#### Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 33 maths textbook solution

The correct option is (b)

Hint:

A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$

(i) Standard limits

$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1, \lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1$

Given:

$f(x)=\left\{\begin{array}{cc} \frac{\log (1+3 x)-\log (1-2 x)}{x} \quad, x \neq 0 \\ k \qquad \quad, x=0 \end{array} \right.$

Step 1: Understand that, if f(x) is continuous at x = 0

\begin{aligned} &\therefore \lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+3 x)-\log (1-2 x)}{x}=k \end{aligned}

$\lim _{x \rightarrow 0} \frac{3 \log (1+3 x)}{3 x}-\frac{2 \log (1-2 x)}{2 x}=k$

Simplifying further,

$3 \lim _{x \rightarrow 0} \frac{\log (1+3 x)}{3 x}+2 \lim _{x \rightarrow 0} \frac{\log (1-2 x)}{-2 x}=k$

Applying formula (i)

\begin{aligned} &\therefore 3 \times 1+2 \times 1=k \\ &k=3+2 \\ &k=5 \end{aligned}

So, correct option is (b)