Let $T$ be left shift, say on $\ell^2(\mathbb{Z}_{\ge 0})$. This is Fredholm with $1$ dimensional kernel and $0$ dimensional cokernel, so it has index $1$. Therefore, $t T + (1-t) I$ cannot be Fredholm for all $t \in [0,1]$. So for which values of $t$ does this fail to be Fredholm?
Doing this by hand looks time-consuming, so is there a simple way of figuring this out that uses more machinery?
For $t \ne 0$, $$ tT+(1-t)I = t\left\{T+\left(\frac{1}{t}-1\right)I\right\} $$ The spectrum of $T$ is the closed unit disk. Therefore the term in braces on the right is invertible whenever $1/t-1 > 1$ or $1-t > t$ or $1 > 2t$ or $t < 1/2$. This gives an index of $0$ for $tT+(1-t)I$ whenever $0 \le t < 1/2$. And $tT+(1-t)I$ is definitely not invertible for $t > 1/2$, but it is Fredholm. At $t=1/2$, you're looking at the operator $\frac{1}{2}(T+I)$, which is not Fredholm because it doesn't have a closed range.