MHT CET · Maths · Continuity and Differentiability
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
- C 5
- D -5
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
\(\mathrm{f}\) is continuous at \(x=0\).
\(\begin{aligned}
\therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\
\therefore \quad \mathrm{k} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)\right) \\
& =\lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}\right) \\
& =3+2=5
\end{aligned}\)
\(\begin{aligned}
\therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\
\therefore \quad \mathrm{k} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)\right) \\
& =\lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}\right) \\
& =3+2=5
\end{aligned}\)
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