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

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

Answers (1)

Answer:

                k=\frac{9}{5}

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}{l} k x+1, \text { if } x \leq 5 \\\\ 3 x-5, \text { if } x>5 \end{array}\right)

Solution:

                f(x)=\left(\begin{array}{l} k x+1, \text { if } x \leq 5 \\\\ 3 x-5, \text { if } x>5 \end{array}\right)

We have

(LHL at x=5 )

                \lim _{x \rightarrow 5^{-}} f(x)=\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} k(5-h)+1=5 k+1

(RHL at x=5 )

                \lim _{x \rightarrow 5^{+}} f(x)=\lim _{h \rightarrow 0} f(5+h)=\lim _{h \rightarrow 0} 3(5+h)-5=10

If f\left ( x \right )  is continuous at x=5 , then

                \begin{aligned} &\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{-}} f(x) \\ &5 k+1=10 \\ &k=\frac{9}{5} \end{aligned}

               

Posted by

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