rightarrow (k-2)x^2+8x+k+4>0$
$\Rightarrow D<0$
$\Rightarrow 64-4(k-2)(k+4)<0$
$\Rightarrow k^2+2k-24>0$
$\Rightarrow (k+6)(k-4)>0$
$\Rightarrow k\in (-\infty ,-6)\cup (4,\infty )$

And k>0
So, option D lies in this range
and $(5{)}^2+10-24=11>0$