#### Explain solution RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 8 maths

The correct option is (b)

Hint:

Use the given formula:

$\text { (i) Standard limit } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$

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

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text { iii }) \lim _{x \rightarrow a}\{f(x) \pm g(x)\}=1 \pm m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}

Given:

$f(x)=\left\{\begin{array}{l} \frac{\log (1+a x)-\log (1-b x)}{x} \\ k \end{array}\right.$        \begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}

And f(x) is continuous at x = 0

Solution:

$\because$ f(x) is continuous at x = 0

$\Rightarrow \lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}=k$

Using formula (ii)

$\Rightarrow a \lim _{x \rightarrow 0} \frac{\log (1+a x)}{a x}+b \lim _{x \rightarrow 0} \frac{\log (1-b x)}{-b x}=k$

Using formula (i)

$\Rightarrow a+b=k$

So, the correct option is (b)