JEE Main 2024 — Quadratic Equation Question with Solution

Given:

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

For quadratic $x^2-8x+32=0, D_1={\left(-8\right)}^2-4\left(32\right)=-64$

Since the discriminant is less than zero and the leading coefficient is positive, this quadratic will always be positive.

Now, solving $a{x}^2+2(a+1)x+9a+4<0$

We know that, for a quadratic to be always negative, the coefficient of $x^2<0, D<0$.

$\Rightarrow a<0$ 

But we want positive values.

So, no positive integral value exist.