Let S be the set of positive integral values of a for which . Then, the number of elements in S is:

Engineering

Mathematics

Inequalities

Question

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

∞

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

ax² + 2(a + 1) x + 9a + 4 < 0 as x² – 8x + 32 is always positive

case – I when a ≠ 0 and a < 0

then D < 0

4(a + 1)² – 4a(9a + 4) < 0

a² + 2a + 1 – 9a² – 4a < 0

0 < 8a² + 2a – 1

$a \in (- \infty, - \frac{1}{2}) \cup (\frac{1}{4}, \infty)$

$So a \in (- \infty, - \frac{1}{2})$

case – II when a = 0

then 2x + 4 < 0 is not always true

So, a ≠ 0

So number of positive integral values of a = 0