Answer
$f(t)=\sqrt {t+1}, a=1$
Work Step by Step
Let $f(t)=\sqrt {t+1}$, we have:
$$f'(1)=\lim_{t\to 1}\frac{\sqrt {t+1}-\sqrt {1+1}}{t-1}=\lim_{t\to 1}\frac{\sqrt {t+1}-\sqrt {2}}{t-1}$$
Thus we have $f(t)=\sqrt {t+1}, a=1$
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