Q57 / 150 · MHT CET 2023 (14 May Shift 2) · Mathematics

The function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right), & x \neq 0 \\ \mathrm{k} & , \quad x=0 \end{array}\right.$ is continuous at $x=0$, then $\mathrm{k}$ is

  • A
    6
  • B
    1
  • Correct answer
    C
    5
  • D
    -5

Solution

$\mathrm{f}$ is continuous at $x=0$. $\begin{aligned} \therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\ \therefore \quad \mathrm{k} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}\right) \\ & =3+2=5 \end{aligned}$

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

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