The first problem asks to find the value of $a$ such that the function $f(x)$ is continuous at $x=1$, where $f(x) = \frac{\sqrt{x^2-x+1}-x}{x-1}$ for $x \neq 1$ and $f(1)=a$.

1. Problem Description

The first problem asks to find the value of $a$ such that the function $f(x)$ is continuous at $x=1$, where

$f(x) = \frac{\sqrt{x^2-x+1}-x}{x-1}$ for $x \neq 1$ and $f(1)=a$.

2. Solution Steps

For $f(x)$ to be continuous at $x=1$, we need $\lim_{x \to 1} f(x) = f(1) = a$. We compute the limit:

$\lim_{x \to 1} \frac{\sqrt{x^2-x+1}-x}{x-1} = \lim_{x \to 1} \frac{(\sqrt{x^2-x+1}-x)(\sqrt{x^2-x+1}+x)}{(x-1)(\sqrt{x^2-x+1}+x)} = \lim_{x \to 1} \frac{(x^2-x+1)-x^2}{(x-1)(\sqrt{x^2-x+1}+x)} = \lim_{x \to 1} \frac{-x+1}{(x-1)(\sqrt{x^2-x+1}+x)} = \lim_{x \to 1} \frac{-(x-1)}{(x-1)(\sqrt{x^2-x+1}+x)} = \lim_{x \to 1} \frac{-1}{\sqrt{x^2-x+1}+x}$.

Now, we can substitute $x=1$:

$\frac{-1}{\sqrt{1^2-1+1}+1} = \frac{-1}{\sqrt{1}+1} = \frac{-1}{1+1} = -\frac{1}{2}$.

Therefore, $a = -\frac{1}{2}$.

3. Final Answer

The value of $a$ is $-\frac{1}{2}$. So the answer is A.