Question:-
The given figure shown a cantilever of span ‘L’ subjected to a concentrated load ‘P’ and a moment ‘M’ at the free end. Deflection at the free end is given by

Option (A)
(PL2/2EI)+(ML2/3EI)
Option (B)
(ML2/2EI)+(PL3/3EI)
Option(C)
(ML2/3EI)+(PL3/2EI)
Option(D)
(ML2/2EI)+(PL3/48EI)
Correct Option:
(B)
Question Solution:
Deflection can be found by using super imposition method. Deflection due to point load (cantilever) d1 = (P|3/3E|) Deflection due to moment at end in cantilever d2 = (M|2/2E|).
Summation of d1 and d2 will give the total deflection of the cantilever. Macaulay method can also be used which is as follow

According to Macaulay at any section x, the bending moment
Mx = E|(d2y/dx2) = - px – Mx0
On integration
E| (dy/dx) = - (Px2/2) – Mx + C1
On integration again
Ely = - (Px3/6) –(Mx2/2) + C1x + C2 at x = L; y =0
(-PL3/6) – (ML2/2) + C1L + C2 = 0
Also at x = L; (dy/dx) = 0
\ C1 = (PL2/2) + ML and C2 = - (PL3/3) –(ML2/2)
\ El.y = -(Px3/6) – (Mx2/2) + {(PL2/2)+ML}x – (PL3/3) – (ML3/2)
At x = 0
El.y = -(PL3/3) – (ML2/2)
\ y =- {(PL3/3El)+(ML2/2El)}