Cho hàm số $y=f\left(x\right)=\left\{\begin{array}{c}\sqrt{x-1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x>2\\ c\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=2\\ 1-x^2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}\hspace{0.17em}}x<2\end{array}\right.$

a) \(f\left( 0 \right) = 1\).

b) Sai. Với \(x > 2\), ta có điều kiện hàm \(y = \sqrt {x - 1} \) là \(x \ge 1\).

Điều này luôn được thỏa mãn với mọi \(x > 2\).

c) Đúng. \(f\left( 2 \right) + f\left( 1 \right) = 1 \Leftrightarrow c + 0 = 1 \Leftrightarrow c = 1\).

Khi đó ta có hàm số \(y = f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\sqrt {x - 1} \,\,\,\,\,\,\,\,{\rm{khi}}\,\,\,x > 2}\\{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,\,x = 2}\\{1 - {x^2}\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x < 2}\end{array}} \right.\).

d) Đúng. Ta có \(t = x + 1 \Rightarrow x = t - 1\).

Khi \(x < 2 \Rightarrow t = x + 1 < 3\). Vậy \(t < 3\) nên ta có \(f\left( t \right) = 1 - {\left( {t - 1} \right)^2} =  - {t^2} + 2t\).