Question

The least integral value of $k$ for which $(k-2)x^2+8x+k+4>0$ for all $x\in R,$ is

• A. $5$
• B. $4$
• C. $3$
• D. none of these

Answer 1

The correct option is A $5$

For $(k-2)x^2+8x+k+4>0$, we need to check the following two cases:
Case 1) $k-2>0$ (concave up curve)
$\Rightarrow k>2$ .....(1)
Case 2) $b^2-4ac<0$ (no intersection on X-axis)
$\Rightarrow 64-4(k-2)(k+4)<0$
$\Rightarrow (k-2)(k+4)-16>0$
$k^2+2k-24>0$
$(k+6)(k-4)>0$
$\Rightarrow k\in (-\infty ,-6)\cup (4,\infty )$ ....(2)
From (1) and (2), it follows $k>4$
Therefore, least integral value is $5$.