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  • Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 31

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 point 12 Question 31

Answers (1)

Answer:

k=\frac{1}{2}

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)=\left\{\begin{array}{c} \frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\\\ k, \text { if } x=2 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{c} \frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\\\ k, \text { if } x=2 \end{array}\right.

If f\left ( x \right )  is continuous at x=2 ,

                \begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=f(2) \end{aligned}

                \lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}=k   [Taking 4 as common]                       \left[\because a^{2}-b^{2}=(a+b)(a-b)\right]

                \begin{aligned} &\lim _{x \rightarrow 2} \frac{4}{\left(2^{x}+4\right)}=k \\ &\frac{4}{\left(2^{2}+4\right)}=k \\ &k=\frac{4}{8} \\ &k=\frac{1}{2} \end{aligned}

               

               

               

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