If the function
Question:

If the function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, & 0 \leq x<1 \\ 2, & x=1 \\ x+1, & 1<x \leq 2\end{array}\right.$ is continuous at $x=1$, then $a-b=$________________

Solution:

The function $f(x)=\left\{\begin{array}{cc}a x^{2}-b, & 0 \leq x<1 \\ 2, & x=1 \\ x+1, & 1<x \leq 2\end{array}\right.$ is continuous at $x=1$.

$\therefore f(1)=\lim _{x \rightarrow 1} f(x)$

$\Rightarrow f(1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$              …..(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 $f(x)=\left\{\begin{array}{cc}a x^{2}-b, & 0 \leq x<1 \\ 2, & x=1 \\ x+1, & 1<x \leq 2\end{array}\right.$ is continuous at $x=1$, then $a-b=$  ____2____.