Let $\mathrm{S}$ be the set of positive integral values of $a$ for which $\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32} < 0, \forall x \in \mathbb{R}$. Then, the number of elements in $\mathrm{S}$ is :
Solution
$x^2-8 x+32>0 \forall x \in R$ as discriminant of this quadratic is $64-4 \times 32<0$
<br/><br/>$\Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R$
<br/><br/>$\Rightarrow$ Only possible when $a<0$ and $D<0$
<br/><br/>$\Rightarrow$ Since $S$ is set of positive
values of $a \Rightarrow S$ is a null set
<br/><br/>$\Rightarrow n(S)=0$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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