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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ and $f(x)$ is continous at $x = 0,$ then the value of $k$ is:
  • A
    $a - b$
  • ✓
    $a + b$
  • C
    $\log\text{a}+\log\text{b}$
  • D
    none of these
✓

Answer

Correct option: B.
$a + b$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{x}}-\frac{\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{ax}}\times\text{a} -\frac{\log(1-\text{bx})}{\text{-bx}}\times(-\text{b})=\text{k}$
$\text{a}+\text{b}=\text{k}$

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