The correct answer is $\boxed{{t\left( {t – 1} \right)} \over 2}u\left( {t – 1} \right)$.
The output of a system is the convolution of the input and the impulse response. The impulse response is a function that describes how the system responds to a single impulse. The input is the function that is applied to the system. The convolution is a mathematical operation that combines two functions.
In this case, the impulse response is $h(t) = tu(t)$. The input is $u(t – 1)$. The convolution of these two functions is given by
$$y(t) = \int_{-\infty}^{\infty} h(t – \tau) u(\tau) d\tau = \int_{-\infty}^{t-1} tu(\tau) d\tau = {t\left( {t – 1} \right)} \over 2}u\left( {t – 1} \right).$$
The output is a function that is zero for $t < -1$, and is equal to ${t\left( {t – 1} \right)} \over 2$ for $t \ge -1$. The output is also a unit step function, delayed by one unit.
The other options are incorrect because they do not match the convolution of the impulse response and the input.