SOLUTION —
$t_{2}=x_{2}-1$
$x_{2}=1+t_{2}$
$x \frac{d x}{d t}=t$
$\Rightarrow \quad\left(\frac{d x}{d t}\right)^{2}+x \frac{d^{2} x}{d t^{2}}=1$
$\Rightarrow \quad \frac{x d^{2} x}{d t^{2}}=1-\left(\frac{d x}{d t}\right)^{2}=1-\left(\frac{t}{x}\right)^{2}$
$=\frac{x^{2}-t^{2}}{x^{2}}=\frac{1}{x^{2}} \Rightarrow a=\frac{d^{2} x}{d t^{2}}=\frac{1}{x^{3}}$
So, The correct option of this question is (B).