Effective Bertini Let $X$ be a smooth complex projective manifold, and $L$ an ample line bundle. By Bertini's Theorem, for every integer $q$ big enough there exists an open dense subset $U_q\subset |qL|$ such that every divisor $D$ in $U_q$ is smooth.
Warm up question: is the complement of $U_q$ always a divisor?
We can define a bigger open subset $V_q\subset |qL|$ as
$$
V_q:=\{D\in |qL| \; \textrm{s. t.} \; (X,\frac{1}{q}D) \; \textrm{ is klt} \}
$$
My question is: what is the dimension of the complement of $V_q$ ? (or at least can we bound its asymptotic in $q$ ? e.g. is it upper-bounded by $aq^{\dim X}$ with $a$ a constant which is strictly smaller than the volume of $L$? )
 A: Regarding the warm up question: No (although for $q$ sufficiently large the answer is yes, as Jason Starr comments.) Let $X = \mathbb{P}^1 \times \mathbb{P}^2$ and let $L= \mathcal{O}(1,1)$. Write homogeneous coordinates on the first factor as $(u:v)$ and on the second factor as $(x:y:z)$. A divisor $D$ in $H^0(L)$ is of the form 
$$u (ax+by+cz) + v (dx+ey+fz)=0.$$
$D$ is singular if and only if the matrix
$$\begin{bmatrix}
a & b& c \\ d & e & f \\ \end{bmatrix}$$
has rank $1$. (If this matrix has rank $2$, $D$ is isomorphic to $\mathbb{P}^2$ blown up at a point, when the matrix drops rank, $D$ turns into the union of a $\mathbb{P}^2$ and a $\mathbb{P}^1 \times \mathbb{P}^1$.) The condition that this matrix drops rank is codimension $2$.
The complement of $U_q$ is called the dual variety to $X$. It is usually (no precise meaning attached) true that the dual of a smooth variety is a divisor. For a while, it was an open problem to characterize smooth projective toric varieties whose duals were not divisors; this was solved by Dickenstein, Feichtner and Sturmfels.
