Question

The least value of n for which $(n-2{)}^2+8x+n+4>si{n}^{-1}(sin12)+co{s}^{-1}(cos12)\forall xϵR$ and $nϵN$ is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The correct option is B

5

$si{n}^{-1}\left(sin12\right)+co{s}^{-1}\left(cos12\right)$
$=si{n}^{-1}\left(sin\right(4\pi -(4\pi -12)\left)\right)+co{s}^{-1}\left(cos\right(4\pi -(4\pi -12)\left)\right)$
$=-(4\pi -12)+(4\pi -12)=0$
So that $(n-2{)}^2+8x+n+4>0\forall xϵR$
$\Rightarrow n-2>0n\Rightarrow n\ge 3$
And $8^2-4(n-2)(n+4)<0$
Or $n^2+2n-24>0$
$n^2+6n-4n-24>0$
$(n+6)-4(n+6)>0$
$(n+6)(n-4)>0\Rightarrow n<-6$ or $n>4\Rightarrow n=5$ as $nϵN$