Step 1: Find the support reactions.
The beam PR rests on a pin at P and a roller at R. It carries only a concentrated moment \(M\) at the midspan point Q.
Since there is no transverse load, vertical equilibrium gives \(R_P + R_R = 0\). So the two reactions are equal and opposite, forming a force couple.
Taking moments about P, this couple must balance the applied moment: \(R_R \cdot L = M\). So \(R_R = M/L\) and \(R_P = -M/L\).
Step 2: Write the bending moment on each side of Q.
Cut the beam at a distance \(x\) from P. Summing moments of the forces to the left of the cut gives the bending moment function.
For \(0 \le x < L/2\), only \(R_P\) acts to the left, so \(M(x) = R_P x\). Just before Q this gives \(M(L/2^-) = (-M/L)(L/2) = -M/2\).
For \(L/2 < x \le L\), the applied moment \(M\) also lies to the left of the cut. So \(M(x) = R_P x + M\).
Just after Q this gives \(M(L/2^+) = -M/2 + M = M/2\). At \(x = L\) it gives \(M(L) = -M + M = 0\), so the moment returns to zero at the roller.
Step 3: Match this to the diagrams.
So the diagram is a straight line from 0 at P, rising to magnitude \(M/2\) just before Q. It then jumps instantly by \(M\) to the opposite sign just after Q, then runs straight back to 0 at R.
Such a jump, equal to the full applied moment at its point of application, is the signature of a concentrated moment load. It splits into \(M/2\) and \(M/2\) here only because Q sits exactly at midspan.
Options B and D show a smooth curve reaching a single value of \(M\) at Q with no jump, which cannot come from a concentrated moment. Option C wrongly shows a nonzero moment at the pin and roller, but a simple support cannot carry any moment.
Final Answer:
Only option A shows zero moment at both supports with the correct \(M/2\) jump at midspan.
\[ \boxed{\text{Option A}} \]
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