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

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

Answers (1)

Answer:

                k=\frac{1}{2}

Hint:

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

 

Given:

                f(x)= \begin{cases}k\left(x^{2}+2\right), & x \leq 0 \\ 3 x+1 & , x>0\end{cases}

Solution:

                f(x)= \begin{cases}k\left(x^{2}+2\right), & x \leq 0 \\ 3 x+1 & , x>0\end{cases}

We have

(LHL at x=0  )

                \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} k\left((-h)^{2}+2\right)=2 k

(RHL at x=0 )

                \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} 3 h+1=1

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

                \begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \\ &2 k=1 \\ &k=\frac{1}{2} \end{aligned}

               

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