Let S be the set of positive integral values of a for which a x 2 + 2 ( a + 1 ) x + 9 a + 4 x 2 - 8 x + 32 < 0 , ? ? ? x ? R . Then, the number of elements is S is [2024]

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Solution

Q.
Let $S$ be the set of positive integral values of $a$ for which $\frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32} < 0, \forall x \in R .$ Then, the number of elements is $S$ is [2024]

1 $\infty$

2 $0$

3 $1$

4 $3$

Ans.
(2)

Given $\frac{a x^{2} + 2 (a + 1) x + 9 a + 4}{x^{2} - 8 x + 32} < 0, \forall x \in R$

For $x^{2} - 8 x + 32,$

$D = b^{2} - 4 a c, D = {(- 8)}^{2} - 4 (1) (32), D = 64 - 128,$

$D = - 64 < 0, a > 0$

So, $x^{2} - 8 x + 32 > 0,$ when

$∴$ $a x^{2} + 2 (a + 1) x + 9 a + 4 < 0, \forall x \in R$

$a < 0 and D < 0$

Here, $S$ is the set of positive integral values of $x,$ which is not possible.

So, the number of elements in $S = 0$

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