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  • Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 30 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 30 Maths Textbook Solution.

Answers (1)

Answer:

                \frac{a+b}{a b}=f(0)

Hint:

f\left ( x \right )  must be defined. The limit of the  f\left ( x \right )  approaches the value  x must exist.

Given:

                f(x)=\frac{\log \left(1+\frac{x}{a}\right)-\log \left(1-\frac{x}{b}\right)}{x}, x \neq 0

If f\left ( x \right ) is continuous at x=0 , then

                \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)-\log \left(1-\frac{x}{b}\right)}{x}\right)=f(0) \end{aligned}

                 \lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)}{\frac{a x}{a}}-\frac{\log \left(1-\frac{x}{b}\right)}{\frac{b x}{b}}\right)=f(0)                                        [Multiplying the denominator of first term by \frac{a}{a} ]       

                                                                                                                                        [Multiplying the denominator of second term by \frac{b}{b} ]

                \frac{1}{a} \lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)}{\frac{x}{a}}\right)-\left(\frac{-1}{b}\right) \lim _{x \rightarrow 0}\left(\frac{\log \left(1-\frac{x}{b}\right)}{\frac{-x}{b}}\right)=f(0)

                                                                                              

                \frac{1}{a} \times 1-\left(\frac{-1}{b}\right) \times 1=f(0)                                                           \left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]

                \begin{aligned} \\\\ &\frac{1}{a}+\frac{1}{b}=f(0) \\\\ &\frac{a+b}{a b}=f(0) \end{aligned}                                                                          

 

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