Let $\det(A)\ne0$ and $\det(A+E)\ne0$. Can we say that, no eigenvalue of pencil $A+tE$ can be equal $1$? Let $A,E\in M_n$ and $t\in[0,1]$ 
and $\det(A)\ne0$ and $\det(A+E)\ne0$.
Can we say that, no eigenvalue of pencil $A+tE$ can be equal $1$?(for all $t\in[0,1]$)
 A: The answer is yes. As the comments and the other answer here show, there is an unfortunate misunderstanding here.
The eigenvalues of a matrix pencil $(A,B)$ (or $A+tB$ if one is sloppy) actually refers not to the eigenvalues of the matrix $A+tB$, but the set of all complex numbers $t$ such that $\det(A+tB)=0$ (or, in an alternative definition, those $t$s such that $\det(A-tB)=0$). Confusing or misleading as it is, the term is unfortunately a standard one in linear algebra literature.
In the OP's case, since the given conditions state that both $\det(A+0E)$ and $\det(A+1E)$ are nonzero, the numbers $0$ and $1$ are, by definition, not the eigenvalues of the matrix pencil $(A,E)$.
A: Without further conditions, I do not think you can say anything along these lines, even if you restrict to the open interval $(0,1)$. It is easy to imagine that $0$ and $1$ might somehow be "edge" cases, but they aren't. Consider $A = \frac{5}{4}I$ and $E=-\frac{1}{2}I$, then for all $t$, $A + tE = (\frac{5}{4}-\frac{t}{2})I$. Picking $t = \frac{1}{2}$, we get $A+tE = I$ which obviously has an eigenvalue of $1$.
