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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
MCQ
If $\lim _{x \rightarrow \infty}\left[\frac{x^3+1}{x^2+1}-(a x+b)\right]=2$, then
  • A
    $a=1$ and $b=1$
  • B
    $a=1$ and $b=-1$
  • ✓
    $a=1$ and $b=-2$
  • D
    $a=1$ and $b=2$
✓

Answer

Correct option: C.
$a=1$ and $b=-2$
(C)
$\lim _{x \rightarrow \infty}\left[\frac{x^3+1}{x^2+1}-(a x+b)\right]=2$
$\Rightarrow \lim _{x \rightarrow \infty}\left(\frac{x^3(1-a)-b x^2-a x+(1-b)}{x^2+1}\right)=2$ ..(i)
Since the limit of the given expression is a finite non-zero number, numerator and denominator are of the same degree.
$\therefore 1-a=0 \Rightarrow a=1$
Putting the value of a in (i), we get
$\lim _{x \rightarrow \infty}\left(\frac{-b x^2-x+1-b}{x^2+1}\right)=2$
$\Rightarrow \lim _{x \rightarrow \infty}\left(\frac{- b -\frac{1}{x}+\frac{1}{x^2}-\frac{ b }{x^2}}{1+\frac{1}{x^2}}\right)=2$
$\Rightarrow- b =2$
$\Rightarrow b =-2$

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