Self-induction

Voltage across an inductor: $V = L\frac{dI}{dt}$ (the induced emf is $-L\frac{dI}{dt}$)
From 0 to $\frac{T}{2}$ the current rises linearly, so $\frac{dI}{dt}$ is a positive constant and V is constant and positive.
From $\frac{T}{2}$ to T the current falls linearly at the same rate, so V is constant and negative with the same magnitude.
So the V–t graph is a rectangular (square) wave: a positive step up to $\frac{T}{2}$, then an equal negative step up to T.

Induced emf in the coil: $e = -L\frac{di}{dt}$, so the emf is proportional to the negative of the slope of the i–t graph.
From 0 to T/4 the current increases linearly (constant positive slope), so the emf is constant and negative.
From T/4 to T/2 the current is constant (zero slope), so the emf is zero.
From T/2 to 3T/4 the current decreases linearly (constant negative slope), so the emf is constant and positive.
From 3T/4 to T the current is again constant, so the emf is zero.
The correct graph is the one showing a constant negative emf from 0 to T/4, zero from T/4 to T/2, a constant positive emf from T/2 to 3T/4 and zero afterwards.