Explanation:

Regular Grammar:
A regular grammar is a formal grammar that generates regular languages. A regular language is a language that can be expressed using a regular expression or a finite automaton.

Given options:
a) L is a set of numbers divisible by 2
- This can be expressed using a regular grammar as it is a regular language.

b) L is a set of binary complement
- This can be expressed using a regular grammar as it is a regular language.

c) L is a set of string with odd number of 0
- This can be expressed using a regular grammar as it is a regular language.

d) L is a set of 0^n1^n
- This language cannot be expressed using a regular grammar.

Explanation for option d:
The language L is a set of strings of the form 0^n1^n, where n is any positive integer. This language is not a regular language, thus it cannot be expressed using a regular grammar. This can be proved using the pumping lemma for regular languages.

Pumping lemma:
The pumping lemma for regular languages states that if a language L is regular, then there exists a constant n such that any string w in L with length greater than or equal to n can be divided into three parts, w = xyz, such that:
- |xy| ≤ n
- |y| ≥ 1
- for all i ≥ 0, xy^iz ∈ L

Using the pumping lemma, we can prove that the language L = {0^n1^n | n ≥ 1} is not a regular language.

Proof:
Assume that L is a regular language. Then, there exists a constant n such that the pumping lemma holds for all strings in L with length greater than or equal to n.

Let w = 0^n1^n be a string in L with length greater than or equal to n. By the pumping lemma, we can divide w into three parts, w = xyz, such that:
- |xy| ≤ n
- |y| ≥ 1
- for all i ≥ 0, xy^iz ∈ L

Since |xy| ≤ n, we know that y contains only 0's. Let y = 0^k, where 1 ≤ k ≤ n. Then, x = 0^(n-k) and z = 1^n.

Now, consider the string xy^2z. This string is of the form 0^(n+k)1^n, which is not in L. Thus, the pumping lemma does not hold for the string w = 0^n1^n, which contradicts our assumption that L is a regular language.

Therefore, the language L = {0^n1^n | n ≥ 1} is not a regular language and cannot be expressed using a regular grammar.