The slope of the tangent to the curve

Given that $x=3 t^{2}+1, y=t^{3}-1$

For $x=1$

$3 t^{2}+1=1$

$\Rightarrow 3 t^{2}=0$

$\Rightarrow t=0$

Now, differentiating both the equations w.r.t. t, we get

$\frac{\mathrm{dx}}{\mathrm{dt}}=6 \mathrm{t}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}=3 \mathrm{t}^{2}$

$\Rightarrow$ Slope of the curve:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$=\frac{3 t^{2}}{6 t}$

$=\frac{1}{2} t$

For $t=0$,

Slope of the curve $=0$

Hence, option B is correct.