Question

The least integral value of $a$ such that $(a-2)x^2+8x+a+4>0,\forall x\in R$ is

• A. $3$
• B. $5$
• C. $4$
• D. $6$

Answer 1

Answer: (B)

$5$

Conditions are $a-2>0,D<0$
$a-2>0\Rightarrow a>\mathrm{2.....}\left(i\right)$
$D<0\Rightarrow 8^2-4(a-2)(a+4)<0$
$\Rightarrow 16-(a^2+2a-8)<0$
$a^2+2a-24>0$
$a^2+6a-4a-24>0$
$(a+6)(a-4)>0$
$\Rightarrow a\in (-\infty ,-6)\cup (4,\infty ).....\left(ii\right)$
Intersection of (i) and (ii) gives $a\in (4,\infty )$
$\Rightarrow$ least integer value of $a=5$