If \(\lim _{x \rightarrow 1} \frac{x^2-\mathrm{a} x+\mathrm{b}}{x-1}=7\), then \(\mathrm{a}+\mathrm{b}\) is equal to

\(\lim _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=7\)
Limit exists if \(x^2-\mathrm{a} x+\mathrm{b}\) at \(x=1\) is 0 .
\(\therefore (1)^2-\mathrm{a}+\mathrm{b}=0 \)
\( \Rightarrow 1-\mathrm{a}+\mathrm{b}=0 \)
\( \Rightarrow \mathrm{a}-\mathrm{b}=1 ...(i)\)
\( \therefore \lim _{x \rightarrow 1} \frac{x^2-(1+\mathrm{b}) x+\mathrm{b}}{x-1}=7 \)
\( \Rightarrow \lim _{x \rightarrow 1} \frac{x^2-x-\mathrm{b} x+\mathrm{b}}{x-1}=7\)
\(\Rightarrow \lim _{x \rightarrow 1} \frac{(x-1)(x-b)}{x-1}=7 \)
\( \Rightarrow 1-\mathrm{b}=7 \) \( \ldots[x \rightarrow 1, x \neq 1, x-1 \neq 0] \)
\( \Rightarrow \mathrm{b}=-6\)
From (i) \(a=-5\)
\(\therefore \quad a+b=-5-6=-11\)