Question

The number of integral values of $a$ for which $4^t-(a-4)2^t+\frac{9a}{4}<0,\forall t\in (1,2)$ is

Answer 1

$4^t-(a-4)2^t+\frac{9a}{4}<0$

Let $2^t=x$
$f\left(x\right)=x^2-(a-4)x+\frac{9a}{4}$
We want $f\left(x\right)<0,\forall x\in (2^1,2^2)$ i.e., $\forall x\in (2,4)$
So the equation must have two distinct roots one less than $2$ and other greater than $4.$


$f\left(2\right)<0\Rightarrow 4-2(a-4)+\frac{9a}{4}<0$
$\Rightarrow a<-48\cdots \left(1\right)$

$f\left(4\right)<0\Rightarrow 16-4(a-4)+\frac{9a}{4}<0\Rightarrow a>\frac{128}{7}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ $a\in \varphi$
Hence, no such integral value of $a$ exists.