Question

If $(n-2)x^2+8x+(n+4)<0$, $\forall x\in R$, find the range of $n.$

• A. $(4,\infty )$
• B. $(2,4)$
• C. $(-\infty ,-6)$
• D. $(-\infty ,2)$

Answer 1

The correct option is C $(-\infty ,-6)$

Given: $(n-2)x^2+8x+(n+4)<0$, $\forall x\in R$
To find: Range of n
Step -1: Draw the graph for $y=(n-2)x^2+8x+(n+4)$.
Step-2: Write the applicable conditions.
Step-3: Solve all the conditions and combine all the results for n


Applicable conditions:
i) $a<0$
ii) $D<0$

Condition: i) $a<0$
$\Rightarrow (n-2)<0\Rightarrow n<2\Rightarrow n\in (-\infty ,2)=A$ (Say)

Condition: ii) $D<0\Rightarrow b^2-4ac<0$
$8^2-4(n-2)(n+4)<0$
$\Rightarrow 64-4(n^2+2n-8)<0$
$\Rightarrow 64-4n^2-8n+32<0$
$\Rightarrow -4n^2-8n+96<0$
$\Rightarrow n^2+2n-24>0$
$\Rightarrow (n+6)(n-4)>0$
$\Rightarrow n\in (-\infty ,-6)\cup (4,\infty )=B$
$\therefore$ Range of $n$ is, $A\cap B=(-\infty ,2)\cap ((-\infty ,-6)\cup (4,\infty ))$
or,$A\cap B=(-\infty ,-6)$
Hence, Range of $n$ is $(-\infty ,-6)$