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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Continuity Excercise 8.1 Question 36 Subquestion (ix)

Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 36 subquestion (ix) Maths Textbook Solution.

Answers (1)

Answer:

                                k=7

Hint:

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

 

Given:

                f(x)=\left(\begin{array}{c|c} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}, & x \neq 2 \\\\ k & , x=2 \end{array}\right)

Solution:

                f(x)=\left(\begin{array}{c|c} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}, & x \neq 2 \\\\ k & , x=2 \end{array}\right)

                f(x)=\left(\begin{array}{cc} \frac{x^{3}+x^{2}-16 x+20}{x^{2}-4 x+4} & , x \neq 2 \\\\ k & , x=2 \end{array}\right)                            \left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]

                f(x)=\left(\begin{array}{c} x+5, x \neq 2 \\ k, x=2 \end{array}\right)

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

                \begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2}(x+5)=k \\ &2+5=k \\ &k=7 \end{aligned}

               

               

               

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