Question

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

• A. $5$
• B. $6$
• C. $7$
• D. $4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $5$
B $6$
C $7$

We know for $a{x}^2+bx+c>0,\forall x\in R$
We must have $a>0$ and $b^2-4ac<0$
Therefore,
$k-2>0$ $\Rightarrow k>2$ ...(1)

$8^2-4(k-2)(k+4)<0$

$\Rightarrow 16-(k^2+2k-8)<0$
$\Rightarrow k^2+2k-24>0$

$\Rightarrow (k+6)(k-4)>0$
$\Rightarrow k\in (-\infty ,-6)\cup (4,\infty )$ ...(2)

Using (1) and (2), we get $k>4$
Hence, integral values of $k$ are $5,6,7,....$