Question

If the inequality $(m-2)x^2+8x+m+4>0$ is satisfied for all $x\in R$, then the least integral $m$ is

• A. $4$
• B. $5$
• C. $7$
• D. $2$

Answer 1

The correct option is B $5$

$(m-2)x^2+8x+m+4>0\forall x\in R$
$\Rightarrow (m-2)>0,D<0$
$\Rightarrow m>2,64-4(m-2)(m+4)<0$

$16-m^2-2m+8<0$
$\Rightarrow m^2+2m-24>0$
$\Rightarrow (m+6)(m-4)>0$
$\Rightarrow m\in (-\infty ,-6)\cup (4,\infty )$
Since $m>2,$ $\therefore m\in (4,\infty )$

Hence, least integral value of $m$ is $5.$