The process state transition diagram of an operating system is as given below.…

Answer: The statement "It uses preemptive scheduling" is FALSE.

Key observation: The diagram does not show any transition from the Running state back to the Ready state caused by the scheduler (for example, a time-slice expiry or preemption).

• What the diagram shows: The only outgoing transitions from Running are to Blocked (waiting for I/O or resource) and to Terminated (Exit). There is a Ready -> Running transition labeled "Schedule."

• Why preemptive scheduling is ruled out: Preemptive scheduling requires an explicit Running -> Ready transition triggered by the scheduler (e.g., time-slice expiration). Since no such transition is shown, preemption is not indicated.

• Why non-preemptive scheduling is supported: Processes leave the CPU only by blocking for I/O or by exiting, which is the behavior of non-preemptive (cooperative) scheduling.

• Multiprogramming: The existence of Ready and Blocked states implies the OS can manage multiple processes concurrently (keeps processes in memory while others run), so the diagram is consistent with a multiprogrammed OS.

• Multi-user: The diagram does not contain information about the number of users. State-transition structure is compatible with both single-user and multi-user systems, so this cannot be inferred from the diagram alone.

Conclusion: Because the diagram lacks a scheduler-driven Running -> Ready transition, the claim that the system uses preemptive scheduling must be FALSE; the other statements are either supported by the diagram (multiprogrammed, non-preemptive) or are not contradicted by it (multi-user).