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

Answers (1)


 The correct option is (b)


 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}


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


\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)

Posted by

Gurleen Kaur

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