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  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 19

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 19

Answers (1)

Answer:

                k=-1 

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,  

                f(x)=\left(\begin{array}{c} \frac{x^{2}-3 x+2}{x-1}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right)

If f\left ( x \right )  is continuous at x=1 , then

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

                \begin{aligned} &\lim _{x \rightarrow 1} \frac{(x-2)(x-1)}{x-1}=k \\\\ &\lim _{x \rightarrow 1}(x-2)=k \\\\ &k=-1 \end{aligned}

               

               

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