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STD 11 - 12. limits — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 12. limits1 Mark
MCQ
Let $[ t ]$ denote the greatest integer $\leq t$. If for some $\lambda \in R -\{0,1\}, \lim \limits_{x \rightarrow 0}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=L,$ then $L$ is equal to
  • A
    $1$
  • ✓
    $2$
  • C
    $\frac{1}{2}$
  • D
    $0$
✓

Answer

Correct option: B.
$2$
b
$\operatorname{LHL}: \lim _{x \rightarrow 0^{-}}\left|\frac{1-x-x}{\lambda-x-1}\right|=\left|\frac{1}{\lambda-1}\right|$

$\operatorname{RHL}: \lim _{x \rightarrow 0^{+}}\left|\frac{1-x+x}{\lambda-x+1}\right|=\left|\frac{1}{\lambda}\right|$

For existence of limitt

$LHL = RHD$

$\Rightarrow \frac{1}{|\lambda-1|}=\frac{1}{|\lambda|} \Rightarrow \lambda=\frac{1}{2}$

$\therefore \quad L=\frac{1}{|\lambda|}=2$

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