Find the values of $k$ , for which the system is stable.

Question

I have a multiple choice question , that says , if $y(t) = (k^2 -3k -4)\log(x) + \sin(x)$ , find the values for which the system is stable ( I guess it means BIBO stable) .

$$a. 1 \ \ \ \ b. 3 \ \ \ \ c.-1 \ \ \ \ d. 5 \quad e. 4 \quad f. 2$$

We are allowed to pick as many choices as we want. I circled every single choice, am I missing something?

Let's suppose $ \exists M \in R $ such that: $|x(t)| < M \quad\forall t \in \mathbb{R}$
$ |y(t)| = |(k-4)(k+1)\log(x) + \sin(x)| \leq |(k-4)(k+1)\log(x)| + 1 $
since $\log(x)$ is defined, then $x(t)>0 \forall t \in \mathbb{R} \rightarrow |x(t)| = x(t) < M$
and since $\log_{10}(x) $ is an increasing function $\rightarrow x(t) < M \rightarrow \log(x(t)) < \log M$
$ |y(t)| \leq |(k-4)(k+1)\log(M)| + 1 $

For every given value of $k$, $|(k-4)(k+1)\log(M)| \in [0,+\infty]$ so I can't see why not , every value is possible

Answer 1

Hint: Show that if $ k^2 - 3k - 4 \neq 0$, then the system is unstable by considering values around $x=0$.

Then do the case of $k^2 - 3k - 4 = 0 $.

Answer 2

Okay after some rethinkig i ended ip with the conclusion:
If $(k^2 -3k 4) = (k-4)(k+1) = 0 \rightarrow k \in {-1,4}$ then $|y|=|sinx|<2$ it is BIBO stable
but if $c=(k^2 -3k 4) \neq 0$ then $|y| = |c||logx(t)| + |sinx|<=|c||logx(t)| +1$
Suppose $x(t) = \frac{1}{10^|t|}<=1 \forall t $ but then ...
$lim_{|y| \rightarrow \infty} = \infty$ it is not BIBO -BOUNDED