If $\lim _{x \rightarrow 1} \frac{x^2-\mathrm{a} x+\mathrm{b}}{x-1}=7$, then $\mathrm{a}+\mathrm{b}$ is…

If $\lim _{x \rightarrow 1} \frac{x^2-\mathrm{a} x+\mathrm{b}}{x-1}=7$, then $\mathrm{a}+\mathrm{b}$ is equal to
  1. -1
  2. 1
  3. -11
  4. 11

Solution

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

Asked in: MHT CET 2024 (04 May Shift 2)

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