Q65 / 150 · MHT CET 2024 (09 May Shift 1) · Mathematics

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$ then

  • A
    $\mathrm{a}=1, \mathrm{~b}=4$
  • Correct answer
    B
    $\mathrm{a}=1, \mathrm{~b}=-4$
  • C
    $\quad \mathrm{a}=2, \mathrm{~b}=-3$
  • D
    $\mathrm{a}=2, \mathrm{~b}=3$

Solution

$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-\mathrm{a} x-\mathrm{b}\right)=4 \\ & \Rightarrow \lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-\mathrm{ax}(x+1)-\mathrm{b}(x+1)}{x+1}\right)=4 \\ & \Rightarrow \lim _{x \rightarrow \infty}\left[\frac{(1-\mathrm{a}) x^2+(1-\mathrm{a}-\mathrm{b}) x+(1-\mathrm{b})}{x+1}\right]=4\end{aligned}$ Limit is finite, it exists when $(1-a)=0$ $\begin{array}{cc} \therefore \quad & 1-a=0 \\ & \Rightarrow a=1 \end{array}$ $\begin{array}{ll}\therefore \quad & \lim _{x \rightarrow \infty}\left[\frac{1-a-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}\right]=4 \\ & \Rightarrow 1-a-b=4 \\ & \Rightarrow \quad \mathrm{~b}=-4 \\ \therefore \quad & a=1, b=-4\end{array}$

Hence, the correct answer is option: 'B'.

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