Halting problem of TM which recognize recursive languages is undecidable?

I am preparing for an exam and I came across this question in one of the tests.

Halting problem of Turing machines which recognize recursive languages is undecidable. (True / False)

The solution given says the statement its False.

But I think the statement is True. Halting problem in itself is undecidable. We can't determine whether the Turing Machine will halt or not.

The test series has a history of wrong answers that's why I am seeking an explanation even if I am wrong.

Edit: I was looking at the answers and I have trouble understanding the explanations. I know that 'By definition, recursive languages are decideable by Turing machines.' I will tell you how I breakdown the question. I am given a TM and told that this TM recognizes a recursive language. How do I know that the TM will halt?

• Seems question is not well formulated by the exam. – Mr. Sigma. Dec 10 '18 at 15:39
• I agree with you. – imagine5am Dec 10 '18 at 15:42
• "I am given a TM and told that this TM recognizes a recursive language." This is redundant. TMs recognize recursive languages by definition. – ubadub Dec 10 '18 at 16:36
• But the TM may fall in infinite loop. Can't it? – imagine5am Dec 10 '18 at 16:42
• @imagine5am no, the whole point of a recursive language- it's very definition- is that the TM won't. Recursive languages are definitionally decidable. – ubadub Dec 10 '18 at 16:57

If the task is to determine whether a given machine $$M$$ halts, when you already know that $$M$$s language is decidable then the answer is always "yes" because every TM that decides a language halts on every input (by definition of "to decide"). Hence, the halting problem on these kind of TMs is decidable.
However, if you want to determine whether the language $$A = \{\langle M \rangle \mid L(M) \text{ is decidable}\}$$ you can find an easy reduction that shows that $$A$$ is undecidable. I do not think this is the task but it was my first thought reading this question, so maybe it is still useful.