JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 3)
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 :
0
$$\infty$$
3
1
Explanation
$x^2-8 x+32>0 \forall x \in R$ as discriminant of this quadratic is $64-4 \times 32<0$
$$ \Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R $$
$\Rightarrow$ Only possible when $a<0$ and $D<0$
$\Rightarrow$ Since $S$ is set of positive values of $a \Rightarrow S$ is a null set
$$ \Rightarrow n(S)=0 $$
$$ \Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R $$
$\Rightarrow$ Only possible when $a<0$ and $D<0$
$\Rightarrow$ Since $S$ is set of positive values of $a \Rightarrow S$ is a null set
$$ \Rightarrow n(S)=0 $$
Comments (0)
