Question

Find the least value of $n$ such that $(n-2)x^2+8x+n+4>0,\forall x\in R$, where $n\in N$.

• A. $3$
• B. $5$
• C. $8$
• D. $4$

Answer 1

Answer: (B)

$5$

$(n-2)x^2+8x+n+4>0;\forall x\in R$
$64-4(n-2)\Rightarrow (n+4)<0$ and $n-2>0$
$\Rightarrow 16-(n^2+2n-8)<0$ and $n>2$
$\Rightarrow n^2+2n-24>0$ and $n>2$
$\Rightarrow (n+6)(n-4)>0$ and $n>2$
$\Rightarrow n>4$ as $n\in N$ and $n>2$
$\Rightarrow n\ge 5$
Hence, the least value of $n$ is $5$.