Q213 / 270 · NDA 2015 (Phase 1) · Mathematics

Given that $\lim _{x \rightarrow \infty}\left(\frac{2+x^2}{1+x}-A x-B\right)=3$. What is the value of A?

  • A
    –1
  • Correct answer
    B
    1
  • C
    2
  • D
    3

Solution

$\begin{aligned} & \lim _{\mathrm{x} \rightarrow \infty}\left(\frac{2+\mathrm{x}^2}{1+\mathrm{x}}-\mathrm{Ax}-\mathrm{B}\right)=3 \\ \Rightarrow & \lim _{\mathrm{x} \rightarrow \infty}\left(\frac{2+\mathrm{x}^2-\mathrm{Ax}-\mathrm{B}-\mathrm{Ax}^2-\mathrm{Bx}}{1+\mathrm{x}}\right)=3 \\ \Rightarrow & \lim _{\mathrm{x} \rightarrow \infty}\left[\frac{(1-\mathrm{A}) \mathrm{x}^2-(\mathrm{A}+\mathrm{B}) \mathrm{x}+2-\mathrm{B}}{1+\mathrm{x}}\right]=3 \end{aligned}$ Applying L'Hospital rule, $2 x(1-A)-(A+B)=3$ Comparing coefficients $\begin{array}{ll} & 2(1-A)=0 \\ \therefore \quad & A=1 \text { and } \\ & -(A+B)=3 \\ \therefore \quad & B=-3-1=-4 \\ & A+B=-3 \\ & A=1 . B=-4 \end{array}$

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

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