If the function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, & 0 \leq x<1 \\ 2, & x=1 \\ x+1, & 1
The function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, & 0 \leq x<1 \\ 2, & x=1 \\ x+1, & 1 $\therefore f(1)=\lim _{x \rightarrow 1} f(x)$ $\Rightarrow f(1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$ .....(1) Now, $f(1)=2$ ....(2) $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(a x^{2}-b\right)=a \times(1)^{2}-b=a-b$ ....(3) $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x+1)=1+1=2$ .....(4) From (1), (2), (3) and (4), we have $2=a-b=2$ $\therefore a-b=2$ Thus, the value of $a-b$ is 2 . If the function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, & 0 \leq x<1 \\ 2, & x=1 \\ x+1, & 1